Spencer slays with sarcasm

Heh. In response to a ridiculous claim making the rounds (I get comment bombed at WUWT daily with that nonsense) which I debunked here: A misinterpreted claim about a NASA press release, CO2, solar flares, and the thermosphere is making the rounds

Dr. Roy Spencer employs some power visual satire, that has truth in it. He writes:

How Can Home Insulation Keep Your House Warmer, When It Cools Your House?!

<sarc> There is an obvious conspiracy from the HVAC and home repair industry, who for years have been telling us to add more insulation to our homes to keep them warmer in winter.

But we all know, from basic thermodynamics, that since insulation conducts heat from the warm interior to the cold outside, it actually COOLS the house.

Go read his entire essay here. <Sarc> on, Roy!

UPDATE: Even Monckton thinks these ideas promoted by slayers/principia/O’Sullivan are ridiculous:

Reply to John O’Sullivan:

One John O’Sullivan has written me a confused and scientifically illiterate “open letter” in which he describes me as a “greenhouse gas promoter”. I do not promote greenhouse gases.

He says I have “carefully styled [my]self ‘science adviser’ to Margaret Thatcher. Others, not I, have used that term. For four years I advised the Prime Minister on various policy matters, including science.

He says I was wrong to say in 1986 that added CO2 in the air would cause some warming. Since 1986 there has been some warming. Some of it may have been caused by CO2.

He says a paper by me admits the “tell-tale greenhouse-effect ‘hot spot’ in the atmosphere isn’t there”. The “hot spot”, which I named, ought to be there whatever the cause of the warming. The IPCC was wrong to assert that it would only arise from greenhouse warming. Its absence indicates either that there has been no warming (confirming the past two decades’ temperature records) or that tropical surface temperatures are inadequately measured.

He misrepresents Professor Richard Lindzen and Dr. Roy Spencer by a series of crude over-simplifications. If he has concerns about their results, he should address his concerns to them, not to me.

He invites me to “throw out” my “shredded blanket effect” of greenhouse gases that “traps” heat. It is Al Gore, not I, who talks of a “blanket” that “traps” heat. Interaction of greenhouse gases with photons at certain absorption wavelengths induces a quantum resonance in the gas molecules, emitting heat directly. It is more like turning on a tiny radiator than trapping heat with a blanket. Therefore, he is wrong to describe CO2 as a “coolant” with respect to global temperature.

He invites me to explain why Al Gore faked a televised experiment. That is a question for Mr. Gore.

He says I am wrong to assert that blackbodies have albedo. Here, he confuses two distinct methods of radiative transfer at a surface: absorption/emission (in which the Earth is a near-blackbody, displacing incoming radiance to the near-infrared in accordance with Wien’s law), and reflection (by which clouds and ice reflect the Sun’s radiance without displacing its incoming wavelengths).

He implicitly attributes Margaret Thatcher’s 1988 speech to the Royal Society about global warming to me. I had ceased to work with her in 1986.

He says that if I checked my history I should discover that it was not until 1981 that scientists were seriously considering CO2’s impact on climate. However, Joseph Fourier had posited the greenhouse effect some 200 years previously; Tyndale had measured the greenhouse effect of various gases at the Royal Institution in London in 1859; Arrhenius had predicted in 1896 that a doubling of CO2 concentration would cause 4-8 K warming, and had revised this estimate to 1.6 K in 1906; Callender had sounded a strong note of alarm in 1938; and numerous scientists, including Manabe&Wetherald (1976) had attempted to determine climate sensitivity before Hansen’s 1981 paper.

He says, with characteristic snide offensiveness, that I “crassly” attribute the “heat-trapping properties of latent heat to a trace gas that is a perfect energy emitter”. On the contrary: in its absorption bands, CO2 absorbs the energy of a photon and emits heat by quantum resonance.

He says the American Meteorological Society found in 1951 that all the long-wave radiation that might otherwise have been absorbed by CO2 was “already absorbed by water vapor”. It is now known that, though that is largely true for the lower troposphere, it is often false for the upper.

The series of elementary errors he here perpetrates, delivered with an unbecoming, cranky arrogance, indicates the need for considerable elementary education on his part. I refer him to Dr. Spencer’s excellent plain-English account of how we know there is a greenhouse effect.

The Viscount Monckton of Brenchley (April 18, 2013)

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April 24, 2013 10:06 am

Here in the UK most houses are built with a double skinned wall with breeze blocks (I think in the USA you call them ash blocks) forming the inner wall, then an air gap, then bricks on the outer wall.
The bricks are porous and absorb water when it rains, which then evaporates when it is windy and dry. This arrangement keeps the inside of the house warm in winter and cooler in summer because of the still air in between the two. This system has worked well for years.
There are companies that will fill the cavity with foam or fibreglass insulation and claim that as a result your house will be warmer. It can also be damper, as the rainwater passes from the outer wall, by capillary action to the breeze blocks and then to the plaster on the inside wall. Mildew then develops and the house is damp, also water vapour from breathing, cooking, showering etc condenses on the coolest parts of the interior of the house, which is usually the parts of the walls that the cavity wall insulation could not get to, causing more mildew and dampness. The house feels cold and damp and is a health hazard to those with asthma.
This is all very expensive to put right and these companies usually go out of business on a regular basis to avoid legal action.
So this article is less sarcastic than Roy thinks!
By the way “insulation conducts heat” is that not an oxymoron or at least a contradiction?

Jim Hodgen
April 24, 2013 10:14 am

One weakness of sarcasm is that it requires honest comparison between norms and the offered information in order to get it. Thus it frequently leaves the target of the pointed information completely clueless…. but a great fallback is the entertainment value for those that have the ability to make the comparison.
Thank you Dr. Spencer for the analogy… and thank you Anthony for the entertainment.

Mark Bofill
April 24, 2013 10:15 am

Cue the Slayers to come in and …explain… this again.
cringe Oh noes, not again!
I’ve quit arguing this point with people. When I get this urge, I remind myself that I’ve found it’s much more productive and fulfilling to beat my head against concrete pylons instead.

Tom J
April 24, 2013 10:16 am

I mean, c’mon, duh. A home is a home, not a greenhouse. Got it? The Earth is a greenhouse. That’s the difference. A greenhouse is for growing plants. A home is for growing people. So the Earth is not a home. Uh, wait a minute. Uh, we grow here too. So the Earth is a home, I guess, but, uh, mmm…
Quick, John Holdren, help me out here.

alex
April 24, 2013 10:22 am

Sarcasm apart. What did the guy wanted to say?
Surely more insulation helps you keeping the house warmer. What was the question?
More clothes you have on, the warmer you feel.

Stuart Elliot
April 24, 2013 10:22 am

All materials conduct heat. Insulation conducts heat, but the heat transfer is slowed compared to any material which insulates less.
The first commenter touches on the problem that arises when moisture is not managed inside a sealed house. The examples he cites are caused by contractors not understanding the right way of insulating a dwelling. It’s not the insulating material wot dunnit, it was the act of changing only one thing in a system that was more complicated than it looked.
Kind of like “climate” “policy”…

April 24, 2013 10:29 am

Insulation in your house prevents draft leakage and convective loss to the outside, by trapping material molecules. This simply makes it easier for the furnace to hold temperature. A gas, CO2, doesn’t trap itself. It doesn’t prevent its own convection and drafting etc. And, such insulation doesn’t raise the temperature above the temperature of the input, of the furnace.
So then the analogy is created that trapping photons, via CO2, has s a similar heat trapping effect, even though trapping photons does occur in a real greenhouse but actually has nothing to do with the temperature inside a real greenhouse, because the temperature inside a real greenhouse is caused by preventing convective cooling when the input is sunshine at, say, +80C at noon. The analogy is then further extended to say that trapped photons can cause a -18C input (the incorrect and absurd flat earth assumption for the solar input), to become +15C, even though this doesn’t actually occur inside a real greenhouse nor with the insulation in your home. In fact, trapped radiation inside a cavity simply produces a blackbody spectrum of the temperature of the source, and this is of course the origin of quantum mechanics; trapped radiation doesn’t raise the temperature of its own source in any case. Photons are bosons and can pile on top of each other without noticing, other than equal amounts of constructive and destructive interference which results in no net change. This is markedly different behaviour from molecules which, when packed on top of each other, can not simply pass through but must find smaller and smaller volumes of space in which to situate themselves, thus increasing pressure and temperature. Shine two flashlight directly into each other and seal them perfectly, and they will not explode or even shine any brighter, unlike what would happen if you were compressing more and more gas into the sealed chamber.
You’re being lied to by these sophists. Either you understand it or you don’t. That being said, it is quite difficult to understand, but within that difficulty is where the sophists create their garbage. 50% (and more) increase of CO2 over the last few decades and no increase in temperature above 1930′s values, or above the warm periods before that, or above the first entire half (or more) of this whole interglacial; ice core records which show that the Earth enters an ice-age when CO2 level is high(!). You’re being lied to by sophists.

Josh C
April 24, 2013 10:34 am

Close!
But the real cause of warming in the house is the insulation traps the CO2’s and that leads to ‘Housing Warming.’

April 24, 2013 10:40 am

What this infra-red thermography shows is convective heat stratification within the home and the thermal exit paths. The higher section of the windows and roof are heated more than the lower section. Home insulation provides a physical barrier to convective flow. The home repair industry does not add ‘radiative barriers’ of Carbon Dioxide to the hot spots to ‘redirect’ the radiative flow. Since the premise is meaningless, there can be no ‘satire’. Roy would benefit from reviewing a sample of satire. I recommend, “Amazing! New! Wrongco Proxy Crock” posted at Canada Free Press.

April 24, 2013 10:41 am

If things behaved this way, if trapping heat could raise the temperature above the temperature creating the heat, then ALL of thermodynamic research in the 1800’s would have revolved around exploiting this and today we would have devices which could create an internal extremely high temperature (to use for doing useful work with) with only a minuscule, tiny input; say, a AAA battery and tiny resistor could be used to generate 5000K inside a shell, and then you could smelt some steel or something with a AAA battery. Instead they created steam engines powered by coal and developed things like the Carnot Cycle. They never ever developed anything about trapping heat to create higher temperature than the input. What they found when they tried to do so, was what was eventually codified into the Laws of Thermodynamics.

April 24, 2013 10:45 am

One can hope that folks will follow anthony’s and roy’s lead here by calling out skydragon nonsense for what it is. The real debate is over how much warming GHGs will cause.
That is the debate. That science is not settled. Skeptics with smart arguments will get published, see Nic lewis and Troy masters. Its the central debate..

Luis Dias
April 24, 2013 10:45 am

Yeah, this sarcasm just went over my head. The argument was very funny, but I really don’t know what the hell was that about.

tadchem
April 24, 2013 10:52 am

The difference between a greenhouse and the Earth is convection. A greenhouse sitting in the sunlight gets hotter because it has no convection – the hot air gets trapped beneath the cover provided by whatever media is used to put a lid on everything.

Stephen Wilde
April 24, 2013 10:54 am

But is CO2 an insulator ?
It is supposed to block outward longwave from the surface but in doing so increases convection, evaporation and radiation to space (by providing a radiative window not supplied by non GHGs).
Evaporation has a net cooling effect of 5 to 1 (enthalpy of vaporisation) and so is a hugely powerful negative system response.
I am not yet convinced that CO2 has any net insulating effect at all once the negative system responses have been accounted for.
Note that I do not deny the thermal characterisatics of GHGs, merely do I question the sign and power of the system response.
The insulated house analogy is entirely inappropriate because one is simply interposing a less effective conductor between the air inside the house and the air outside the house.

April 24, 2013 10:58 am

Kind of like the “thermos Joke”:
http://www.funniestcleanjokes.com/joke/funny-thermos-joke

Paul Westhaver
April 24, 2013 11:02 am

I can actually work that one out from first principles. Having studied a great deal of thermodynamics, I am equipped to tell you that Roy’s statement is 100% true.
The angels are in the details.
If you have an electric water heater, how thick do you make the insulation jacket?
As Roy points out, making the insulation thicker does not mean that you will reduce the total heat loss.
Heat loss happens by conduction, convection and radiation. The models for heat loss for an electric water heater are very well known and accurate and reliable. As you increase the diameter of the heater you increase the surface area, thereby increasing the conduction, the convection and the radiation even though you’ve increased the resistance of heat flow from the tank. Insulation is a conductor… in fact.
I recall it was a question on one of my Thermo II course exams.
Common sense does not prevail. Understanding the underlying models and experimental results wins. Real science is great!

Paul Westhaver
April 24, 2013 11:06 am

🙂

Paul Westhaver
April 24, 2013 11:10 am

It is a problem of diminishing return. At what point does the cost of insulation yield a saving that is less than the cost of insulation. It depends on the cost of heat and the cost of insulation.

April 24, 2013 11:36 am

Let’s say you have a pot of water that is being boiled (~373 K) and a block of ice (~273 K) radiating towards each other.
Correct: P = εσA(Th⁴ – Tc⁴)
Incorrect: P = εσATh⁴ – P = εσATc⁴
Doing it the right way gives 782.63~ W/m^2 from the pot to the ice block.
Doing it the wrong way gives 1097.59 W/m^2 from the pot to the ice, and 314.96 W/m^2 from the ice to the pot, so now the pot is receiving input+314.96 W/m^2, and we then equate the output to the input to say the pot is receiving 1097.59+314.96=1412.55 W/m^2!
So now we claim it has to emit 1412.55 W/m^2 which equates to a temperature of 397.28 K!
So just by putting a block of ice next to a pot of boiling water we raised the temperature of the boiling water by 24 K!
It is the difference between this:
P = εσA(Th⁴ – Tc⁴)
and this:
P = εσA(Th⁴ + Tc⁴)

April 24, 2013 11:47 am

If your home is heated from the outside by a big lamp, adding insulation will make it cooler.

John West
April 24, 2013 11:55 am

Joseph E Postma says:
”if trapping heat could raise the temperature above the temperature creating the heat”
I don’t recall that ever being said:
http://www.asterism.org/tutorials/tut37%20Radiative%20Cooling.pdf
Let’s say I turn the heat on to some body for 12 hours out of every 24, obviously, if I add insulation the body will cool slower during the off heat 12 hours and could be warmer at the initiation of heat for the next heat on 12 hours; in this manner of reducing heat loss the body gets warmer not that the insulation directly heats the body.

A C Osborn
April 24, 2013 11:58 am

crosspatch says:
April 24, 2013 at 11:47 am
If your home is heated from the outside by a big lamp, adding insulation will make it cooler.
That is precisely what I asked Roy Spencer, his response was
The house would be slower to heat if that heat was applied to its outside, it would warm quicker if it the source of heat was inside. In the case of the climate system, most sunlight is absorbed at the surface, so it is “heated from within”.”
So CO2 does not act as an Insulator at all because if it did it would prevent the Radiation from heating surface in the first place.
So we are back to the Back Radiation issue.
The House is a crap analogy and he knows it, I am glad Anthony finds it funny, but for all the wrong reasons.

April 24, 2013 12:02 pm

Joe Postma, yes you probably CAN design a system for generating 5,000K temperatures using just a AAA battery. All you need to do is resistance heat a very small object with little heat capacity, and insulate it very well. The problem would be the insulation, because it, too,would heat up. Remember, 10,000,000 K temperatures are created at the core of the Sun with heating rates less than what the human body produces (per unit mass).

DaveG
April 24, 2013 12:04 pm

Roy,
Many thanks for coming out all guns blazing on this. It should be greatly entertaining as well as informative to all who follow this issue. What PSI finds amusing is how you have tripped yourself up by your blatant contradictions. e.g. You say:
“One of the first things you discover when putting numbers to the problem is the overriding importance of infrared radiative absorption and emission to explaining the atmospheric temperature profile. These IR flows would not occur without the presence of “greenhouse gases”, which simply means gases which absorb and emit IR radiation. Without those gases, there would be no way for the atmosphere to cool to outer space in the presence of continuous convective heat transport from the surface.
Indeed, it is the “greenhouse effect” which destabilizes the atmosphere, leading to convective overturning. Without it, there would not be weather as we know it. The net effect of greenhouse gases is to warm the lowest layers, and to cool the upper layers.
The greenhouse effect thus continuously “tries” to produce a lapse rate much steeper than the adiabatic lapse rate, but convective overturning occurs before that can happen, cooling the lower troposphere and warming the upper troposphere through a net convective transport of heat from lower layers to upper layers.”
http://www.drroyspencer.com/2011/12/why-atmospheric-pressure-cannot-explain-the-elevated-surface-temperature-of-the-earth/
So, Roy, what you’re claiming is that the enormous heat from GHGs leads to convective heat-mixing, which makes the lapse rate shallow. But the RADIATIVE GHE would make the lapse very steep.
Good lord. That’s just awful. You are saying that without GHG’s, the atmosphere wouldn’t be able to cool???!!!! Don’t you mean it would be HOTTER without GHG’s??
Convection occurs automatically because warm air is less dense…duh! It gets heated and then rises. GHG’s don’t cause convection…how could 0.04% force the rest? So backwards.
And then why is the lapse rate for dry air SIMPLY what it is if you calculate it using U = gh + CpT. Where’s the GHG effect? How come when you factor in condensation heat release to the equation, then you just get exactly the wet rate? You are referring to an effect on the lapse rate that is non-existent.

April 24, 2013 12:11 pm

Roy Spencer says: “yes you probably CAN design a system for generating 5,000K temperatures using just a AAA battery”
Wow. That is of course ridiculous. Actually let me rephrase that.
Please make it so and prove it. 🙂

April 24, 2013 12:12 pm

DaveG has just posted a comment that is actually from John O’Sullivan, who Anthony has banned.
But I will provide the same think I posted on my blog to John’s Comment:
“Without greenhouse gases, the UPPER atmosphere would indeed be much warmer, but the LOWER atmosphere would be much cooler. Quit confusing the two, like you did with your recent misinterpretation of the NASA upper-atmosphere study. It’s been almost 50 years since Manabe and Strickler (1964) showed how the upper atmosphere is cooled by CO2, which is PART OF the greenhouse effect. Nothing new there.
Adding more insulation to your house makes the outside of your house cooler, and the inside warmer (in winter). Greenhouse gases (thermodynamically like insulation in your house) reduce the rate at which heat flows from higher temperatures to lower temperatures, thus making the warm side warmer, and the cool side cooler.”

April 24, 2013 12:15 pm

Roy Spencer says: “Remember, 10,000,000 K temperatures are created at the core of the Sun with heating rates less than what the human body produces (per unit mass).”
Good god man. You forgot about gravity. Fusion is not the GHE.
Let me repeat this for everybody: Nuclear fusion is not the GHE. The solar core and nuclear fusion have no relevance or support to the supposed GHE. It is arguments like this which should destroy any credibility to the faith…oh wait. 🙂

April 24, 2013 12:16 pm

Remember, 10,000,000 K temperatures are created at the core of the Sun with heating rates less than what the human body produces (per unit mass).” ~Roy
Uh, the power leaving the core is low because everything is around the same temperature deep within the Sun.
You wouldn’t want a chunk of fusing core material dropped next to you, it wouldn’t be fun for anyone but Superman, who could shrug off a thermonuclear bomb.
____________________________
For the record, the temperatures aren’t created by the heating rates you’re talking about, the temperatures are created by the mass of the rest of the star compressing the core material enough for it to undergo fusion.
If a hunk of stellar material were replaced with non-fusing material at around the same temperature as your body it would be crushed rather dramatically until it reached a temperature where radiation pressure could support the rest of the star above it.

Gary Hladik
April 24, 2013 12:19 pm

Anthony, you weren’t getting enough traffic at WUWT, so you bring this up again? 🙂
Max™ says (April 24, 2013 at 11:36 am): “Let’s say you have a pot of water that is being boiled (~373 K) and a block of ice (~273 K) radiating towards each other.”
Max, your thought experiment is quite complicated, involving phase changes, atmospheric cooling & heating via conduction/convection, etc.
What’s your take on the simpler thought experiment described here:
http://www.drroyspencer.com/2010/07/yes-virginia-cooler-objects-can-make-warmer-objects-even-warmer-still/

A C Osborn
April 24, 2013 12:26 pm

Gary Hladik says: “What’s your take on the simpler thought experiment described here:”
Only someone who believes in the Back Radiation theme would say that because it gets cooler in the box than the surrounding atmosphere that Proves Back Radiation because without it the box would go down to 3 degrees K.
To most people it would suggest that the back Radiation is actually cooling the box.
So what was your own take on it, or did you just accept Roy’s conclusion without actually thinking about it?

April 24, 2013 12:29 pm

“Adding more insulation to your house makes the outside of your house cooler, and the inside warmer (in winter).”
Preventing drafts is not what a gas can do…a gas is a gas and is drafty. CO2 doesn’t prevent drafts. Putting something beside a heater simply heats up the thing…not the heater. Heat simply conducts into the thing.

A C Osborn
April 24, 2013 12:35 pm

Steven Mosher says:
April 24, 2013 at 10:45 am
The real debate is over how much warming GHGs will cause.
That is the debate. That science is not settled. Skeptics with smart arguments will get published, see Nic lewis and Troy masters. Its the central debate..
Absolute B***Sh** as usual, if you can’t get an absolute measure of how much is natural and how much is supposed CO2 or other GHGs you can never ever work out the sensitivy, it is all complete guesswork, hence the “pause” in warming, which by any other name is quickly going in to cooling.
Even Anthony has to show all the Cold Records being set, but where are the explanations, where is it being shouted from the MSM roof tops.
Even the Met Office has had to back track yet again.

A C Osborn
April 24, 2013 12:43 pm

Isn’t it amazing what difference a few years can make, it was all about CO2 is the control knob, it overpowers everything.
Now we have it is Coal burning causing cooling, the Oceans eating the heat, it,s black dust, anything but CO2 doesn’t do what they said it does.

Gary Hladik
April 24, 2013 12:46 pm

[snip we don’t link to that book promoting website -mod]

Stephen Richards
April 24, 2013 12:56 pm

Roy’s example is appalling as many of the comments here have shown.
REPLY: no, what is appalling is the principia fools that believe this rubbish about the greenhouse effect being “bogus” as they say right on their web page.

This sort of stuff is the rational skeptics worst enemy. Like arguing with vaxxers – Anthony

Alberta Slim
April 24, 2013 1:00 pm

Two questions;
1. What happens if the house has the R15 [or whatever] spun glass insulation taken out and R15 sealed bags of CO2 put in?
Will the house [assuming constant heat supply] cool off faster or slower?
2. I thought gasses[atmosphere] expanded to absorb extra heat.
The greenhouse gets hotter because the air cannot expand.
What stopping the atmosphere from expanding?

Jim S
April 24, 2013 1:00 pm

My understanding is that when you “add” insulation to a wall cavity, it’s the billion pockets of isolated, trapped air (gasses) that slow down the convection of heat from one side of the wall to the other. The insulating material actually serves as a “thermal bridge” that allows “higher” conductive heat-loss across the cavity – but the conductive heat-loss is more than off-set by the reduction in convective heat-loss.
The presence of the solid, insulation material actually speeds up heat-loss.

Gary Hladik
April 24, 2013 1:07 pm

Joseph E Postma says (April 24, 2013 at 10:41 am): “…say, a AAA battery and tiny resistor could be used to generate 5000K inside a shell, and then you could smelt some steel or something with a AAA battery.”
This reminds me of Willis Eschenbach’s “steel greenhouse”:
http://wattsupwiththat.com/2009/11/17/the-steel-greenhouse/
http://wattsupwiththat.com/2013/02/06/the-r-w-wood-experiment/

more soylent green!
April 24, 2013 1:12 pm

More insulation keeps my house cooler in the summer, when I’m running the air conditioner, or warmer in the winter, when I’m running the heat.
So adding more insulationing to my home can and does keep it cooler.

Gary Hladik
April 24, 2013 1:17 pm

A C Osborn (April 24, 2013 at 12:26 pm), I think you’re confusing the “Yes, Virginia” thought experiment with something else. The “Yes, Virginia” thought experiment is the one with the chilled vacuum chamber containing two facing plates, one heated, one not.
I have thought quite a bit about the “Yes, Virginia” thought experiment, which I think distills the issue of so-called “back radiation” down to its essence, in contrast to Max’s messy thought experiment. I’ve also thought a lot about the criticism of this thought experiment by the so-called “slayers”, and I find their arguments…unconvincing, to put it politely.
Mods: My apologies for linking to a prohibited site. I’ll try to remember that in future.

April 24, 2013 1:18 pm

Sorry if this is a double post.
The solar input occurs only over a single hemisphere. The intensity distribution of this input goes as the cosine function about the solar zenith. The integrated average value of this intensity has a temperature value of +49C. Solar power has an average heating power of +49C continuously on the sunlit hemisphere. This is very different from the flat-earth approximation of -18C. This is a rational consideration, because +49C continuous input is physically real, and will have a very different effect on the system than a global -18C. +49C can do a lot of things that -18C can’t.

davidmhoffer
April 24, 2013 1:19 pm

Joseph E Postma says:
April 24, 2013 at 10:29 am
Insulation in your house prevents draft leakage and convective loss to the outside, by trapping material molecules
>>>>>>>>>>>>>>>>>>
Wow. Three absolutely false statements in a single sentence. Way to go Joe.
REPLY: Yeah, I wanna see those molecule traps. Joe needs to take a time out before he claims the sun revolves around the Earth or something – Anthony

April 24, 2013 1:21 pm

Max is correct I suppose. Nobody never measured 324W coming from the night sky AFAIK. And whoever believes, that this “back ratiation” from “greenhouse gases” by itself increases our night time temperature from -150C (like on the Moon) to +15 here, well, keep believing. Simple thermal retention of the N2+O2 explains bulk of the (wrongly calculated) “+33K” myth. It is as foolish as forever trying to calculate doubling sensitivity from CO2 and latest positive phase of AMO. You get some numbers in all cases; but it is just virtual number.

April 24, 2013 1:21 pm

Jim S, it doesn’t matter to the point being made whether we are talking about conduction, convection, or radiation. ANYTHING that reduces heat loss from a heated object can increase its temperature. And that “anything” is usually at a COOLER temperature than the heated object itself. No laws of thermodynamics are broken, as alleged by the slayers.

April 24, 2013 1:28 pm

By trapping material molecules I was of course referring to preventing drafts, which prevents warm air escape to the outside. Warm air is composed of molecules. Preventing drafts is of course one of the main functions of insulation, in your house.

davidmhoffer
April 24, 2013 1:33 pm

Juraj V. says:
April 24, 2013 at 1:21 pm
Max is correct I suppose. Nobody never measured 324W coming from the night sky AFAIK.
>>>>>>>>>>>>>>>
I suggest you read what has been measured:
http://wattsupwiththat.com/2011/03/10/visualizing-the-greenhouse-effect-emission-spectra/

April 24, 2013 1:35 pm

” it doesn’t matter to the point being made whether we are talking about conduction, convection, or radiation. ANYTHING that reduces heat loss from a heated object can increase its temperature.”
It only increases “the” temperature, it doesn’t increase the temperature of the source. Real greenhouses aren’t heated to higher than the source, nor does the insulation in a house cause an increase of the temperature of the gas flame in the furnace.
Here’s a physics textbook example of trapping radiation from an active source inside a cavity.
http://books.google.no/books?id=PfadZy35Wh0C&pg=PA442&lpg=PA442&dq=blackbody+sphere+surrounded+shell+radiation&source=bl&ots=TDbus0Dwu4&sig=3Aj5S6SlUB55MY9ry_MBXzTBm84&hl=no&sa=X&ei=d6VxUfOHC8Os4ASGsYCACQ&ved=0CEUQ6AEwBDgU#v=onepage&q=blackbody%20sphere%20surrounded%20shell%20radiation&f=false
Problem 27.: “A sphere of radius R is maintained at a surface temperature T by an internal heat source (Figure 3). The sphere is surrounded by a thin concentric shell of radius 2R. Both objects absorb and emit as blackbodies. Show that the temperature of the shell is T/(8^1/4) = 0.595 T. (Hint: Both the inner and outer surfaces of the shell emit as blackbodies.)”
The surface temperature of the sphere is maintained by its internal heat source. And that’s that. There is no “then the sphere has to heat up some more because radiation is trapped” etc etc. Of course, trapping photons in a cavity from a thermal source only produces a blackbody spectrum; it doesn’t cause the photons to change frequency to higher temperature. Photons are bosons and can pile on top of each other without caring, aside from equal constructive and destructive interference which results in no net change. This is very different than the behaviour of trapping more and more molecules into a confined space.
REPLY: and none of this matters – Anthony

Gary Hladik
April 24, 2013 1:35 pm

Alberta Slim says (April 24, 2013 at 1:00 pm): “Two questions;
1. What happens if the house has the R15 [or whatever] spun glass insulation taken out and R15 sealed bags of CO2 put in?”
If the R-value of the new material is the same as the old, shouldn’t it insulate the same?
http://en.wikipedia.org/wiki/R-value_(insulation)
“2. I thought gasses[atmosphere] expanded to absorb extra heat.
The greenhouse gets hotter because the air cannot expand.
What stopping the atmosphere from expanding?”
Nothing. Like regular houses, greenhouses generally aren’t airtight. Air convection is inhibited, not prevented entirely. Note that heating your house with, say, a gas heater doesn’t raise the house internal pressure and make your ears pop.

April 24, 2013 1:37 pm

ALL:
It is important to remember that there is NO WAY to determine the temperature of anything based upon the rate of energy input alone, for example the Earth absorbing an average of ~240 W/m2 from the Sun. Temperature is a function of BOTH energy input (typically not temperature dependent) AND energy loss (typically VERY temperature dependent), neglecting issues related to heat capacity which mainly affect the time required for the system to equilibrate. The temperature of anything heated will increase until the rate of energy *loss* equals the rate of energy *gain*. So, temperature can be increased by increasing INPUT, or decreasing OUTPUT.

April 24, 2013 1:42 pm

“The temperature of anything heated will increase until the rate of energy *loss* equals the rate of energy *gain*”
When energy input equals energy output, then the system is in equilibrium. In this state the system will not be hotter than the input. Of course, the input isn’t -18C, but is +49C, on the Earth, and, the Earth isn’t hotter than this. In the textbook example of trying to control the output radiatively, this only creates a blackbody spectrum inside the cavity which is created to do that. When you shine a flashlight at a mirror, into a mirror, or two flashlights at each other with a good seal between the, it doesn’t cause the flashlights to shine brighter. Equal temperatures can not heat each other.
REPLY: Joe, take a time out. This is becoming absurd with your thread bombing – Anthony

Sidney Somes
April 24, 2013 1:44 pm

The position that Co2 causes heating at the Earth’s surface seems to be contradicted by millions of years of temperature records showing that temperature precedes Co2 increases, by about 800 years. The so-called “runaway greenhouse effect” would have happened millions of years ago when Co2 was ten times the level that it is today. So perhaps our greenhouse physics is missing something, like DATA.

April 24, 2013 1:49 pm

Ok, have it be two black bodies of arbitrary material, large enough that the area we’re interested in is radiating towards the other body, one at 373 K and one at 273 K, radiating towards each other.
I’m confident that the radiation field between them can be described by the following equation:
P = εσA(Th⁴ – Tc⁴)
rather than:
P = εσA(Th⁴ + Tc⁴)
max@Funktastic:~$ calgebra
>>> (5.6703*10^-8)*(373^4 – 273^4)
782.63220248
>>> (5.6703*10^-8)*(373^4 + 273^4)
1412.55397668
Both of those can not be right, and if the colder black body raises the temperature of the warmer black body then the second one is correct, which would be roughly the same power as that emitted by a 297.28 K black body.
If the radiation from the colder body is added to that from the warmer body, that is akin to a block of ice raising the temperature of a pot of nearly boiling water by over 24 K!

April 24, 2013 1:49 pm

Anthony I am not thread bombing, I am responding to comments, and on topic. I am being perfectly pleasant. Cheers.

John West
April 24, 2013 1:54 pm

Obviously we’re still having trouble with the concept of NET heat exchange despite Max’s efforts:
The surface radiates as described by the Stephan-Boltzmann law:
Surface radiation = εσ T^ 4
The atmosphere radiates as described by an empirically derived equation:
Down-welling atmospheric radiation = (1+ KC^2) x8.78 E – 13T^5.852 x RH^0.07195
Therefore the NET heat loss from radiation of the surface is:
NET Radiation = [εσ T^ 4] – [(1+ KC^2) x8.78 E – 13T^5.852 x RH^0.07195]
Therefore, as long as [(1+ KC^2) x8.78 E – 13T^5.852 x RH^0.07195] is greater than 0, the NET radiation MUST be LESS THAN the “gross” (or Stephan-Boltzmann calculated) radiation that would be the NET if there were no radiation from the atmosphere (i.e. no GHE). This is measurable, known, understood stuff; like the world is round(ish).
Please read these 5 pages for details:
http://www.asterism.org/tutorials/tut37%20Radiative%20Cooling.pdf

Robert Austin
April 24, 2013 1:58 pm

The best way to understand the role of greenhouse gases is to do a thought experiment as to the temperature structure without greenhouse gases including water. I such a case, ignoring airborne aerosols, cooling of the earth would be occur only by black or grey body radiation from the earth’s surface. So the energy balance equation would show a much colder earth’s surface and there would be no significant lapse rate structure to the atmosphere. On the other hand, a significant portion of outgoing energy is radiated to space from the top of the troposphere in an atmosphere with greenhouse gases and the temperature at the top of the troposphere is much lower than the surface temperature. So the greenhouse gas containing atmosphere is like an insulating wall in that a greenhouse wall radiates from the exterior surface of the wall whereas a non greenhouse wall would be akin to a thin membrane wall essentially radiating from the interior wall surface.

CodeTech
April 24, 2013 2:06 pm

I have triple glazed windows in my house. The gas they put inside the windows is Nitrogen. Seems to me if CO2 was a great insulator, they’d be using that instead…

jono1066
April 24, 2013 2:06 pm

Sorry but I couldnt help it.
take house wall with a cavity down the middle, standard thermal calcs for houses says `and then ignore the thermal R value (U value in good old Blighty) of the outer wall` this is specifically due to the radiatiive value of the surface and the convective value of the air in the cavity once heated above ambient and the normal movement of air through the cavity due to the house `breathing`. This is true in both directions, bricks heated from the outside by the sun or from the house heated from inside. Just think of it as a plate heat exchanger. Reducing the radiative value (emmissivity) of the cavity surface (polish it) and preventing to a great degree simple convection/pressure driven air flow allows the walls (lets say normal heating from the inside) to change the temperature gradient across them, hence the cavity side of the ash block increase in temperature towards but never past the temperature inside the house, Thermal stabilisation will occur at some point with the cavity side of the inner wall always below the temperature inside the house. Convection is the biggest problem for double glased windows hence the normal limit of 19 to 22m between inner and outer pane after which thermal resistance goes down. Styrofoam beads approx 3 to 5mm diameter are very highly packed closed cell honeycombe nature and convection is very difficult to initiate within such a small volume CO2 filled eps would not noticeably change the insulating/resistance value. Adding insulation to your house allows you to maintain a specified and stable thermal gradient from any one side to the other with a lower thermal input. Incorrectly applied insulation works directly and absolutely measurably through the `placebo` effect.
Did I get the job ?

Kelvin Vaughan
April 24, 2013 2:07 pm

Just build 2 greenhouses of identical size. Enhance the CO2 in one to 1000ppm. Keep them at the same temperature over a year. Check if the enhanced CO2 greenhouse used less KWh over the year.

davidmhoffer
April 24, 2013 2:11 pm

Kelvin Vaughan says:
April 24, 2013 at 2:07 pm
Just build 2 greenhouses of identical size. Enhance the CO2 in one to 1000ppm. Keep them at the same temperature over a year. Check if the enhanced CO2 greenhouse used less KWh over the year.
>>>>>>>>>>>>>>>>>
Sure, that would provide measurable data. Of course to get measurable data large enough to actually measure with even very accurate instrumentation, you’d need to build the greenhouses several kilometers tall.

April 24, 2013 2:14 pm

In the case of the climate system, most sunlight is absorbed at the surface, so it is “heated from within”.”
So CO2 does not act as an Insulator at all because if it did it would prevent the Radiation from heating surface in the first place.

But the sun does emit infrared that does get blocked from reaching the surface by “greenhouse” gases. It works in both directions.

April 24, 2013 2:15 pm

Just in case anyone is curious, with my cleaned up example above (two black bodies, 373 K and 273 K, respectively), the radiation field from the warmer to colder body is roughly the same as that emitted by a black body at 342.758 K, and while I can’t quite justify it as being certain, I would not be surprised if the two bodies reached equilibrium around that temperature, i.e. with the warmer body being cooled and the colder body being warmed.
Seems reasonable enough for the pot of nearly boiling water + block of ice as well, getting warm water wouldn’t seem odd, but getting hotter water than you started with by adding ice would be a very interesting result, wouldn’t it?

April 24, 2013 2:16 pm

And it is true that per unit of mass, the sun produces very little heat. It’s about that produced by the average compost heap. It’s just a very big compost heap and so it produces a lot of total heat that takes a long time to get to the surface.

davidmhoffer
April 24, 2013 2:21 pm

CodeTech says:
April 24, 2013 at 2:06 pm
I have triple glazed windows in my house. The gas they put inside the windows is Nitrogen. Seems to me if CO2 was a great insulator, they’d be using that instead…
>>>>>>>>>>>>>>>>>
Seems to me they’d put in whatever gas experimentation proved provided the best result. Probably argon btw, not nitrogen. What they used and why they used it says absolutely zero about co2 as an insulator other then they found something they think works better.

Selgovae
April 24, 2013 2:29 pm

Gary Hladik mentions Roy Spencer’s “Yes, Virginia” thought experiment. I have a question about this that’s been been bugging me for some time. Perhaps someone can put me out of my misery. There appear to be two views about “back radiation”: that energy can radiate from a cooler material (atom I guess) to a warmer one or that it can’t. I’d always held the latter view, for no particular reason other than that’s how I thought things worked. But in either case, isn’t the net effect more or less the same? This is how I interpret things:
Case 1: Radiation flows between both warmer and cooler plates. The effect is the result of the net radiation flow, and so the warmer plate becomes warmer than it would have been if the cooler plate wasn’t there.
Case 2: Radiation only flows from the warmer plate to the cooler plate. But at some point, the atoms at the surface of the cooler plate reach the temperature of the warmer plate. The warmer plate can no longer radiate to the cooler plate until the cooler plate’s surface atoms cool again, presumably by a combination of conduction within the plate and radiation from the other side. Those times when the warmer plate can’t radiate lower the rate of heat loss from the warmer plate, and so the warmer plate becomes warmer than it would have been if the cooler plate wasn’t there.
Gently, please!

Gary Hladik
April 24, 2013 2:32 pm

Max™ says (April 24, 2013 at 1:49 pm): “Ok, have it be two black bodies of arbitrary material, large enough that the area we’re interested in is radiating towards the other body, one at 373 K and one at 273 K, radiating towards each other.”
Max, what’s your take on the “Yes, Virginia” thought experiment?
http://www.drroyspencer.com/2010/07/yes-virginia-cooler-objects-can-make-warmer-objects-even-warmer-still/

Greg House
April 24, 2013 2:39 pm

“Spencer slays with sarcasm”
=======================================================
Sarcasm is indeed there, but there is also a problem with the scientific point.
First, the analogies with “cold keeps warm”, blankets, houses etc. a very misleading. A blanket would keep us cold, not warm, if the air temperature outside the blanket is higher, than our body temperature.
Second, even on the theoretical level the concept of “back radiation warming” is physically absurd and therefore impossible, because if the source has been kept initially at a constant temperature, then “back radiation warming” will inevitably lead to an endless mutual warming of the source of radiation and the thing that provides back radiation, without an additional input of energy, which is physically absurd. And everybody can make a simple experiment: just stand in front of the mirror and enjoy the “back radiation warming”. Must be 33C or more, if the “greenhouse effect” as presented by the IPCC exists. Please, not too close to the mirror, to avoid the effect of suppressed convection.
Coming back to the theoretical level, it goes like that. You have initially a body kept at a certain temperature by it’s internal source of energy. Now you put another colder body at the absolute zero temperature, let us say, in vacuum close to the warm body.
The warmer body will start warming the colder body immediately. Then, according to the “back radiation warming” concept, the back radiation from the colder body will increase the temperature of the warmer body. Actually, already on this stage we should start screaming and crying “how come?”, but let us proceed. So, the now even warmer warm body will warm the colder body even stronger, and the colder body will repay by sending even more back radiation to the warmer body, thus increasing it’s temperature even further. The warmer body will get warmer again. So will the colder body in turn. And so on.
This is the mutual endless warming without any additional input of energy I meant previously, and I hope it is easy to understand how physically absurd it is. This proves that the concept of “back radiation warming” is physically absurd. The “greenhouse effect” as presented by the IPCC can not exist.
REPLY: No, it doesn’t. This argumentum ad infinitum that the slayers push about “cold can’t heat warm” is the absurd part, and nothing but a strawman argument. The greenhouse effect (a misnomer) isn’t so much about reheating the atmosphere with reradiated LWIR from top to below, it is about slowing the progress of LWIR to the top of the atmosphere. Without CO2 or other GHG’s the LWIR would proceed quickly to space, and the Earth would cool faster, and have a lower average temperature at night. The “backradiation” has a limited scope of effect, but it is there as part of the trasnfer process. GHG’s are not unidirectional, they are omnidirectional in the LWIR process.
GHG’s act as a LWIR transfer regulator from the surface to the top of the atmosphere where it is radiated into space. They slow the transfer. Without GHG’s nightitme temperature would drop quickly as LWIR is lost directly to space.
A simple proof of this has to do with the differences between nighttime temperatures between moist and dry climates at the same latitude/altitude where water vapor is the dominant GHG. See Knappenberger et al 1996 for example.
Dry desert climes cool faster at night and have a greater diurnal temperature variation than moist climes. Water vapor is the difference. CO2 has a smaller effect, but an effect nonetheless. – Anthony

Mark Bofill
April 24, 2013 2:49 pm

Selgovae says:
April 24, 2013 at 2:29 pm
————–
Well, look at it this way. Radiation has no way of knowing the temperature of bodies out in existence someplace it may eventually encounter. The sun shines outwards in all directions equally (possibly some random variations, I’m not a solar expert, but darn close I think), regardless of whether or not there is a hotter surface out there someplace in space that some of its radiation may eventually encounter.
So yes, I’d expect radiation to propagate from a source through space with complete disregard for the temperature of the object it may eventually hit.

Noelene
April 24, 2013 3:08 pm

So there is a consensus on the greenhouse effect?

Kristian
April 24, 2013 3:15 pm

davidmhoffer says, April 24, 2013 at 2:21 pm:
“Seems to me they’d put in whatever gas experimentation proved provided the best result. Probably argon btw, not nitrogen. What they used and why they used it says absolutely zero about co2 as an insulator other then they found something they think works better.”
http://gaia.lbl.gov/btech/papers/29389.pdf
Interesting quotes:
“Low-emittance coatings are much more effective at reducing infrared radiation heat transfer than IR absorbing gasses. Gasses for gas-filling should be chosen for their low conductivity and high kinematic viscosity in order to effectively reduce conductive/convective heat transfer. The effective use of infrared absorbing gasses is thus limited to horizontal windows heated from above, or to thin gaps where low-emittance coatings cannot be used.”
“(…) even though the gas absorption/emission dampens the natural convection, the absorbing gasses being used as gas-fills have lower kinematic viscosities than air and some of the other low-conductivity gasses (argon, krypton) being used in windows (fig. 6). And, from Glaser’s results for vertical windows it can be seen that the convective transfer becomes significant at around 9 mm for SF6, while there is practically no convective transfer through an air-filled window at gapwidths up to 20 mm under these conditions. In fact, air outperforms SF6 at gapwidths greater than 9 mm in a vertical window and the benefits from infrared absorption by SF6 have been negated by the magnitude of the convection.
“Fig. 3 shows that the effect of the infrared radiation properties of CO2 is unnoticable (…)”

April 24, 2013 3:18 pm

“Max, what’s your take on the “Yes, Virginia” thought experiment?” ~Greg
The heated bar will raise the temperature of the unheated bar, adding another object to be warmed reduces the energy density, there is no way it can result in a higher temperature without creating energy.
150 Fahrenheit is 338.7 K and would emit 746.2~ W/m^2 as a black body
100 Fahrenheit is 310.9 K and would emit 529.7~ W/m^2 as a black body
max@Funktastic:~$ calgebra
>>> (5.6703*10^-8)*(338.7^4)
746.219894665
>>> (5.6703*10^-8)*(310.9^4)
529.771907497
>>> (5.6703*10^-8)*(338.7^4 – 310.9^4)
216.447987168
>>> (5.6703*10^-8)*(338.7^4 + 310.9^4)
1275.99180216
From the heated bar:
746.2 to shell, 216.4 to unheated bar
From the unheated bar:
529.7 to the shell, -216.4 to heated bar
If the power supply is constant then consider the case where the unpowered bar begins at the same temperature as the powered bar.
I see no reason why a body at given temperature should raise the temperature of another body at the same temperature, so accordingly I see no reason why a cooler body should raise the temperature of a warmer body.

OldWeirdHarold
April 24, 2013 3:24 pm

“Roy Spencer says:
April 24, 2013 at 12:02 pm
Joe Postma, yes you probably CAN design a system for generating 5,000K temperatures using just a AAA battery. All you need to do is resistance heat a very small object with little heat capacity, and insulate it very well. The problem would be the insulation, because it, too,would heat up.”
=====
It’s called an incandescent flashlight.

Reply to  OldWeirdHarold
April 24, 2013 3:29 pm

“It’s called an incandescent flashlight.”
It wasn’t about a AAA battery lighting a flashlight filament to 5000K. There’s nothing wrong with that. It is about the resistor (lets use the filament) becoming even hotter still if you shine the flashlight at a mirror. This does not increase the brightness (temperature) of the filament. Trapping the radiation does not increase the temperature of the filament.

Gary Hladik
April 24, 2013 3:24 pm

Selgovae says (April 24, 2013 at 2:29 pm): “This is how I interpret things: [snip]”
Mark covered the basic problem above (i.e. postulating clairvoyance by inanimate–or animate–objects), but here are a couple more thoughts on your case 2:
a) The temperature of the unheated plate (“plate 2”) never reaches that of the electrically heated plate (“plate 1”). Plate 2 is receiving energy from plate 1 on one side but radiating energy from two sides.
b) Plate 1’s radiation is based on its temperature
http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law
so if its radiation decreases, its temp must have decreased. But as you know the temp doesn’t decrease, so your postulated decrease in radiation can’t happen.

pat
April 24, 2013 3:26 pm

Spencer “with sarcasm”, Ban Ki-Moon without satire!
25 April: Bloomberg: UN Says Clean Energy Funding Too Low to Halt Climate Harm
UN Secretary General Ban Ki-moon said the funds flowing to renewable power and efficiency aren’t sufficient to avert environmental calamities and that investors must move more quickly to back new energy technologies.
“Climate change is a threat to economies large and small and to the stability of the ***global financial system***,” Ban said today in a speech at a Bloomberg New Energy Finance conference in New York. “The climate clock is ticking. The longer we delay the greater the cost. We only have one planet Earth. We have no plan B.”…
The UN has led an effort to bring more than 190 nations together to develop carbon-reduction policies, and Ban said he received a personal commitment from U.S. President Barack Obama two weeks ago to work toward a binding climate agreement in 2015.
“He’s assured me that the U.S. will lead by example,” Ban said…
He told bankers and fund managers gathered at the conference they must “lead by example” and that he’s doing the same.
“I have been dutifully, faithfully turning off lights in my hotel,” Ban said. “Sometimes it’s very difficult these days, all different hotels have a very different system of lighting. Normally I stay in a suite so there are many, many lights.”…
http://www.bloomberg.com/news/2013-04-24/un-says-clean-energy-funding-too-low-to-halt-climate-harm.html
——————————————————————————–

davidmhoffer
April 24, 2013 3:29 pm

Kristian;
“Fig. 3 shows that the effect of the infrared radiation properties of CO2 is unnoticable (…)”
>>>>>>>>>>>>>>>>
I wouldn’t expect it to be noticeable. Not at that gap width. The effect we’re talking about in terms of the atmosphere is from an air column 14 km high. Further, the article you quote refers to suppression of convection and conduction as being the primary factors for the application, and so they recommend materials that perform the best against those parameters.
Again, not using CO2 in a window has nothing to do with how it behaves in a 14 km high atmospheric air column and everything to do with what delivers the best result in a window. I could just as easily argue that they don’t build cargo ship hulls out of wood as proof that wood doesn’t float. It does float, there’s just other ways to float your boat that are more effective.

Greg House
April 24, 2013 3:32 pm

REPLY: No, it doesn’t. This argumentum ad infinitum that the slayers push about “cold can’t heat warm” is the absurd part, and nothing but a strawman argument. The greenhouse effect (a misnomer) isn’t so much about reheating the atmosphere with reradiated LWIR from top to below, it is about slowing the progress of LWIR to the top of the atmosphere. Without CO2 or other GHG’s the LWIR would proceed quickly to space, and the Earth would cool faster, and have a lower average temperature at night.
GHG’s act as a LWIR transfer regulator from the surface to the top of the atmosphere where it is radiated into space. They slow the transfer. […] – Anthony

==============================================================
Anthony, about “the slayers”, I do not refer to any slayers, I did not even read their book, however I read some of their articles. I do not care who exactly said what and what else right or wrong they said etc., I only like good argumentation. Only the quality of the argumentation counts.
Your second point is about what “greenhouse effect” means. This issue is a little bit confusing, and not because it is a misnomer, but because there are a few different versions around. When I say “the greenhouse effect as presented by the IPCC”, I do it specifically to avoid confusion, like in my previous comment. The point is, that the “greenhouse effect” as presented by the IPCC is exactly about warming by back radiation: http://www.ipcc.ch/publications_and_data/ar4/wg1/en/faq-1-3.html. Then I say that the warming by back radiation is physically impossible, therefore the “greenhouse effect” as presented by the IPCC does not exist.
If some people have other version of “greenhouse effect”, then we could discuss it, but my point is that the “greenhouse effect” as presented by the IPCC does not exist. I hope you can agree with me on that.

April 24, 2013 3:36 pm

” It’s called an incandescent flashlight.” ~OldWeirdHarold
Tungsten filaments reach around 3,600 K, not quite 5,000 K.
I don’t know of a material that will survive at 5,000 K and remain solid, the surface of the sun is around 5,700~6,000 K for comparison.

April 24, 2013 3:42 pm

Kelvin Vaughan says:
April 24, 2013 at 2:07 pm
Just build 2 greenhouses of identical size. Enhance the CO2 in one to 1000ppm. Keep them at the same temperature over a year. Check if the enhanced CO2 greenhouse used less KWh over the year.
Wrong.
The “greenhouse’ effect has nothing whatsoever to do with these types of experiments.
Energy leaves earth Via one route and one route only: radiation to space.
Earth radiates to space from the ERL.
When you add GHGs you raise the ERL.
When you raise the ERL the earth radiates to space more slowly.
That results in a surface that is warmer than it would be otherwise.
This effect cannot be tests with woods like experiments EVER because the effect depends upon changing the ERL and no experimental set up or lab can duplicate the conditions required: the full atmospheric column.
in 20 or so simple slides
http://www.aos.wisc.edu/~aos121br/radn/radn/sld001.htm

April 24, 2013 3:42 pm

Noelene:
Your post at April 24, 2013 at 3:08 pm goes to the crux of this thread’s subject. It asks

So there is a consensus on the greenhouse effect?

The answer is YES but the consensus is not what the media tends to present.
Anyone who understands radiative physics knows
the radiative properties of greenhouse gases (GHGs; mostly water vapour but also carbon dioxide, methane, etc.) in the air provide a greenhouse effect (GE) without which the Earth would be much colder.
That is the overwhelming consensus and is irrefutable physics.
However, that fact does not – of itself – mean additional GHGs in the air will discernibly raise global temperature. Any such rise in global temperature will depend on the feedbacks in the climate system. Those who believe in anthropogenic (i.e. man-made) global warming (AGW) think feedbacks are strongly positive so will greatly increase any such temperature rise. But AGW-skeptics (or climate realists) either think the feedbacks are weakly positive so will have little effect, or they are negative so will reduce any such temperature rise.
If the feedbacks are weakly positive or negative then AGW would be so small as to be indiscernible because natural climate variability is much larger.
In other words, the argument about AGW is really an argument about the feedbacks.
However, there are a few people (often known as ‘Dragon Slayers’ or ‘Slayers’) who claim that radiative physics is wrong so the GE does not exist. Of course, all science is open to amendment – that is the nature of science – but the assertions made by the Slayers are preposterous.
Unfortunately, the Slayers are vociferous, and the ludicrous nature of their assertions brings discredit (i.e. guilt by assumed association) on all AGW-skeptics.
Spencer is an AGW-skeptic, and his sarcasm is aimed at the silly notions of the Slayers. Hence, the title of this thread is “Spencer slays with sarcasm”.
I hope this clarifies the matter.
Richard

April 24, 2013 3:50 pm

“When you add GHGs you raise the ERL.
When you raise the ERL the earth radiates to space more slowly.”
Missing hot spot. Earth is radiating more these days, not less. For something warming to be emitting less radiation is a violation of the Stefan-Boltzmann Law. Only if you reduce the emissivity can the temperature be raised. CO2 does not reduce emissivity. O2 and N2 have zero emissivity and therefore trap and hold heat much better than any other gas and hold temperature much better than otherwise. CO2 collisionally absorbs energy from O2 and N2 and then causes it to be lost.

Alberta Slim
April 24, 2013 3:52 pm

Greg House says:
April 24, 2013 at 2:39 pm
“Spencer slays with sarcasm”…………………………
I totally agreewith Greg. “Back-welling” or “back-radiation” from CO2 is not possible.
It may help slow thing down. Also, no one [Dr Spencer?] has told me why the atmosphere will
not expand with extra heat from the earth’s suface. This would satisfy the “Conservation of Energy” with no increase in Temperature
Question 3. If in the thermograph, above, the triple glazed windows were filled with CO2,
what color would we observe? If CO2 is “backwelling” into the house the color
should be other than red. No???

Greg House
April 24, 2013 4:07 pm

Alberta Slim says (April 24, 2013 at 3:52 pm): “I totally agreewith Greg. “Back-welling” or “back-radiation” from CO2 is not possible.”
===========================================================
I did not say that “back-radiation” from CO2 was not possible. My point was that back radiation would not affect the temperature of the source.

Gary Hladik
April 24, 2013 4:12 pm

Max™ says (April 24, 2013 at 3:18 pm): “I see no reason why a body at given temperature should raise the temperature of another body at the same temperature, so accordingly I see no reason why a cooler body should raise the temperature of a warmer body.”
Thanks for taking a look at the “Yes, Virginia” thought experiment, Max. Your initial numbers are close to what “Nullius In Verba” calculated for the one-plate case at a certain Site-Which-Must-Not-Be-Linked, although he went on to calculate the equilibrium temperature of the two-plate system, in which the temp of the heated bar is higher.
Radiative physics as illustrated by the “Yes, Virginia” thought experiment is fundamental to modern science, see numerous textbook examples in the comment thread under Willis’s R W Wood Experiment article:
http://wattsupwiththat.com/2013/02/06/the-r-w-wood-experiment/
In fact, if the “Yes, Virginia” version of radiative physics is overturned, it would be worth a Nobel Prize at least, and would earn wealth, fame, and adulation for whoever brought about this scientific revolution. All from performing a relatively simple thought experiment “for real” and getting a different result from what Dr. Spencer predicts.
My question is, since this is a relatively simple experiment to do (no particle accelerators required), why have none of the so-called “slayers” performed it (or one like it) and taken their place alongside Newton, Planck, Einstein, and Hawking?
Any thoughts, Max?

KevinK
April 24, 2013 4:15 pm

Max wrote (re “Yes Virginia…..”);
“If the power supply is constant then consider the case where the unpowered bar begins at the same temperature as the powered bar.”
NOTE the word “power supply”, this is distinct from an “energy supply”, hence the distinct
nomenclature.
Yes a power supply will cause (under the very best of thought experiment conditions) the first bar to reach a higher temperature since it’s view factor to the colder vacuum chambers wall is reduced. HOWEVER it will consume more energy from the power supply while doing so. The power supply will have to provide enough energy to “charge” both metals bars with heat. It will draw more current from the energy supply (your friendly local electric company). Yes, I know electric utilities are often called XYZ electric POWER Company, but they are in fact selling you energy. If you do Dr. Spencer’s thought experiment you should expect a higher energy bill.
If you replace the power supply with a battery (a store of energy, or energy supply) the first bar will reach a higher temperature (again under the very best of thought experiment conditions), but more energy will be drawn from the energy supply (since it has to “charge” both bars with heat). Thus the battery discharges more quickly. So the bars get warmer for a while, then start cooling down when the energy supply is exhausted. Which results in no increase in the average temperature.
1 bar results in a lower temperature for a longer time, 2 bars results in a higher temperature for a shorter time, same amount of energy used in both cases. No trapped heat, no extra energy, no net energy gains.
The sunlight arriving during one day is represented by the initial charge in the battery, once it’s gone, it’s gone….
UNITS do matter when doing energy budgets (w/m^2 are units of power density, NOT energy).
Engineers that routinely calculate “net energy gains” either get fired or bumped up into management.
Cheers, Kevin.

Jim Brock
April 24, 2013 4:15 pm

Well, insulation does in a sense cool the house. It does transmit heat from the hot (inside) side of the insulation to the cool (outside) of the house. Some transmits more, some less. That is what the R factor measures. BUT it transmits less heat than the uninsulated version. This keeps the house warmer than the alternative, but still cooler than if it did not conduct heat at all (the posited but non-existent perfect insulator). Even vacuum bottles transfer heat from the hot side to the cool interior, just slower than the alternative.

Greg House
April 24, 2013 4:16 pm

Steven Mosher says (April 24, 2013 at 3:42 pm): “Earth radiates to space from the ERL. When you add GHGs you raise the ERL. When you raise the ERL the earth radiates to space more slowly. …”
=============================================================
This is not what the IPCC considers to be the “greenhouse effect” and what they base their policy recommendations upon. They mean that “greenhouse gases” warm the surface by back radiation. Apparently they do not support your “effect”.

JimF
April 24, 2013 4:18 pm

There certainly seems to be a lot of animosity in the comment train. I enjoyed Postma’s paper “The Model Atmospheric Greenhouse Effect” because I learned a lot about black body radiation, etc., and I fully subscribe to his idea that the Sun’s energy hitting the Earth’s surface isn’t something divided by four, but the whole thing (at some point on the surface).
My guess – based on my experience in geological arguments – is that both sides of this argument are blowing hot air at least in part. It further seems to me that a great deal of what passes for knowledge is simply stating one’s case, louder with each iteration. CAGW is utter BS in my mind, and the condemnation of CO2 by many is dangerous. If one believes the assessments of the amount of CO2 in the Earth’s atmosphere over time, and understands its role in the life as we now know it of this planet, then we are in a precarious position. Much of the CO2 endowment of Earth seemingly is now in a state that cannot benefit plants, and that ain’t good.

davidmhoffer
April 24, 2013 4:33 pm

Greg House;
They mean that “greenhouse gases” warm the surface by back radiation. Apparently they do not support your “effect”.
>>>>>>>>>>>>>>>>>>>>>
Back radiation increases the ERL. Same thing explained from a different perspective.

Gary Hladik
April 24, 2013 4:38 pm

KevinK says (April 24, 2013 at 4:15 pm): “If you replace the power supply with a battery (a store of energy, or energy supply) the first bar will reach a higher temperature (again under the very best of thought experiment conditions), but more energy will be drawn from the energy supply (since it has to “charge” both bars with heat). Thus the battery discharges more quickly.”
So if instead of using an electric outlet or a battery to heat the first plate, we use an embedded radioactive heat source
http://en.wikipedia.org/wiki/Radioisotope_thermoelectric_generator
the plutonium (for example) will decay more rapidly when the second plate is introduced?

davidmhoffer
April 24, 2013 4:38 pm

JimF says:
April 24, 2013 at 4:18 pm
There certainly seems to be a lot of animosity in the comment train.
>>>>>>>>>>>>>>>>
It is indeed unfortunate that the “information age” has given way to the “misinformation age”. In today’s world, it is mind boggling to me that the CAGW nonsense can live side by side with the “back radiation doesn’t warm things” nonsense. The actual physics seems to take a back seat to both arguments, and yes that generates some animosity.

richard verney
April 24, 2013 4:45 pm

crosspatch says:
April 24, 2013 at 2:16 pm
And it is true that per unit of mass, the sun produces very little heat. It’s about that produced by the average compost heap. It’s just a very big compost heap and so it produces a lot of total heat that takes a long time to get to the surface
///////////////////////////
Is that so? Per unit of mass USED? The issue is not how much mass the sun has, but rather how much of that mass it uses to produce the energy in which we all bask.
Is it not the position that sun only uses a little of its mass to produce its energy and thats why its still burning after 4.5 billion years and will burn for another 4.5 biliion years or so?

Greg House
April 24, 2013 4:45 pm

davidmhoffer says (April 24, 2013 at 4:33 pm): “Back radiation increases the ERL.”
========================================================
Again, this is not what the IPCC supports. You can invent whatever effect you want, but only the IPCC’s one is politically relevant, because governments and agencies refer to the IPCC reports and recommendations, not to davidmhoffer or Steven Mosher.
And again, the IPCC “greenhouse effect” is physically impossible, as I demonstrated earlier on this thread, so their proposed policy of cutting CO2 emissions is not justified by their reports. I do not think we should ignore that.

April 24, 2013 4:50 pm

UNITS do matter when doing energy budgets (w/m^2 are units of power density, NOT energy).” ~KevinK
In discussions of radiation W/m^2 is intensity, if we’re going to nitpick. Watts are power, or the rate at which energy is supplied per second, Joules are energy or work done.
Again though I will go back to the case of a heated or powered bar at 338 K (~150 Fahrenheit) and an unpowered bar at 338 K.
Will the presence of another body at the same temperature cause the temperature of the first body to increase?

April 24, 2013 4:53 pm

“and the ludicrous nature of their assertions”
It is not ludicrous to state that the distribution of solar energy is a cosine function about the solar zenith, and that this distribution has an integrated average of +49C, as the input. What is ludicrous is that people deny that the Earth is round and think that +49C vs. -18C doesn’t matter. I do detect some ludicrousness when it is said that the mere statement of these facts is supposed to be ludicrous.
The cosine function is real. An integrated average is real. Sunshine is really hot. You can’t average energy input into a geometry where it doesn’t exist. These are all rational facts. It is ludicrous to ridicule those of us who state them and want to explore what effect and changes it has on the usual assumptions. As we have shown, the changes are not insignificant. Both the wet and dry lapse rate can be calculated precisely without any reference to GHG radiation, for example. Latent heat itself holds the temperature higher than it would otherwise be, for example. This is science. It is ludicrous to ignore it.

Downdraft
April 24, 2013 4:53 pm

Which brings to mind the age old question, “How does a Thermos bottle know whether you put cold or hot things in it, and that it is supposed to keep cold things cold and hot things hot?” Maybe the Slayers can work on that one next.

Gary Hladik
April 24, 2013 4:55 pm

Greg House says (April 24, 2013 at 4:45 pm): “Again, this is not what the IPCC supports.”
Hi, Greg! I figured you’d show up sooner or later. Welcome.
Now it’s a party! 🙂
“And again, the IPCC “greenhouse effect” is physically impossible, as I demonstrated earlier on this thread…”
Anthony refuted your argument verbally, scienceofdoom refutes it mathematically here
http://scienceofdoom.com/2010/05/08/radiation-basics-and-the-imaginary-second-law-of-thermodynamics/

tjfolkerts
April 24, 2013 4:58 pm

@ Greg April 24, 2013 at 2:39 pm
“So, you have initially a body kept at a certain temperature by it’s internal source of energy.”
Just to be more definite, make it a blackbody sphere far from any star, so the surrounding are already ~ 0 K. Furthermore, let’s make that heater 240 W for each square meter, so the temperature is 255 K (like the numbers for earth).
“Now you put another colder body at the absolute zero temperature, let us say, in vacuum close to the warm body.”.
Let’s make that cold object a thin shell completely surrounding the sphere.
“The warmer body will start warming the colder body immediately.”
Yes, but all this time, the cool outer shell will be radiating less than 240 W/m^2 out to space (because it is colder than 255 K). This means energy will be building up inside the shell, warming the shell. Eventually the shell will warm to 255 K. At this point the shell will be radiating 240 W/m^2 and a steady-state will be achieved. (Actually, the shell must be slightly less than 255 K since it will have slightly larger surface area, but this is a minor correction).
“Actually, already on this stage we should start screaming and crying “how come [the inner object warms, and not just the shell around it]?”
I would be screaming “how could you doubt the inner object will warm?”.
Suppose the inner sphere did NOT warm above 255 K. The shell and the sphere around it would both be 255 K. But heat (240 W/m^2) would be flowing with no temperature gradient. Why in the world would 240 W/m^2 of heat be flowing from one object at 255 K to another object at 255 K?
Max was right with
P = εσA(Th⁴ – Tc⁴)
There must be a difference between Th (the inner sphere) and Tc (the shell) for this 240 W/m^2 of power to flow. Th will be 302 K in this particular case.
There is no “run-away” effect.
* The inner sphere warms from 255 K to 302 K.
* The outer shell warms from 0 K to 255 K.
* 240 W/m^2 of heat moves from the inner sphere to the outer shell
* 240 W/m^2 of heat moves from the outer shell to space.
* No laws of thermodynamics are broken.

Reply to  tjfolkerts
April 24, 2013 5:04 pm

“No laws of thermodynamics are broken.”
Except that the source is now twice as hot as its own heat input. Of course, we have seen the textbook examples which show that this does not occur. A shell surrounding a heated sphere simply gets heated by the sphere. This does not cause the sphere to heat up some more. This was shown in worked problems in physics textbooks which I am sure you saw. The supposition of trapped radiation causing self-heating was refuted.
REPLY: Best advice: stop digging – Anthony

davidmhoffer
April 24, 2013 4:59 pm

Greg House;
And again, the IPCC “greenhouse effect” is physically impossible,
>>>>>>>>>>>>>>>
The only thing physically impossible is educating you, as a rather long list of PhD physicists, engineers, and chemists recognized around the world for their knowledge have repeatedly demonstrated on this blog.

Selgovae
April 24, 2013 4:59 pm

Mark, Gary, thanks for the responses.
I don’t get the clairvoyance argument. That model suggest photons are being fired off like tennis balls and that they ‘travel’ in space until they find a destination. That’s at odds with my mental image of relativity and the speed of light. Surely they know where they’re going (in a sense ‘going nowhere’ as they have no distance to travel or time to take), and therefore can’t leave until they have a destination.
“The temperature of the unheated plate (“plate 2″) never reaches that of the electrically heated plate (“plate 1″).”
Is that true? I’m thinking only of atoms on the surface nearest to plate 1. Won’t some of them temporarily reach the same temperature as plate1, thus slowing down the emission from plate 1 until they cool down again?

steveta_uk
April 24, 2013 5:01 pm

Max, your assumptions were wrong.
You started with a block of ice and a pot of boiling water. So you know the temps at the outset, 0C and 100C.
You then try and add energy radiated from the ice to the boiling water, and assume this means the pot is even hotter.
Why? That energy is already included in the energy being radiated by the boiling pot. The energy output from the pot is exactly as you calculated. If you know that this must include the energy from the block of ice (again, as you calculated, with the somewhat unphysical assumption that all that energy goes the the pot, and all other directions are at absolute 0) then you could work out the reduced heating requirement to keep the pot at 100C when the ice is present, compared to open space.
I.e. as per Dr Spencer, the block of ice has reduced the heating requirements of the pot.

davidmhoffer
April 24, 2013 5:02 pm

tjfolkerts;
There is no “run-away” effect.
* The inner sphere warms from 255 K to 302 K.
* The outer shell warms from 0 K to 255 K.
* 240 W/m^2 of heat moves from the inner sphere to the outer shell
* 240 W/m^2 of heat moves from the outer shell to space.
* No laws of thermodynamics are broken.
>>>>>>>>>>>>>>>
Thank you sir for injecting some reality into the discussion.

richard verney
April 24, 2013 5:10 pm

I can remember an old design of 3kw electric fire.
It consisted of three 1 Kw elements running horizontally, ie., one of the elements was positioned midway between the upper and lower elements. The back of the fire was a concave mirror. It had a simple grill over the front of a not sufficiently tightly spaced grill to stop prying fingers! The fire could be switched low ( the bottome 1kw element), meduium (2 of the 1kw elements) or high (all 3 of the 1kw elements).
When switched on high all 3 elements were on. The middle element would recieve ‘heat’ from being sandwiched between the upper and lower elements, and of course, in addition, that reflected from the concave mirror. The middle element never looked noticably hotter than the other two elements (ie., it did not appear white hot and the other element red hot, I recall that they all had a similar glow).
So the question is why did the middle element not get noticably hotter and take on a much brighter appearance?

Greg House
April 24, 2013 5:11 pm

davidmhoffer says (April 24, 2013 at 4:38 pm): “In today’s world, it is mind boggling to me that the CAGW nonsense can live side by side with the “back radiation doesn’t warm things” nonsense.”
============================================================
Who said “back radiation doesn’t warm things”??:evil:
It is “back radiation doesn’t affect the temperature of the source“, davidmhoffer, the source.

CodeTech
April 24, 2013 5:17 pm

Actually, the only reason I mentioned the gas being used in my windows was to be sarcastic. Here, they use nitrogen for the same reason they put nitrogen in car tires. It’s not as prone to pressure changes at different temperatures, and it’s dry.
In fact, the best gas to use in windows would be NO gas – a vacuum. Unfortunately that would tend to cause the glass to bow inward.
When it comes to windows in a winter climate the “radiative” factor is essentially meaningless. We’re rarely trying to keep solar energy out, and for that my windows have a tinted film. The idea is to have dead layers that slow thermal transfer.

steveta_uk
April 24, 2013 5:21 pm

CodeTech, they could make double glazing work even better by adding some glass wool as well as dry nitrogen between the panes to stop convection.
Might make it difficult to see out, I suppose.

April 24, 2013 5:24 pm

richard verney says:
April 24, 2013 at 5:10 pm
*post snipped due to exceeding the awesome quota*
So the question is why did the middle element not get noticably hotter and take on a much brighter appearance?

Good point, richard, why doesn’t the middle element get hotter/brighter?
” Max, your assumptions were wrong.
You started with a block of ice and a pot of boiling water. So you know the temps at the outset, 0C and 100C.
You then try and add energy radiated from the ice to the boiling water, and assume this means the pot is even hotter.”
~steveta_uk
Are you referring to my post at 1:39 PM?
Ok, have it be two black bodies of arbitrary material, large enough that the area we’re interested in is radiating towards the other body, one at 373 K and one at 273 K, radiating towards each other.
I’m confident that the radiation field between them can be described by the following equation:
P = εσA(Th⁴ – Tc⁴)

rather than:
P = εσA(Th⁴ + Tc⁴)
max@Funktastic:~$ calgebra
>>> (5.6703*10^-8)*(373^4 – 273^4)
782.63220248
>>> (5.6703*10^-8)*(373^4 + 273^4)
1412.55397668
Both of those can not be right, and if the colder black body raises the temperature of the warmer black body then the second one is correct, which would be roughly the same power as that emitted by a 297.28 K black body.
If the radiation from the colder body is added to that from the warmer body, that is akin to a block of ice raising the temperature of a pot of nearly boiling water by over 24 K!

For clarification, I am stating that this is the correct calculation:
P = εσA(Th⁴ – Tc⁴)
As follows:
>>> (5.6703*10^-8)*(373^4 – 273^4)
782.63220248

tjfolkerts
April 24, 2013 5:29 pm

TIM: “No laws of thermodynamics are broken.”
JOE: “Except that the source is now twice as hot as its own heat input.”
No!
The “heat input” is not “255 K”. The heat input is “240 W/m^2” to the shell and this was fixed in the example. This input will be whatever temperature is needed (within some broad engineering limits). (Similarly, sunlight will provide energy to earth’s surface, not matter what the surface temperature might already be.)
If the “heat input” were held at 255 K by some thermostat, then the power required to hold the inner shell @ 255 K would drop. Specifically, the power required would drop to 240/2 W/m^2 = 120 W/m^2 and the shell would only warm to 214 K.
There is no law of thermodynamics that says “a heater with a fixed power will be the same temperature no matter what ‘heat load” you impose.”

tjfolkerts
April 24, 2013 5:32 pm

DANG! That should have read “The heat input is “240 W/m^2″ to the INNER SPHERE… “ in my last post!

joeldshore
April 24, 2013 5:32 pm

Joseph Postma says:

You can’t average energy input into a geometry where it doesn’t exist.

This is just silly sophistry. If you don’t understand the concept of an average, you don’t have to use it. Just compute energy balance using Total energy in = Total energy out.

Both the wet and dry lapse rate can be calculated precisely without any reference to GHG radiation, for example.

Nice strawman. That is because the wet and dry adiabatic lapse rates represent stability conditions: A lapse rate larger than the appropriate adiabatic lapse rate is unstable to convection, which then is very efficient at lowering the lapse rate back down to the adiabatic lapse rate. Go read any elementary book on atmospheric science and stop pretending that the few things you say that are correct are actually new.

Mark Bofill
April 24, 2013 5:36 pm

Selgovae,
If you want to believe that about electromagnetic radiation, that it doesn’t radiate without some sort of ‘quantum entanglement’ handshake with a receiver at some future space time coordinate, be my guest I guess. I can’t disprove that, am not aware of any theoretical foundation or empirical data to support the idea, and have no particular interest in the notion.
Another way to tackle it might be this. Suppose you shine a flashlight at the sun. Without having performed the experiment myself, I’m utterly confident it would radiate, even though the surface of the sun is considerably hotter than the filament of the bulb. Take the experiment to space if you’ve got an issue with the atmosphere, I’m sure the results will be the same.

April 24, 2013 5:38 pm

“The heat input is “240 W/m^2”
Well this isn’t really relevant to the planet Earth in any case because the input is not 240 W/m^2 but 610 W/m^2, as an integrated average over the projection factor on the hemisphere. So, any of these argument starting at 240 W/m^2 are irrelevant, because they don’t correspond to reality. Of course, the atmosphere isn’t a source of heat and so all the arguments created to try to make it a source of heat or a cause of heating are wrong. The error originates when we assume that sunshine is -18C instead of +49C.
Again, we have seen the text book examples which show that the interior sphere does not become heated. We do not find this in worked textbook examples. If you have a constant power input creating thermal radiation inside a cavity, this just produces a blackbody spectrum, it doesn’t cause a shift in the frequency spectrum of the radiation and thus higher temperature.

davidmhoffer
April 24, 2013 5:39 pm

Greg House;
Who said “back radiation doesn’t warm things”??:evil:
It is “back radiation doesn’t affect the temperature of the source“, davidmhoffer, the source.
>>>>>>>>>>>>>>>>
Ah yes, the source has a little sign at the front door that says “If you have ever been here before, leave your joules of energy at the door”. The sign is probably easy to find, it is right next to Joe P*stma’s “molecule traps”.

KevinK
April 24, 2013 5:44 pm

Max,
Thanks for that kind (cough cough, condescending) explanation of the most basic units of physics (Watts (not Anthony) and Joules). I do note that those namesakes did not deem it necessary to trademark their names, maybe a little less hubris back in the day, oh how I yearn for simpler times.
You wrote;
“Will the presence of another body at the same temperature cause the temperature of the first body to increase?”
Certainly a totally fair question;
If the first body is NOT supplied energy from an external source, and it is radiating towards a colder object (Roy’s imaginary vacuum chamber walls) then NO the presence of a second body (at the same temperature or lower) WILL NOT cause the temperature of the first body to increase
HOWEVER, If the first body is supplied energy (not power) from an external source, and it is radiating towards a colder object, then YES anything that changes the “view factor” MAY cause its temperature to increase. I use the qualifier MAY because there are other secondary factors to consider, for example if a conductive path exists that can cool the second body faster than the first body can warm it.
So like many “simple questions” it is necessary to ask follow up questions to understand ALL the conditions present before pronouncing the final outcome of what is still just a thought experiment. That’s why I prefer real hard observations before accepting any Hypothesis based on the say so of MAX™, or Roy, or Anthony (not the historical one), or Lord Monckton.
BTW; as an engineer, “nitpicking” on my part keeps the people that use my products alive and uninjured. And I am proud to be one of the finest pickers of nits in these parts.
Cheers, Kevin

April 24, 2013 5:45 pm

Molecule traps are ridiculed but radiation traps are ok. Strange. When you stop a draft in your house, you’re stopping molecules – the gas – from moving around. This prevents heat loss, it prevents warm gas (molecules) from escaping with the draft to outside the house. Trapping radiation inside a cavity, however, just makes the radiation constructively and destructively interfere with itself, causing no net change; it actually creates a blackbody spectrum at the temperature of the source.

Gary Hladik
April 24, 2013 5:45 pm

Selgovae says (April 24, 2013 at 4:59 pm): “I don’t get the clairvoyance argument. That model suggest photons are being fired off like tennis balls and that they ‘travel’ in space until they find a destination. That’s at odds with my mental image of relativity and the speed of light. Surely they know where they’re going (in a sense ‘going nowhere’ as they have no distance to travel or time to take), and therefore can’t leave until they have a destination.”
Take a look at the Hubble Extreme Deep Field composite
http://en.wikipedia.org/wiki/Hubble_Extreme_Deep_Field
The light from some of those galaxies “left” about 13.2 billion years ago. If those photons stayed put until they had a destination (the Hubble telescope), some of them could see over 13 billion years into the future, which is what I call “clairvoyance”! 🙂
“I’m thinking only of atoms on the surface nearest to plate 1. Won’t some of them temporarily reach the same temperature as plate1, thus slowing down the emission from plate 1 until they cool down again?”
As a layman I’m uncomfortable discussing the temperature of individual atoms, thinking instead of temperature as a macroscopic property. But note that even the plate 2 layer of atoms facing plate 1 is getting radiated energy from one direction and losing it in two: by radiation toward plate 1 and by conduction to the rest of plate 2.
Besides, talk about clairvoyance! Now a photon can’t leave until it knows where its destination is and what the destination temperature is! Photon emission must be a nightmare process:
“This is Photon Traffic Control. Photon 10244759800321, destination carbon atom 8876635549, flight duration 983,376,455,987.44776 years, has been delayed indefinitely. Target temperature will be a balmy 7.02 degrees Kelvin at arrival time, which unfortunately is 0.01 degrees warmer than we are here. Photon 7466453553…”
The really sad thing about being a photon is that you never get to vacation in a warmer spot…

April 24, 2013 5:52 pm

Photons do not actually experience time or space in the way we expect of material entities. At the speed of light there is no lapse of time and all space is shrunk to zero length. It is an interesting relativistic thing to think about. Photons are in a sense in contact with everything at all “times”. Just apply the Lorentz equations with v = c. You get zero time and zero distance. This is life as a photon.
In any case, you can shine a flashlight at the Sun and the the beam will be there. But the flashlight doesn’t heat the Sun. Only hotter things heat cooler things.
Anthony, I am truly sorry if you think I thread bomb. That’s not what I am trying to do. I am just trying to state some things in a thread which regards me and reply to comments which are useful to make some points. I write fast, true, and can respond quickly, but it is not meant to “bomb”. Best regards.

Selgovae
April 24, 2013 5:52 pm

Mark,
“Suppose you shine a flashlight at the sun. ”
Under my notion, that causes some experimental difficulties. How do I measure whether the flashlight is radiating at the sun? I’m not confident at all about what actually happens. But I like the ‘handshake with a receiver at some future space time coordinate’ idea. Although I’m not sure ‘future’ is the right word, perhaps ‘different’.
Anyway, that’s probably enough for my feeble brain today. Thanks for the comments.

April 24, 2013 5:53 pm

“How do I measure whether the flashlight is radiating at the sun?”
That is a very good point!!! Insightful. Anything you put between you and the Sun to measure becomes the target. And with the strange nature of photons….hmmm. Smart fellow 🙂

KevinK
April 24, 2013 5:58 pm

Gary Hladik wrote;
“So if instead of using an electric outlet or a battery to heat the first plate, we use an embedded radioactive heat source
(redacted wiki link here)
the plutonium (for example) will decay more rapidly when the second plate is introduced?”
No, a radioactive heat source is not the same thing as an electric (actually electro-chemical) storage battery. Names matter, you can’t just willy nilly throw in different parts in your thought experiment and claim the same outcome. A radioactive heat source is infinite (for the time frames of interest) and is more like a battery that never discharges (something the current US DOE has inappropriate dreams about).
Different thought experiment, different expected outcome.
Cheers, Kevin.

Ivan
April 24, 2013 6:00 pm

So, this is actually an attack on Mishcolczy if I am not wrong? I think the only sensible reaction that any normal skeptical person will have here is: Wow, let’s see what got them all worked up so much?

Gary Hladik
April 24, 2013 6:00 pm

richard verney says (April 24, 2013 at 5:10 pm): “So the question is why did the middle element not get noticably hotter and take on a much brighter appearance?”
Have you actually done the calculations so you would know if the difference should be “noticeable”? Please show your work.

Gary Hladik
April 24, 2013 6:03 pm

KevinK says (April 24, 2013 at 5:58 pm): “Different thought experiment, different expected outcome.”
So if it’s radioactively heated, you wouldn’t expect plate 1 to get warmer when plate 2 is introduced?

April 24, 2013 6:13 pm

“If you don’t understand the concept of an average, you don’t have to use it. Just compute energy balance using Total energy in = Total energy out.”
An average applies to where something exists, to what is real. Especially with power inputs. The power from the sun is not ever averaged over the entire globe, but only a hemisphere. This is what you average, and you can do it using calculus – an integrated average. There is no integrated average for solar energy over the globe because that doesn’t exist. You have to do the average correctly. 240 W/m^2 is the output, not the input. Yes the input and output balance in total energy, but not in terms of power flux, not in terms of the work that they can do. The input flux can do way more work than the output flux. The input flux is much higher than the output flux. This is very important, because -18C or 240 W/m^2 input can’t melt ice into water, while the actual real-time solar input and physically real integrated average (of +49C) can do so. It makes a big difference. It is a very simple fact that real sunlight can melt ice into water all by itself, while -18C or 240 W/m2 never could. Then, with the high heat capacity of water, it can stay liquid for a very long time once the sun has melted it, much longer than a single night. Then the latent heat also acts as a barrier to dropping back below zero. This all holds the temperature higher. These are interesting scientific facts.
The wet and dry lapse rates are calculated perfectly without regard to GHG radiation. Convection is of course natural and occurs whenever the near-surface air is heated. Certainly the system will tend to whatever lapse rate is appropriate given the water vapor concentration and strength of gravity (the latter of which doesn’t change of course).

Mark Bofill
April 24, 2013 6:17 pm

How do you know it’s radiating? Measure the power consumption I’d say.
Look, if you want to turn this into a philosophical question, you can’t prove much of anything. Decartes and his malicious demon trying to trick him, or whatever the argument was. That’s fine, but I don’t think it’s got much to do with science. Stick with Occams Razor instead, it gets you places.

April 24, 2013 6:21 pm

I do note that those namesakes did not deem it necessary to trademark their names, maybe a little less hubris back in the day, oh how I yearn for simpler times.” ~KevinK
My last name has a T, M, and vowel sounds, and I noticed many sites will allow unicode for usernames, as Max is often taken, I use this one. Strictly speaking you read it as “Max Thyme”, so it is little different than your Kevin K moniker.
I condescend when condescended towards, seems fair enough right?
Anyways, enough snippiness, we are both adults, my point was raised because of various examples I find in textbooks, such as this one which do not support the idea that adding a passively heated object would raise the temperature of an actively heated object.
A classical example is a star in a dust cloud, the dust cloud will be receiving radiation from the star and will thus be warmed, correct?
There are some who would claim that the dust cloud would be warmed until it emits the same total amount to space as the star would have, and there are some who would claim that the star will emit approximately half of what it receives inwards and outwards, rather than the same intensity it receives in both directions.
I suppose I should show my hand and call though: if adding an object the same temperature at some distance could or would raise the temperature of an actively heated object, what would happen if the second object were brought into contact with the first?

April 24, 2013 6:23 pm

“How do you know it’s radiating? Measure the power consumption I’d say.
Look, if you want to turn this into a philosophical question”
It is quite a real problem, not just philosophical. You provided a good answer in terms of power consumption. That’s as philosophical as required.

Rosco
April 24, 2013 6:43 pm

Roy Spencer wrote:
“ALL
It is important to remember that there is NO WAY to determine the temperature of anything based upon the rate of energy input alone, for example the Earth absorbing an average of ~240 W/m2 from the Sun. Temperature is a function of BOTH energy input (typically not temperature dependent) AND energy loss (typically VERY temperature dependent), neglecting issues related to heat capacity which mainly affect the time required for the system to equilibrate. The temperature of anything heated will increase until the rate of energy *loss* equals the rate of energy *gain*. So, temperature can be increased by increasing INPUT, or decreasing OUTPUT.”
I find many things wrong with this statement.
1. One assumes that the 240 W/sq m “average” over a sphere is derived from 960 W/sq m over a disk – this is the fundamental basis of all the initial mathematical derivations of the greenhouse theory I have seen.
Using Stefan-Boltzmann – 240 W/sq m equates to a maximum temperature of about 255 K.
Using Stefan-Boltzmann – 960 W/sq m equates to a maximum temperature of about 360 K
Which temperature more accurately predicts the potential heating power of the Sun ?
Remember that one “disk shaped” hemisphere of the Earth is always illuminated – it is simply the portion of the Earth that changes form day to night.
2. Your statement appears to imply that you can heat something up beyond the power of the input radiation by reducing output. This is in contradiction of the results of Planck et all and is reflected in the Stefan-boltzmann equation and its application by climate science.
3. I remember that you once described a post I made where I stated my belief that the period of rotation of a planet was significant as :-
“This post is a good example of how some people have polluted the minds of others with pseudo-science. ”
I’m not sure I’ve polluted anyone’s mind – they are free to take it or leave it – as I am free to find little credibility in many of your arguments.
Are you now saying that the period of a planet is now crucial ?
The Moon and Mercury both represent heated surfaces radiating to space. Both have significant temperature swings from high to extremely high down to about 100 K – perhaps lower.
The Moon has been mapped and loses “heat” at about 1 K per hour average from noon to sunset – a time of about about 177 Earth hours to decrease from about 390 K to about 220 K and a further 120 K, maybe more, in the 354 Earth hour long night.
The Earth obviously loses less energy in 12 hours of night than the Moon in about 354 Earth hours or Mercury in its 2111 odd Earth hour long night.
Is not this the same as saying the Earth’s output is decreased as compared to the others ?
Venus is a totally different kettle of fish but I find it amazing that climate science can keep a straight face claiming Venus’ 730 K or 16,101 W/sq m – yes 16.1 KW/sq m – can be caused by the albedo adjusted solar input of 132 W /sq m – lecture 3 on the Greenhouse Effect UCLA.
4. When you say “So, temperature can be increased by increasing INPUT, or decreasing OUTPUT.” I assume you refer to the “constant” input of solar radiation.
Even so the Stefan-Boltzmann determines that the temperature is determined by the radiation input so the question remains – can one increase the radiative input by adding the radiation from a cold atmosphere to the solar radiation to create a higher temperature ?
If so which radiative input – the more realistic 960 W/sq m during the day exclusively or the mythical averaged 240 W/sq m ignoring day and night ?
Remember a hemisphere of the Earth is always in day – under the 960 W/sq m (not my figures by the way but simply the albedo reduced solar constant over a disk as used in calculating the “effective radiating temperature of Earth) – it is simply rotating which part is under the Sun at its zenith.
I find “averaging” the solar input as meaningless for any real purpose – the Sun is either heating the surfaces of a part of the Earth during the day or not during the night.
Adding support to this is the real observation that my solar panels only produce during the day and they require an input of approximately 1000 W/sq metre to produce anything like their rated capacity.
At 240 W/sq m illumination at 15.6% efficiency (quoted for 1000 W/sq m – significantly less for lower power but why argue) and 1.2 sq m panels my 14 panel array would produce about 630 W per hour.
I know they produce a “feed in” of up to 7 KW hrs a day and supply up to 5 Kw hours of reduced energy from the grid.
I don’t have a 19 hour long day at 240 W/sq m solar input – we rarely have more than 14 hours sunlight a day being almost within the tropics.
Where I live the clear sky solar power at noon is calculated approximately to be Cos 27 degrees S x 1367 x 0.7 or about 852 W/sq m.
842 x 1.2sq m per panel x 14 panels results in a maximum of 2.2 Kw per hour and requires 6 to 8 hours of varying levels of solar radiation due to the angle of the Sun.
This result seems much more realistic to me.
Anyway – that’s my 2 cents worth.

CodeTech
April 24, 2013 6:49 pm

By the way, if you shine your flashlight at the sun you won’t reach it. You need to lead. Slightly to the right. But your beam WILL hit the sun. And be just as effective as the rest of the light bouncing off the planet at increasing the sun’s overall temperature… unless you have a mighty bright flashlight.

tjfolkerts
April 24, 2013 6:53 pm

Joe Postma says: Well this isn’t really relevant to the planet Earth …
Well, the discussion is specifically relevant to the situation described above (introduced April 24, 2013 at 2:39 pm and refine April 24, 2013 at 4:58 pm). Quite clearly this is not the earth. But to the extent that correct physics is relevant everywhere, it is relevant to earth.
” … the input is not 240 W/m^2 but 610 W/m^2, as an integrated average over the projection factor on the hemisphere. “
No, the input from the sun (accounting for albedo) is ~ 0.7 * 1370 W/m^2 = ~ 960 W/m^2
Averaged over a hemisphere, that is 480 W/m^2
Averaged over a whole sphere that is 240 W/m^2
Your 610 W/m^2 is averaged over a CYLINDER, not a hemisphere (or equivalently, it only applies at the equator, but no where else). You are too big by a factor of 4/π for a hemisphere.
“The error originates when we assume that sunshine is -18C instead of +49C.
Sunshine has (approximately) the spectrum of a ~ 5800 K blackbody. As such, the only truly meaningful “temperature” of the photons is 5800 K. Assuming either -18C or 49C is wrong.
* Focus sunlight onto an object from all sides and that object will be 5800 K (but sunlight can never be focused to warm the object above 5800 K because that would violate the 2nd Law of Thermodynamics).
* If that sunlight comes from only a small part of the sky (ie a disk ~ 0.5 degrees across as seen at 1 AU, with the rest of the sky @ 2.7 K) ) and is spread over a hemisphere with albedo 0.7, the object will be ~ 88 C
* if is is spread out over a cylinder, the object will be ~ 49 C
* if is is spread out over a sphere, the the object will be ~ -18 C
Sunlight is 5800 K. Geometry and albedo combine to limit its warming ability. (But if you make the rest of the sky warmer than 2.7 K, the combined effect of this sunlight and “atmospheric light” will combine to make the new temperature -18 C < T < 5800 K, ie somewhere between the temperature of the cool atmosphere and the warm sun).
“Again, we have seen the text book examples which show that the interior sphere does not become heated.”
Here is one “textbook example” (Sorry I don;t have the reference for the original).
http://i341.photobucket.com/albums/o396/maxarutaru/bunnytaru/lol%20censorship/heatsourceinshellatequilibrium_zps1c9d662a.png
This will get a little subtle, so hopefully people will not get confused — the situation in this textbook and the situation Greg introduced are SIMILAR, but not the same.
The “black sphere” (eg a star) in this example DOES stay the same temperature BUT that was part of the assumptions for this example. What DOES change is the power input. This star maintains the same temperature but with HALF the power when the shell is added.
That means that if the power was “turned back up” to the original value (corresponding to Greg;’s example), the interior would necessarily warm up.
Sorry, but this particular textbook supports me and not you. Do you have a “textbook example” you can cite that supports your claim?

joeldshore
April 24, 2013 6:55 pm

Joseph Postma says:

Well this isn’t really relevant to the planet Earth in any case because the input is not 240 W/m^2 but 610 W/m^2, as an integrated average over the projection factor on the hemisphere. ..
The error originates when we assume that sunshine is -18C instead of +49C.

Cut the sophistry, Joe. The Earth absorbs an amount of energy each second equal to (pi*R^2)*L_solar*(1-albedo)*(1 second) where the albedo is about 0.3, the solar constant L_solar = 1361 W/m^2, and the radius of the Earth is R = 6371 km. That makes for 1.21 x 10^17 Joules.
If the Earth were at an average temperature of 49 C, its emission of energy each second would be [at least because of Holder’s Inequality] (4*pi*R^2)*sigma*T^4*(1 second) where T = 322 K and the Stefan-Boltzmann constant sigma = 5.67 x 10^-8 W/[m^2*K^4]. That makes for 3.11 x 10^17 Joules. [If instead we use the correct average temperature of 255 K (-18 deg C), we get an emission of 1.22 x 10^17 Joules, which within the accuracy of the calculation, gives energy balance.]
In other words, you are spewing nonsense in an apparent attempt to confuse people who don’t know any better.

Reply to  joeldshore
April 24, 2013 8:07 pm

“That makes for 1.21 x 10^17 Joules.”
And these Joules are absorbed over a hemisphere with the intensity distribution going as the cosine of the solar zenith angle. This is reality.
“If the Earth were at an average temperature of 49 C”
I did not ever say that the Earth was at +49C. I said this is the equivalent temperature value of the average input. This too is reality. This is very different from the usual assumption of a -18C or 240 W/m2 input. It makes a big difference in what the input can and can not do. The total energy is conserved, the input is equal to the output in terms of total energy, but not in terms of power, not in terms of what the input and output power flux can do. The input can do a lot more work than the output. The input can generate very high temperatures on the surface. The atmosphere is then heated by the surface conductively/convectively and radiatively. That’s all that needs to happen because the input is so hot. Hot heats cold. You can go outside and feel the hot input on a sunny day. Latent heat carries a lot of this input around the planet via water vapor and liquid water.

April 24, 2013 7:07 pm

Here’s the original, Tim: that up there is just a snapshot I took to embed.
There was nothing about the power input being changed in that example, go to problem 1026 to see another examination of this problem.
As for the integration over a sphere, you’re treating his 610 value as a projection, it is integrated from the peak value at the subsolar point (1366~ before albedo) to the value at the edges (0~) of the terminator.
After albedo it works out to around 476 W/m^2, or as I prefer to express it, 1.22×10^17 Joules absorbed by one hemisphere, and ~5.11×10^16 Joules emitted by each hemisphere.

tjfolkerts
April 24, 2013 7:10 pm

@ Max April 24, 2013 at 6:21 pm
Your “textbook example” is similar to the one I just discussed. It says that to maintain a constant temperature, an object needs less power by a factor of
R^2 / (R^2 + r^2)
If the radius of the shell (R) is about the same as the radius of the sphere (r), this becomes 1/2. So only 1/2 as much power is needed to keep the inner sphere warm with the shell in place as was required without the shell. This example has the advantage of also showing what would happen is the shell was significantly larger that the sphere inside.
And as I just said, if the power was “turned back up” to the original value (corresponding to Greg’s example), the interior would necessarily warm up above its original temperature.

KevinK
April 24, 2013 7:12 pm

Max, fair enough, I took the “TM” as a slight bit of “argument from authority”, my apologies, I did not realize it was a simple contraction from a common name.
And my generic observation about using the right units was not “aimed” at anyone specific, just a “look out that you don’t step in that same mud puddle I just soiled my pants with” kind of observation. I have been tripped up many times when I mismatched the correct units for a calculation. One of the most valuable lessons back in college was to “reconcile your units” always, always, always…..
So, in summary, I believe that the proponents of the “GHE” have some valid points; for one, a photon leaves its source without regard to where it’s going. And yes a colder object (or gas cloud) can affect the temperature of a warmer object (ONLY by slowing the velocity of energy flowing away from the warmer object, that’s what the fiberglass in the walls of your house accomplishes)).
But the “deniers” of the GHE also have some valid points; the proponents of the GHE hypothesis have not demonstrated that the “effect” actually reduces the velocity of heat/energy (they alternate between IR EMR and molecular rate of vibration (i.e. temperature)) through the system. In fact I maintain that the “GHE” only causes heat/energy to make multiple passes through the system at the speed of light (there are several flavors of the “speed of light”, the fastest is in a vacuum, and the slowest is in the “flint” glasses) which is known to be MUCH faster than the speed of heat through any gases (GHG or otherwise) or the Oceans or Icesheets or soil in the complex Earth/Atmosphere system.
The much ballyhooed “GHE” only delays the transit of heat/IR by a few tens of milliseconds. That is why you can measure it with an IR sensor with no time reference (Dr. Spencer’s “look I can measure it” posting) and not realize that it is simply “passing through” at the speed of light and causes no ”Higher Equilibrium Temperature”.
Cheers, Kevin.

April 24, 2013 7:19 pm

“Do you have a textbook example?”
Here are some textbooks examples of how a continuous source of heat as a sphere trapped inside a shell doesn’t cause an increase of temperature of there sphere (with continuous input):
http://books.google.co.in/books?id=dQGC0ifkE34C&pg=PA24&lpg=PA24&dq=concentric+sphere+black+body&source=bl&ots=Zh6N1e35jc&sig=m-7nVWch4_zv-l3ISR5k7bluSUQ&hl=en&sa=X&ei=_ldYUd7EFZLU9ATVzYHoDw&redir_esc=y#v=onepage&q&f=false
Also problem *1023* above it. Oh, I see that’s the one you referred to. The problem says nothing about halving the input to the sphere, nor does the other one. You’re making that up. The problem specifically refers to a continuous input which means the sphere is heated to a constant temperature. You could think of it as an infinite heat sink. There is nothing in the two examples about the input being reduced – the input is constant.
“Averaged over a whole sphere that is 240 W/m^2”
You can not average the input over a sphere because the input does not occur over a sphere. The input does not occur over a sphere, it only ever occurs over a hemisphere. The 610 W/m^2 average input is the integrated average intensity over the hemisphere. The energy comes in over a hemisphere, and so this is the correct and only meaningful way to average it, as an integrated average over that geometry.
“the only truly meaningful “temperature” of the photons is 5800 K. Assuming either -18C or 49C is wrong.”
Tell that to the GHE crowd. Of course, we are speaking about the heating potential of sunlight at the distance of the Earth and given its input distribution on the Earth. Which results in the integrated average as discussed. Of course the raw heating potential from sunlight for a flat blackbody is +121C.
The only way to get higher temperatures than this from sunlight at the Earth is to re-condense it using a magnifying glass. GHG’s do not re-condense incoming sunlight, hence they can not create 5000K out of sunlight. GHG’s only interact with the outgoing IR, and as you have pointed out they can not create a higher temperature than their source – the surface of the Earth.
Of course, the actual input is +49C as a temperature and this is higher than the surface temperature. Total energy balances out but the input itself is much stronger than the output – the input can heat the Earth to high temperature, melt ice etc etc.

April 24, 2013 7:27 pm

Steven Mosher [April 24, 2013 at 3:42 pm] says:
Wrong.
The “greenhouse’ effect has nothing whatsoever to do with these types of experiments.
Energy leaves earth Via one route and one route only: radiation to space.
Earth radiates to space from the ERL.
When you add GHGs you raise the ERL.
When you raise the ERL the earth radiates to space more slowly.
That results in a surface that is warmer than it would be otherwise.
This effect cannot be [seen? measured?] tests with woods like experiments EVER because the effect depends upon changing the ERL and no experimental set up or lab can duplicate the conditions required: the full atmospheric column.
in 20 or so simple slides

http://www.aos.wisc.edu/~aos121br/radn/radn/sld001.htm

Well that’s awfully convenient. Greenhouse Theory hypothesis cannot be experimentally reproduced so we’ll just have to take the word of climate kooks instead. Experiments replaced by models and slideshows. Yep. The natural conclusion of the pop-science movement ever since Sagan, Ehrlich, Hansen and Mann lowered the bar to unimaginable levels.
P.S. some C02 related experiments can be reproduced Steven …

LdB
April 24, 2013 7:27 pm

Roy I am so glad that scientists on the AGW side of climate science is taking on the pseudoscience junkies and there is no greenhouse effect garbage it had been worrying me for sometime.
The fact all these pseudoscience types ignore is the ability to challenge it is way out the league of climate science stupidity because it meshed in Planck’s law which forms part of the foundations of Quantum Mechanics.
Planck already worked out classic physics of Boltzmann, Wein and Kirchoff were simplifications based on classic physics. Using quantum mechanics he showed you could derived the classic physics law it is an important piece of science history like E=MC2.
A reasonable short version link on the subject:
http://csep10.phys.utk.edu/astr162/lect/light/radiation.html
Wikipedia carries the full version of derivations of the classic laws from Planck’s laws
http://en.wikipedia.org/wiki/Planck%27s_law_of_black-body_radiation
The dragon slayers like all pseudoscience lunatics miss the fact we know QM is right we have tested it extensively and continue to test it everyday of every year and to date it has never been shown wrong.
I have stated this before we also know that the greenhouse gases are quantum active because all of the gases involved can also be used as the medium in a standard laser tube (CO2, N2O, Water Vapour). If you want absolute proof that you can optically pump the gases involved it doesn’t get more basic than a laser because there are only 3 requirements for a laser tube
A laser is constructed from three principal parts: (http://en.wikipedia.org/wiki/Laser_construction)
An energy source (usually referred to as the pump or pump source),
A gain medium or laser medium, and
Two or more mirrors that form an optical resonator.
So dragonslayers there is your absolute proof that the substances must be a gain medium because they can be used as a laser medium.
All of this is way beyond doubt and argument especially when your argument omits the fact that your alternative explaination is falsified by Quantum mechanics.

KevinK
April 24, 2013 7:29 pm

Whoops, I inadvertently typed the “D” word, apparently my latest comment is “trapped”…………….. ha ha ha
I still don’t see much “slaying” going on here, Dr. Spencer’s “The Sun/Earth system is just like your House/Furnace” analogy is (for lack of a better match) still a severely “urine deprived” analogy.
Cheers, Kevin

Gary Hladik
April 24, 2013 7:30 pm

tjfolkerts says (April 24, 2013 at 7:10 pm): ‘And as I just said, if the power was “turned back up” to the original value (corresponding to Greg’s example), the interior would necessarily warm up above its original temperature.’
Aaaaaand we’re back to the steel greenhouse:
http://wattsupwiththat.com/2009/11/17/the-steel-greenhouse/

Genghis
April 24, 2013 7:37 pm

Roy Spencer says:
April 24, 2013 at 1:37 pm
ALL:
It is important to remember that there is NO WAY to determine the temperature of anything based upon the rate of energy input alone, for example the Earth absorbing an average of ~240 W/m2 from the Sun. Temperature is a function of BOTH energy input (typically not temperature dependent) AND energy loss (typically VERY temperature dependent), neglecting issues related to heat capacity which mainly affect the time required for the system to equilibrate. The temperature of anything heated will increase until the rate of energy *loss* equals the rate of energy *gain*. So, temperature can be increased by increasing INPUT, or decreasing OUTPUT.
———
Kirchhoffs law says that emission = absorption. Absorption at the TOA equals apx 341 W/M ^2 which if we plug into the S-B equation gives us 5˚ C for any albedo value given Kirchhoff’s law.
It just so happens that the average ocean temperature by mass is apx equal to 5˚C and the atmospheric temperature by mass is apx equal to 5˚C. Which clearly demonstrates the accuracy of Kirchhoff’s law and the S-B equation. 341 W/M^2 in = 341 W/M’2 out, which essentially never changes! Which means that the average temperature never changes either!
But we aren’t really talking about average temperatures here are we? We are talking about transient boundary conditions where the surface of the oceans temp varies from 0 to 30˚ and the temperature of the air immediately over the ocean has the same temperature.
But we aren’t really talking about that either are we? In fact the whole AGW meme is over the response time, the rate at which temperature lags radiation inputs.. And that is the question isn’t it? Because the lag time can range from a few seconds for a photon at 15 microns to travel out of the atmosphere impeded by CO2 ( and yes warming the atmosphere temporarily) to millenia if the photon is headed into the dark recesses of the ocean.
While we don’t know the exact response times, we do have clues don’t we? We know that CO2 levels have been much higher in the past and there were no hockey sticks, so we can falsify Hansen’s AGW theory out of hand. That doesn’t mean the GHG effect isn’t real, it is, we are just a little uncertain about its limits.

KevinK
April 24, 2013 7:58 pm

Gary Haldik wrote;
KevinK says (April 24, 2013 at 5:58 pm): “Different thought experiment, different expected outcome.”
“So if it’s radioactively heated, you wouldn’t expect plate 1 to get warmer when plate 2 is introduced?”
Frankly I have not thought about that one yet, few of us folks out in the real world have access to radioactive heat sources, although I do scan ebay once in a while to see if I can get rid of this d—n utility company that wants my money in exchange for their energy.
But if it is radioactively heated (an effective “unlimited” heat source, anybody else remember back in the 50s and 60s when electricity was going to be “too cheap to meter” ?) I would expect that the “cycle time” concerns would go away. And then plate 1 and plate 2 (and all the unicorns in the neighborhood) would all be warmer and cozier. So as soon as the Sun stays UP all day the GHE should show up. I could of course model that for say about $2 million US (oh wait I forgot about all the travel to warm exotic places to WARN everybody about the terrible dangers presented by warm cozy unicorns) Better make that $1 billion.
Cheers, Kevin.

Reed Coray
April 24, 2013 8:02 pm

Dr. Spencer, whom I greatly admire, wrote:
Roy Spencer says: April 24, 2013 at 12:12 pm
Greenhouse gases (thermodynamically like insulation in your house) reduce the rate at which heat flows from higher temperatures to lower temperatures, thus making the warm side warmer, and the cool side cooler.”

I disagree with the above statement. I am going to apply what you wrote to a vacuum thermos bottle. Place coffee in a thermos bottle and place the thermos bottle in a heat sink at a fixed temperature lower than the temperature of the coffee. The thermos bottle has a “warm side” (the wall of the chamber that holds the coffee–i.e., the inside wall) and a “cool side” (the wall in contact with the heat sink–i.e., the outside wall). In a vacuum thermos bottle, the volume between these two “walls” is empty–i.e., is a vacuum. According to what you wrote, relative to a vacuum thermos bottle inserting a greenhouse gas into the vacuum space will “reduce the rate at which heat flows from the higher temperature to the lower temperature.” If true, the performance of the thermos bottle (i.e., the time required for the temperature of the coffee to reach the temperature of the heat sink) is enhanced (lengthened) by the presence of the GHG. I don’t believe it. If this were true, then why does anyone manufacture vacuum thermos bottles? Wouldn’t CO2 (a greenhouse gas) thermos bottles be preferred?
Although you didn’t explicitly say so, if what you say is true, shouldn’t increasing the amount of GHG placed in this space result in improved thermos bottle efficiency? Do you really believe that relative to a vacuum thermos bottle, a thermos bottle with high pressure CO2 gas (or for that matter, CO2 gas at any pressure) in the region between the warmer (inside) wall and the cooler (outside) wall will reduce the rate thermal energy leaves the coffee, and therefore reduce the rate at which the temperature of the coffee declines?

joeldshore
April 24, 2013 8:03 pm

Genghis says:

Kirchhoffs law says that emission = absorption. Absorption at the TOA equals apx 341 W/M ^2 which if we plug into the S-B equation gives us 5˚ C for any albedo value given Kirchhoff’s law.

(1) No Kirchhoff’s Law says emissivity = absorptivity, not emission = absorption.
(2) Kirchhoff”s Law is true only at a fixed wavelength. An object can have a completely different absorptivity for radiation from the sun, which is primarily at visible and near-IR and UV wavelengths, as it has emissivity for radiation at terrestrial temperatures, which is primarily at far-IR wavelengths.
(3) The 341 W/m^2 value is the amount that is incident at the TOA, not the amount that is absorbed.
And, no, we are not talking about “lag times”. We are talking about the ***rate*** at which energy is emitted and absorbed.
It is really painful to read all of these misconceptions. It is fine to not understand things, but please don’t trumpet your misunderstandings as wisdom.

April 24, 2013 8:09 pm

“explaination is falsified by Quantum mechanics”
A laser is not a demonstration of the GHE. QM in fact shows that radiation trapped inside a cavity with a continuous source simply creates a blackbody spectrum. This is the very origin of QM, of Planck’s Law. Trapped radiation does not change its own frequency distribution and hence its temperature.

April 24, 2013 8:09 pm

So dragonslayers there is your absolute proof that the substances must be a gain medium because they can be used as a laser medium.” ~LdB
So… your argument is that because you can excite CO2 molecules and then stimulate them to emit photons… therefore greenhouse effect?
>.>

davidmhoffer
April 24, 2013 8:10 pm

LdB;
The dragon slayers like all pseudoscience lunatics miss the fact we know QM is right we have tested it extensively and continue to test it everyday of every year and to date it has never been shown wrong.
>>>>>>>>>>>>>>>>
Well said. Every day engineers design and build everything from mundane kitchen appliances to nuclear reactors using the exact laws of physics applied in exactly the same way that Spencer and tjfolkerts and Joel Shore and others are explaining. The things they design work. When they don’t, it is always because there was a design error. There has never been a case otherwise. P*stma and company would have us believe otherwise.

joeldshore
April 24, 2013 8:13 pm

Reed Coray says:

Although you didn’t explicitly say so, if what you say is true, shouldn’t increasing the amount of GHG placed in this space result in improved thermos bottle efficiency?

No. Different effects dominate in different circumstances. Heat transfer is a process that can involve conduction, convection, and radiation. For the Earth, the only significant communication of energy out into space is via radiation. For a thermos bottle, conduction and convection of heat are more important (and CO2 probably doesn’t have much effect on radiation because the effect on reducing radiation from the Earth is a subtle interplay between the absorption and re-emission of radiation by greenhouse gases and the temperature structure of the atmosphere which controls how much they emit). Hence, in a thermos, conduction and convection would be increased by putting CO2 in and radiation would hardly be affected at all.

KevinK
April 24, 2013 8:18 pm

LdB;
OH GOD LORD, not the atmosphere acts like a laser argument AGAIN…..
So dragonslayers there is your absolute proof that the substances must be a gain medium because they can be used as a laser medium.
Absolute proof that the substances must be a gain medium………
Where to start;
1) The “gain medium” in a laser DOES NOT, EVER, NEVER, NO WAY, NO HOW provide energy gain, PERIOD, END OF SENTENCE… (moderator please add lots of periods if you have nothing else to do)
2) EVERY laser known to mankind CONSUMES FAR MORE ENERGY INPUT than it produces in coherent EMR (light) output, typically with a ratio of about 5%.
3) For example, a typical semiconductor laser diode produces say 5 milliwatts of coherent optical power and consumes something like 50 milliamps times 2.5 volts or 125 milliwatts of electrical power, for an efficiency of 5/125 = 4%. Go to DigiKey™ or RadioShack™ and check the specs for an “off the shelf” laser diode, these are typical values.
SHOW US THE OPTICAL RESONANT CAVITY IN THE ATMOSPHERE……… (again lots more periods needed here)
It turns out that making optical resonant cavities is in fact VERY DIFFICULT (hence there are very few “garage shops” that make lasers).
A gas CAN be a gain medium (power gain, NOT energy gain) when properly confined and excited, but this is HARD to do.
Cheers, Kevin.

LdB
April 24, 2013 8:19 pm

@Reed Coray
You can’t use classic physics to try and understand any of this CLASSIC PHYSICS is wrong or at best a gross simplification you are dealing with a quantum effect.
The anti-greenhouse position is basically saying a non gain medium does not exist it was an argument that the advent of the laser disproved. There are mediums that have a gain factor because of quantum mechanics.
You can’t talk your way around this with classic physics garbage its a simple test can a medium have a gain factor QM said yes it can and the laser was invented.
The fact CO2 laser exist tells you that CO2 has a medium gain there is no other option or the CO2 laser wouldn’t work.
It is that simple and basic and no amount of stupid classic physics with thermos flasks is going to save you … simple question posed is CO2 a quantum gain medium answer yes …. end of story.
No possibly way to talk around it unless you want to alternatively explain how a laser works.

peterg
April 24, 2013 8:22 pm

If there was a gas layer where the major energy inputs were ozone absorbing shortwave (solar radiation, especially UV) and convective inputs from below, and the energy outputs were primarily CO2 molecules radiating long wave to outer space, then extra CO2 could well have a cooling effect, because more CO2 molecules would cause more radiation. This effect would probably be in the nature of a negative feedback rather than an overall cooling.

April 24, 2013 8:37 pm

The fact CO2 laser exist tells you that CO2 has a medium gain there is no other option or the CO2 laser wouldn’t work.” ~LdB
Ok, are you really equating the process of exciting CO2 molecules and stimulating emission of photons… to the description of the greenhouse effect?

Tim Folkerts
April 24, 2013 8:38 pm

Joe says: “Here are some textbooks examples of how a continuous source of heat as a sphere trapped inside a shell doesn’t cause an increase of temperature of there sphere … There is nothing in the two examples about the input being reduced – the input is constant. “
SIGH.
No, those examples have a constant TEMPERATURE with a DECREASING heat source –exactly the opposite of what you are claiming. The whole POINT is to calculate the decrease in power required. The heat flow required to hold a constant temperature drops (from 240 W/m^2 to 120 W/m^2 for a thin shell close around the inner sphere held at 255 K using my specific numbers). If the power was turned back up to 240 W/m^2, the inner sphere would warm up (to 302 K from 255 K using my numbers).
” The 610 W/m^2 average input is the integrated average intensity over the hemisphere.”
No. That is still the integrated average at the equator. You are over-estimating the heating power of the sun. (Interestingly, this will STILL only get the equator up to a 24 hr average temperature ~ -2 C. That is warmer than -18C, but not nearly warm enough to account for the surface temperature.)
“The only way to get higher temperatures than this from sunlight at the Earth is to re-condense it using a magnifying glass.”
An, but we don’t have to rely on just sunlight to warm things. The very hot sun @ 5800K only covers about 0.001% of the sky. If the rest of the celestial sphere were @ 2.7 K, then the we get those temperatures we both know well (~ 361 K @ noon @ the equator (with vanishingly small heat capacity) or ~ 255 K averaged over the whole earth over 24 hr).
But we could put something ELSE over the other 99.999% of the celestial sphere surrounding us. Even something quite cool that covers ~ 100,000 times as large of a solid angle as the sun will integrate out to be a significant amount of added power to the ground.
And of course, that “something else” is clouds and GHGs that can radiate some energy IN ADDITION TO the sunlight.

Scott
April 24, 2013 8:38 pm

Thanks so much for posting the link to this Anthony. For those of you who wonder why Roy would write such a post, go look at the comments on most of his articles for the last several months–they’re basically dominated by people saying the GHE doesn’t exist…even when the topic has little to do with the GHE. It’s very unfortunate because Roy has some VERY good material at his site. His blog entry immediately before the one linked here is, in my opinion, the number one argument skeptics have on their side…yet the comments were overrun by GHE naysayers, really ruining the discussion. 🙁
-Scott

April 24, 2013 8:43 pm

Tim, I’m not sure if you’re getting the same thing from those examples, are you talking about problem 1023 and 1026 from the links I gave?
Oh, and LdB: http://upload.wikimedia.org/wikipedia/commons/4/48/Commercial_laser_lines.svg
Xenon, Neon, Helium, Argon, Flourine, CO2 and various other elements can be used in a laser, would we have a greenhouse effect if we replaced CO2 with Ruby or Sapphire? (joking of course, but if you wanted to compare it more fairly, work out what a thin shell of ruby or sapphire around the planet would do)

Mooloo
April 24, 2013 8:47 pm

It is “back radiation doesn’t affect the temperature of the source“, davidmhoffer, the source.
Sources don’t have a temperature. They have an energy flow. (That’s what being a source means. Otherwise they would just be a hot thing.)
Their temperature is determined by the ability of the heat to get away. If I put a electric element into a block of copper and a similar mass block of concrete, the concrete will reach a higher temperature at equilibrium. Because it’s an insulator and copper is a conductor. The energy flow in will be the same, but the concrete will need to reach a higher temperature before its energy flow out matches the energy put in.
If you cannot tell the difference between temperature and energy you should leave any thoughts about discussing physics behind.
If I understand the sky-dragon position, a cold mirror should not reflect radiation back at the source. After all, by their “logic” it is only the temperature difference that matters. So how is it that mirroring is an effective means of insulation?

April 24, 2013 8:52 pm

“those examples have a constant TEMPERATURE with a DECREASING heat source ”
They say nothing about a decreasing heat source. You are making things up. They can be thought of as infinite heat sources (usually called a sink). You are changing the meaning of the questions. Radiation trapped inside a cavity doesn’t cause the cavity to heat up above the temperature of the source in any case, it just creates a blackbody spectrum at the temperature of the source.
“No. That is still the integrated average at the equator.”
It is the integrated average over the entire hemisphere. Which is the same as over the equator because an equatorial line is symmetric rotated about the solar zenith. You can literally feel hot sunlight and you can feel very hot rocks, surface, etc., which is created by the sunlight. This then heats the air, and the air circulates the heat about. Then you have the natural lapse rate making sure that the bottom of the atmosphere is the hottest part of the atmosphere, and the bottom of the atmosphere is not representative of the whole ensemble temperature in any case. The majority of the gas (N2 and O2) has very low emissivity and so can hold a much higher temperature than otherwise. Plus the heating only occurs mainly at the bottom of the atmosphere.
“that “something else” is clouds and GHGs that can radiate some energy”
GHG’s do not radiate at 5800K nor even at 286K. They never radiate at a warmer temperature than the surface and so can not heat the surface. GHG’s do not increase the sky-area of solar radiation and so they do not contribute to the solar spectrum at 5800K or the heating from such.

LdB
April 24, 2013 8:56 pm

@Max
Yes if we had ruby and sapphire in the atmosphere yes they too would greenhouse gases.
Here is a crash coarse in QM and the stupidity that is classic physics.
The classic physics view of temperature is it is the sort of the speed of the molecules in the substance. QM scientists 100 years ago realized that is a really bad approximation it is a number of complex quantum properties that make up an effect namely it appears to make a column of liquid expand in a thin tube. See the funny thing in QM there is not even a property called temperature because it is a mixture of different quantum properties.
Now under classic physics absolute zero is the temperature at which all motion of molecules stops and you can’t go any colder that’s not quite so under QM and here is the write up in nature magazine
http://www.nature.com/news/quantum-gas-goes-below-absolute-zero-1.12146
So before you start lecturing me about you massive understanding of temperature and how it works perhaps you can explain how we get a temperature colder than absolute zero to show us all you really really understand temperature and your classic physics works in all world cases.
See the problem classic physics has it is wrong has been for 100 years.
There isn’t an easy way to understand greenhouse effect under classic physics because its a quantum effect and nor can you explain a laser or temperatures below absolute zero using classic physics.

davidmhoffer
April 24, 2013 9:00 pm

Scott;
His blog entry immediately before the one linked here is, in my opinion, the number one argument skeptics have on their side…yet the comments were overrun by GHE naysayers, really ruining the discussion. 🙁
>>>>>>>>>>>>>>>>
You just triggered a thought. Dr Spencer helped design, implement and operate a network of satellites that accurately (verified by weather baloons, weather stations, etc) measures temperatures all over the globe and at many altitudes. Yet he is besieged by nay sayers insisting he doesn’t understand atmospheric physics. How do they suppose he got the satellites working correctly? A gigantic cosmic coincidence?

LdB
April 24, 2013 9:06 pm

To all you dragonslayers who want to keep arguing this stupidity here is a straight forward effect to explain away using your great physics.
We call it laser cooling that is cooling using something you would consider hot
http://en.wikipedia.org/wiki/Laser_cooling
It is routinely done now as the normal way to bring things down to absolute zero and on very large scales
http://web.mit.edu/newsoffice/2007/super-cool.html
So use your classic physics and explain what is happening .. small open challenge.

April 24, 2013 9:07 pm

Negative temperatures aren’t news, LdB, and I’m not operating from a classical position.
I’m not sure why you think I’m somehow ignorant of quantum mechanics, but I could also ask you which quantum paradigm you’re working within.
Spoiler alert: just because someone disagrees with you, it doesn’t mean they are ignorant of modern physics.

tjfolkerts
April 24, 2013 9:07 pm

Max asks: “Tim, I’m not sure if you’re getting the same thing from those examples, are you talking about problem 1023 and 1026 from the links I gave?”
Yes, I’m talking about the same examples. Specifically, 1026 says …
* The object is “at absolute temperature T” and is held at that temperature throughout the problem.
* The goal is to “find the factor by which this radiation shield reduces the rate of cooling”. Now “reduces the rate of cooling” is a bit ambiguous, but then the solution starts out with “the rate of energy loss”, so what they are finding is reduction in heat loss by radiation– which would be equal to the input of heat to maintain the constant temperature T.
* The only thing that is changing is that the temperature of the surroundings goes from T(0) to T(1). We have been assuming that T(0) was ~ 0 K, but that is not necessary.

Rosco
April 24, 2013 9:13 pm

Per problem 1023 from the link above –
for the radiation the limit approaches 0.5 for R approximately equal to r and it approaches 1 if R >>> r and the.
At least this “slays” Willis’ steel greenhouse completely !
The shell is always at a lower temperature than the enclosed object with the maximum being 84 %
In fact I am still not sure there isn’t something wrong with the whole concept of using a “nearby” heat shield.
This model has extra heat shields eventually resulting in almost no radiation escaping –
1/2 + 1/4 + 1/8 + 1/16 and we are already up to 93.75 %.
But at least it is progressing away from infinite energy by adding more shells unlike Willis’ proposal.

Greg House
April 24, 2013 9:13 pm

REPLY: …The greenhouse effect … is about slowing the progress of LWIR to the top of the atmosphere. Without CO2 or other GHG’s the LWIR would proceed quickly to space, and the Earth would cool faster, and have a lower average temperature at night. The “backradiation” has a limited scope of effect, but it is there as part of the trasnfer process. … GHG’s act as a LWIR transfer regulator from the surface to the top of the atmosphere where it is radiated into space. They slow the transfer. Without GHG’s nightitme temperature would drop quickly as LWIR is lost directly to space. […] – Anthony
===============================================================
OK, Anthony, as I said, this is not exactly the IPCC concept, but I would like to address it anyway. For the sake of clarity, when I say “Earth” I mean the solid Earth. So, there is the Earth and the atmosphere around it.
Now, the first problem in your approach is the idea that the radiative cooling of the Earth surface (let us say at night to simplify it) is somehow connected to what happens to the radiation after it leaves the Earth surface, like meeting some “greenhouse gases”. Well, if we ignore conduction/convection, then it is simple: what is gone is gone. The radiation is gone and the surface cools accordingly. If the radiation meets something else and warms it, fine, but this fact alone does not affect the temperature of the surface. To affect the temperature of the surface there must be additional energy coming to it.
Now, at the first glance, the “greenhouse gases” are doing exactly that: providing additional energy by back radiating towards the Earth. The problem is, however, that back radiation can not warm the source, this is physically impossible. So we have this situation: back radiation is there, but no effect on the temperature of the source is possible.
That’s it, actually. How long and hard the way of the radiation from the surface to space is, with all this bouncing or delays or whatever, does not matter. All this happens outside the surface and gives the surface no additional energy.
Sometimes at this point people start asking “but where does the back radiation go?” or “how it is possible that it does not warm?”. My answer usually is like “why should I care and explain everything?” I only care about effect on the temperature of the source, and this is exactly something that leads to an absurd, as I demonstrated earlier, so this is sufficient. We can, of course, start speculating about it and make up theories, but why should we? Anyway, it is clear, that the well known IPCC radiation arithmetic has nothing to do with reality. The radiation simply does not obey the laws of arithmetic. It is a different, special thing. All this talk about photons “not knowing” anything is misleading. If you like, the fact is that something we call “photons”, despite being a countable noun, does not obey the laws of arithmetic and can not affect the temperature of the source. This is the reality. No “greenhouse effect”.

LdB
April 24, 2013 9:20 pm

@max
The explain laser cooling which is the essence of your stupid argument that hot can’t cool by definition laser cooling does exactly that.
And if you understand that you can have temperatures colder than absolute zero then your realize that classic physics can be wrong because it is a bad simplification and therefore you should not be surprised that classic physics can’t be used to understand the greenhouse effect because its a quantum problem … that is if you really aren’t ignorant.

April 24, 2013 9:23 pm

* The object is “at absolute temperature T” and is held at that temperature throughout the problem.
Uh? Problem 1026 says no such thing.
Verbatim:
A spherical black body of radius r at absolute temperature T is surrounded by a thin spherical and concentric shell of radius R, black on both sides. Show that the factor by which this radiation shield reduces the rate of cooling of the body (consider space between spheres evacuated, with no thermal conduction losses) is given by the following expression: aR^2/(R^2 + br^2), and find the numerical coefficients a and b.Problem 1026

April 24, 2013 9:28 pm

LdB, I’m not engaging in childishness with you, sorry bucko.

April 24, 2013 9:29 pm

“So how is it that mirroring is an effective means of insulation?”
So what you are saying is that if you took a whole bunch of mirrors and had them focus on a point you could cook something right?
At night too?

TomR,Worc,MA,USA
April 24, 2013 9:32 pm

Noelene says:
April 24, 2013 at 3:08 pm
So there is a consensus on the greenhouse effect?
===========================================
I see what you did there.
Not that I agree with you, but well done.
TR

LdB
April 24, 2013 9:42 pm

@Max
It isn’t a childish argument its the fact the dragonslayers make at the centre of the argument that heat can’t cool.
Problem is laser cooling does exactly that it is hot put your hand infront of the beam if you want to see how hot and it cools to almost absolute zero.
In there world it isn’t possible but it happens a fact they like you have to ignore because they can’t explain it and there whole argument goes up in smoke.
The hard sciences are that explicit about this topic on in AGW does this sort of garbage get air time.

gymnosperm
April 24, 2013 9:56 pm

Roy Spencer says:
April 24, 2013 at 12:12 pm re: carbon dioxide and insulation
===================================================
The effect is the same but the mechanism is completely different. Insulation does not absorb and radiate photons. The IR photons inside a house don’t even make it past the paint on the sheetrock, much less reach the insulation. Funny how the ocean comes to mind.
House insulation is just a down sleeping bag. A tiny bit of thermal mass and a huge barrier to convection.

Gary Hladik
April 24, 2013 10:21 pm

KevinK says (April 24, 2013 at 7:58 pm): “But if it is radioactively heated (an effective “unlimited” heat source, anybody else remember back in the 50s and 60s when electricity was going to be “too cheap to meter” ?) I would expect that the “cycle time” concerns would go away. And then plate 1 and plate 2 (and all the unicorns in the neighborhood) would all be warmer and cozier.”
Thanks, Kevin, for that, er, picturesque response.
Back to the battery-operated plate 1: Why do you say the battery output would increase when plate 2 is introduced?

Konrad
April 24, 2013 10:26 pm

Dr Spencer and other lukewarmers are wrong. Radiative gases cool our atmosphere at all concentrations above 0.0ppm. The problem is not with physics of radiative gases, but with fluid dynamics and gas conduction
For the record I am not a “Slayer”, my following comments are based on my own empirical experiments. These experiments disprove the radiative green house effect. I have described simple versions that other readers can build and run for themselves here –
http://wattsupwiththat.com/2013/04/05/a-comparison-of-the-earths-climate-sensitivity-to-changes-in-the-nature-of-the-initial-forcing/#comment-1267231
I would urge other readers to try these experiments for themselves. There is far to much cut&paste and linking on climate blogs and not enough empirical work. These experiments relate to the “Do nots” of climate modelling –
A. Do not model the “earth” as a combined land/ocean/gas “thingy”
B. Do not model the atmosphere as a single body or layer
C. Do not model the sun as a ¼ power constant source without diurnal cycle
D. Do not model conductive flux to and from the surface and atmosphere based on surface Tav
E. Do not model a static atmosphere without moving gases
F. Do not model a moving atmosphere without Gravity
G. Do not model the surface as a combined land/ocean “thingy”
H. Do not apply SB equations to a moving gaseous atmosphere (the davidmhoffer rule 😉
Does the surface emit IR? Yes.
Does the atmosphere absorb IR and heat? Yes.
Does the atmosphere radiate some IR back to the surface slowing it’s cooling rate? Yes.
Is there a radiative GHE? Yes.
Is the NET effect of radiative gases in our atmosphere warming? no.
To understand the role of radiative gases in the atmosphere, the conditions in a non radiative atmosphere should first be considered. Radiative gases are critical to strong vertical convective circulation in the troposphere. It should be noted that above this level there are few radiative gases and little vertical convective circulation. The role of energy loss at altitude in convective circulation is demonstrated in Experiment 3. It should also be noted that pneumatic heating and cooling of rising and descending air masses does not represent energy gain or loss. IR emission to space is the primary way gases at altitude lose energy.
If there were no radiative gases in our atmosphere, gases would still be heated by surface conduction. These gases would convect to altitude, but they could not lose energy and buoyancy and descend. Full convective circulation would stall. Experiment 4 demonstrates that a gas column heated at the base and cooled at the top has a lower average temperature than a gas column heated and cooled at separate locations at the base. In Experiment 4 box 1 has a higher “surface” temperature and a lower gas temperature. In box 2 the average “surface” temperature is lower yet the average gas temperature is higher. Box 2 models a non radiative atmosphere. Experiment 4 also demonstrates why an atmosphere heated by surface conduction will have its temperature set by surface Tmax not surface Tav.
Experiment 5 demonstrates that for a moving gaseous atmosphere in a gravity field, the surface is better at conductively heating the atmosphere than it is at conductively cooling it.
Dr Spencer has not correctly modelled the critical role radiative gases play in convective circulation or the importance of this to atmospheric temperatures. The land surface under a non radiative atmosphere may have a lower Tav, however the moving gases in the atmosphere above will have a higher average temperature than present. Build experiment 4 big enough (limiting equalisation through gas conduction) and you can find out just how dramatic the difference would be. If atmospheric temperatures are higher for a non radiative atmosphere, then radiative gases must act to cool the atmosphere at all concentrations above 0.0ppm.

Gary Hladik
April 24, 2013 10:34 pm

Reed Coray says (April 24, 2013 at 8:02 pm): “Although you didn’t explicitly say so, if what you say is true, shouldn’t increasing the amount of GHG placed in this space result in improved thermos bottle efficiency?”
No. By introducing a gas to the space between layers, you’ve added conductive cooling to radiative cooling of the inner shell, which would overwhelm any reduction in radiative cooling by such a thin layer of gas.
Think of the top of our atmosphere as the inner layer of the vacuum thermos bottle. With nothing but a vacuum “outside”, the only significant cooling mechanism is radiation. Convection and conduction are major energy transport mechanisms within the atmosphere, but don’t function in a vacuum.

peterg
April 24, 2013 10:52 pm

I do find all this denial of basic radiative physics fairly boring. The water vapour cloud feedback loop cannot be discussed without presuming the basics of radiation physics, and when assertions based on these laws get challenged, the discussion becomes a bit pointless.

Gary Hladik
April 24, 2013 10:56 pm

Max™ says (April 24, 2013 at 9:23 pm): “Uh? Problem 1026 says no such thing.”
Out of curiosity, if the problem says there is a constant non-zero power input to the inner sphere, and the shell reduces the rate of cooling by (for example) roughly 20%, what do you think would happen to the sphere’s equilibrium temperature after the shell is added? Would it be higher, lower, or the same?

April 24, 2013 11:05 pm

I would be interested in anyone’s comment on this:
I think the classical 33° analysis falls apart because it assumes albedo is the input and solves for the greenhouse effect. This is not how heat is managed on this planet. Albedo is the output. Try it this way instead:
Start with the albedo of areas including everything but clouds first. Include areas with mist / haze, but not clouds (no small task). This will be considerably darker and the SB temperature much higher. Solve for the SB equilibrium temperature. What did you get? A:__________
Note: This is what the situation would be if the earth were at a radius from the sun farther than it is, not including the effect of increasing icecaps. We are too close to the sun to not employ evaporative cooling which is the main mechanism of heat regulation. This really needs to be done as an integral over one rotation.
The clouds are naturally produced to balance the rest of the energy input such that the surface temperature lands where it is. The first answer (A) is the natural temperature of the earth without the main cloud effects.
Now take the measured surface temperature (B). A-B is the cooling effect of water vapor + the warming effect of all other GHG’s. We don’t know the exact magnitude of either, just the net from this exercise.
But we can measure the effect of water vapor’s GHG effect in many areas (like buoys) where we measure LW, T and humidity at the same location all the time. So we know what it is with varying temperatures and humidity which we can uses to estimate humidity’s role on the received LW radiation from the atmosphere, and deduce what “all other” effects total to. I don’t think anyone has so far integrated the moisture data well enough to describe what local effects (storms) are occurring based on the shapes of the curves over time.
Once people give up on the absurd idea that albedo is an input to the equation, the rest of the results should be fairly easy to estimate. Water is going to be the major net driver in the tropics (due to reflection and carrying water droplets aloft where they can radiate out with far less optical thickness), but water vapor will also be a warmer in other conditions. CO2 will be shown to be almost meaningless in the face of H20′s effect.
Albedo is the earth’s output, it is a reaction to the input (solar energy), and internal variation.
So let’s try it. Using the 20% that NASA estimates is reflected from clouds, the results should look like this:comment image
As you can see, treating albedo as an output (since clouds are generated to reflect and transport heat to space), now you only need to account for 16.6°K after the effect of cloud reflection alone is accounted for. It’s a start and I suspect this is one of the major links that supports why sensitivity data is without question coming in at half or less of the flawed classical analysis. Note also, this is not a feedback effect. It is entirely the result of the initial albedo assumption being wrong.comment image

Alec M
April 24, 2013 11:11 pm

I wrote a comment on the original Spencer article describing how most people misinterpret the S-B equation in Climate Alchemy. It’s because the atmosphere is semi-transparent so cannot be considered a grey body. This is one of the three mistakes made by Houghton. The others are to fail to understand you can’t apply the two-stream approximation at an optical discontinuity and the Earth does not emit IR as if it were an isolated black body in a vacuum.
Also, there is no net surface IR from the Earth’s surface in the main GHG IR bands. This means there can be no CO2-AGW from this cause. There is no ‘back radiation’ [because for a normal temperature gradient it is annihilated at the surface]. Pierrehumbert does a good job of saving the blushes of Climate Alchemy by his argument about the ‘CO2 bite’ in OLR. However, this heating, ~3 W for doubled CO2 is offset by a process involving clouds. I’m writing a paper on this.
The Earth’s climate is controlled by a negative feedback control system involving the optical properties of clouds so temperature is independent of CO2 concentration. Forget surface IR, that is mostly irrelevant for CO2.

Konrad
April 24, 2013 11:25 pm

peterg says:
April 24, 2013 at 10:52 pm
“I do find all this denial of basic radiative physics fairly boring.”
———————————————————————————–
The reason the AGW hypothesis fails is not because of any problem with radiative physics. The radiative physics is fine* The problem is that the fluid dynamics of the atmosphere and the surface to atmosphere conductive flux were modelled incorrectly.
Radiative gases are critical to tropospheric convective circulation, without which our atmosphere would heat dramatically due to surface conduction.
The reality of an atmosphere in a gravity field in which the gases are free to move is that the surface is far better at conductively heating the atmosphere than it is at cooling it.
*Except for that little mistake of assuming that the oceans could have their cooling rate slowed by incident LWIR. Not a huge mistake, only 71% of the earth’s surface, probably nothing to worry about 😉

Gary Hladik
April 24, 2013 11:26 pm

Rosco says (April 24, 2013 at 9:13 pm): “At least this “slays” Willis’ steel greenhouse completely !”
Eh? The problems described are in complete agreement with the “Steel Greenhouse” article. Far from “slaying” the steel greenhouse, this reference work vindicates Willis completely.
Check out Figure 1 here:
http://wattsupwiththat.com/2009/11/17/the-steel-greenhouse/
The shell (approximately the same size as the planet) radiates–gasp!–exactly half as much as the planet, just as problem 1023 says it should! Coincidence? 🙂

Gary Hladik
April 24, 2013 11:39 pm

Konrad says (April 24, 2013 at 10:26 pm): “Dr Spencer and other lukewarmers are wrong. Radiative gases cool our atmosphere at all concentrations above 0.0ppm.”
So that would mean that, just as absence of positive correlation between temperature and CO2 argues against a significant warming effect, the absence of a negative correlation between temperature and CO2 argues against a significant cooling effect?

davidmhoffer
April 24, 2013 11:39 pm

Konrad;
The problem is that the fluid dynamics of the atmosphere and the surface to atmosphere conductive flux were modelled incorrectly.
>>>>>>>>>>>>>>>>
So you’re saying that the satellite temperature data is wrong?

Pierre-Normand
April 24, 2013 11:54 pm

Greg House said: “Again, this is not what the IPCC supports. You can invent whatever effect you want, but only the IPCC’s one is politically relevant, because governments and agencies refer to the IPCC reports and recommendations, not to davidmhoffer or Steven Mosher.”
The IPCC AR4 WG1 glossary defines the Greenhouse Effect thus: “Greenhouse gases effectively absorb thermal infrared radiation, emitted by the Earth’s surface, by the atmosphere itself due to the same gases, and by clouds. Atmospheric radiation is emitted to all sides, including downward to the Earth’s surface. Thus greenhouse gases trap heat within the surface-troposphere system. This is called the greenhouse effect. Thermal infrared radiation in the troposphere is strongly coupled to the temperature of the atmosphere at the altitude at which it is emitted. In the troposphere, the temperature generally decreases with height. Effectively, infrared radiation emitted to space originates from an altitude with a temperature of, on average, –19°C, in balance with the net incoming solar radiation, whereas the Earth’s surface is kept at a much higher temperature of, on average, +14°C. An increase in the concentration of greenhouse gases leads to an increased infrared opacity of the atmosphere, and therefore to an effective radiation into space from a higher altitude at a lower temperature. This causes a radiative forcing that leads to an enhancement of the greenhouse effect, the so-called enhanced greenhouse effect.”
This seems to be exactly what Mosher and Hoffer have been saying. The backradiation and the ERL arguments are equivalent and both are explicitly endorsed by the IPCC.

Kristian
April 24, 2013 11:59 pm

tjfolkerts says, April 24, 2013 at 9:07 pm:
“Yes, I’m talking about the same examples. Specifically, 1026 says …”
http://books.google.no/books?id=dQGC0ifkE34C&pg=PA24&lpg=PA24&dq=concentric+sphere+black+body&source=bl&ots=Zh6N1e35jc&sig=m-7nVWch4_zv-l3ISR5k7bluSUQ&hl=en&sa=X&ei=_ldYUd7EFZLU9ATVzYHoDw&redir_esc=y#v=onepage&q=concentric%20sphere%20black%20body&f=false
What Problem 1026 is telling us is specifically that if R ~ r, the Q = Q’ + Q”. This is the exact equivalent to the Carnot cycle: W = Qh – Qc, where Q=Qh, Q”=Qc and Q’=W:
http://i1172.photobucket.com/albums/r565/Keyell/Carnot_zps4049e783.jpg
At any point in time, the heat flux leaving the surface of the sphere (Q) corresponds directly to (depends on) its specific emission temperature, which in turn depends fully and solely on (is set by) the heat flux provided to it by its heat source (so Q equals the heat input).
Q then goes half into warming the shell and keeping up its equilibrated temperature (Q’), half from the new layer to space (Q”): Q = Q’ + Q”. The entire flux comes from the sphere. It is simply split up upon absorption by the shell.
No need to invent secondary heating of the sphere to achieve radiative balance, beyond what the input from its heat source can manage on its own. Before the shell was emplaced, the power from the sphere’s heat source went into warming the sphere only (one object). After, the same amount of power goes into warming the sphere AND the shell (two separate objects). Hence the smaller flux to space. It’s that simple … Q is not reduced. The radiative flux to the cold reservoir (the surroundings) (before, Q, after, Q”) is reduced. This does not increase the surface temperature of the sphere. It maintains the temperature of the shell (Q’).
If the sphere were held at a constant temperature it would be because it received a constant energy supply from its heat source. With a variable heat source, the temperature would not be constant.
Let’s compare Problem 1026 with Problem 1023:
http://i341.photobucket.com/albums/o396/maxarutaru/bunnytaru/lol%20censorship/heatsourceinshellatequilibrium_zps1c9d662a.png
1023 specifically asks: “(…) what is the effect of the shell on the total power radiated to the surroundings?
As you can see, J (Q) doesn’t change. J is simply replaced by J1 as ‘the total power radiated to the surroundings’ as the shell is put around the sphere. The only thing that changes is that ‘the total power radiated to the surroundings’ is halved. The rest goes into heating and maintaining the temperature of the shell, something that J didn’t have to do before the shell was emplaced. Perfect radiative balance: Q = Q’ + Q” — J = (J – J1) + J1.
The sphere/shell is also specifically likened to a star and a surrounding dust cloud (b). A star is pretty constantly heated, isn’t it?
There are more examples of course showing the exact same thing, that the source of the shell’s or the insulating layer’s incoming (warming) radiative heat flux (across a vacuum) does not heat up some more when surrounded or covered by the shell/insulating layer. The ‘insulating effect’ simply amounts to reducing ‘the total power radiated to the surroundings’ (Q –> Q”). There is no energy ‘piling up’ at the surface of the sphere/inner layer. Q (J) remains unchanged.
But we can get back to those later …

LdB
April 25, 2013 12:29 am

@Konrad
The greenhouse effect can heat or cool depends on the energy input and outputs and I am not sure any scientist ever said anything different.
I can also make a laser beam which has an extremely high temperature heat or cool things depending on how we up the relationship and frequency of the beam..
The problem with classic physics it is familiar only with heat only heating and so you struggle with the problem because your understanding of temperature is sort of simplistic and wrong. Roy is trying to bring a QM effect back into sort of classic physics so you can understand it but at some point you can’t.
Try understanding laser cooling and reading about it because sort of shows your problem and understanding of heat and temperature.

Konrad
April 25, 2013 12:35 am

@Gary
The cooling effects of changing CO2 levels from 0.03 to 0.04 percent would be unmeasurable with current technology.

Gary Hladik
April 25, 2013 12:54 am

Konrad says (April 25, 2013 at 12:35 am): “The cooling effects of changing CO2 levels from 0.03 to 0.04 percent would be unmeasurable with current technology.”
Heh. I suppose the same could be said of the alleged warming effect. 🙂

April 25, 2013 1:15 am

LrB, the childishness was you slinging terms like ignorant around when you don’t seem to recognize the problem with your own argument.
A laser is not a spontaneous transfer from hot to cold, thermal radiation is a spontaneous transfer from hot to cold.
A laser involves doing work, and the production of waste heat.
Look into the Clausius statements of the laws of thermodynamics, as well as Carnot efficiency before calling others ignorant, perhaps?
___________
The shell (approximately the same size as the planet) radiates–gasp!–exactly half as much as the planet, just as problem 1023 says it should! Coincidence? 🙂” ~Gary
The shell in 1023 does this without the implication that the inner body would increase in temperature.
Essentially it demonstrates that the steel greenhouse involves these steps:
1. Get an internally heated body
2. Add a shell around it which radiates half as much inward and outward as the inner body
3. A wizard does something
4. The temperature of the internally heated body increases

Rosco
April 25, 2013 1:36 am

I cannot agree that radiation would quickly escape to space if GHGs did not slow the process.
It is inconceivable that greenhouse gases at minus 18 degrees C and trace concentrations could radiate as powerfully as the solid hot Moon surfaces.
NASA have clearly demonstrated that the heated Moon’s surface radiating effectively into a vacuum loses energy at a slow rate. From 390 K at lunar noon to 220 K at lunar sunset is a 170 K change in temperature in approximately 177 Earth hours – a rate of 1K per hour.
The Moon cools far more slowly once the Sun sets – from 220 K to about 100 K in 354 Earth hours.
Although there is less data for Mercury, the night on Mercury is some 2111 Earth hours long so cooling from ~740 K to ~90 K is easily understood.
There is no quick reduction in temperature of heated surfaces radiating into a vacuum.
Surely we observe a similar phenomenon here on Earth ?
The poles receive the least direct Solar radiation of anywhere on Earth and it is the fact that zero direct solar radiation is received for some months that is surely responsible for the extreme cold.
These are observable facts – there is no need of convoluted arguments to explain simple truths.
Anywhere in the Universe that is cold is so because of low levels of radiation. There is evidence that cooling from a warm state to a cold state by radiating to space takes significant time – NASA says so.
I believe them.
Besides the vacuum space at the Earth’s orbit cannot by any stretch of the imagination be thought of as cold with ~1370 W/sq m of solar radiation continually emitted by the Sun.
NASA prove the cold Earth orbit space myth in their “When we left the Earth” series when the solar panels and “parasol” – NASAs words – fitted to Skylab failed to deploy.
The astronauts that arrived to fix the problems found the temperature inside Skylab exceeded 70 degrees C.

Selgovae
April 25, 2013 1:49 am

Joseph E Postma wrote, “In any case, you can shine a flashlight at the Sun and the the beam will be there. But the flashlight doesn’t heat the Sun.”
Joseph, I have trouble with the idea that “the beam will still be there”. Does that imply that the source will lose energy (now in the beam?) and thus cool. In which case, where does the energy go? I find it easier to envisage that the beam will not be there and thus the source gets warmer as it sheds energy at a lower rate.
BTW, I’m not sure that shining a flashlight at the sun is a good thought experiment. I’d assume that some energy would probably be absorbed by the sun, just based on the color of the light from the flashlight. Pointing a finger at the sun might be a better example.

Konrad
April 25, 2013 1:53 am

@David
There are few problems with Dr Spencers method of measuring the microwave emissions from O2 molecules. It is far more accurate than surface station data.
This system may however start to over read as global cooling occurs over solar cycle 25. Reflection from increasing snow and ice cover may be an issue. However correlation with radiosonde balloons during overpass is an eazy fix.

Rosco
April 25, 2013 1:56 am

The other thing I want to comment on is the explanation given by people about what happens when you place some sort of insulation over a heat source – commonly putting a lid on a pot of boiling water.
They claim the back-radiation increases the heat.
Well I say bunkum – you can agree or disagree but at least consider the following :-
Exposed to the conduction/convection of the atmosphere no heat source will ever achieve its heating potential. Reduce that energy loss by effective insulation and the heat source will achieve near its potential. Radiation loss reduction may or may not contribute to this effect.
We know, or should know, that “trapping” warmed air close to our bodies, by reducing convective loss of that warmed air, is the principal manner in which we stay warm.
I see people saying stupid things like the back-radiation from the ice of igloos keeps the occupants warm.
I say rubbish.
The igloo traps the warmer air inside, maybe even warming it enough to melt the ice a little. The occupants do not shed all their clothing and romp inside naked.
There is absolute proof about the fact that reducing convective heat loss is the principal means by which we stay warm – radiation trapping has little to nothing to contribute.
Animals have only one mechanism to keep warm – their fur or feathers. There is no radiation barrier to prevent radiative losses – none.
Animal fur reduces the loss of the warmed air by reducing convection – there is no other mechanism possible.
We know this is true as we understand the phenomenon of wind chill which acts to destroy the trapping of the warmed air.
As a qualified Environmental Health Officer I know I am on firm ground in discussing simple biology.

LdB
April 25, 2013 2:23 am

@Max
And now laser cooling is not spontaneous .. please describe how you arrived at this great piece of physics another of your great theories. You and Konrad look more and more ignorant the more you comment.
If you want to talk about work so lets put the two on even footing you have this big massive energy source in the sky we call it a sun it produces energy. I have a laser unit plug into a power socket so we both have a power source to do work they are equivalent get it functionally identical.
Now going down to the exchange level the transfer in laser cooling has to be instantaneous or you couldn’t get to zero degree kelvin that is in your classic garbage world how you define ABSOLUTE ZERO. Any non instant transfer would mean you would have motion try making an exchange of energies with a collision and have absolutely no movement.
The problem I have with you whackjobs is the same as Roy you are running up against some impossible physics for anything in classic physics to challenge because QM showed that classic physics is wrong 100 years ago. Instead of listening like all pseudoscience junkies instead of listening that you keep digging looking more and more stupid as you go.

April 25, 2013 2:34 am

If animals were relying on reducing radiative losses, to further support Rosco’s point, then evolution would probably begin producing nearly perfect mirrors, as that is by far the best radiative insulation.

April 25, 2013 2:40 am

Again, LrB, I live in a universe governed by relativistic and quantum physics, I know they are experimentally confirmed, and I know I do not live in a classical universe, learn how to read, kiddo.
Let me bold it for you:
I am not operating within a classical paradigm, the universe only resembles a classical one in the weak field regime, at small scales it is quantum, at large and fast scales it is relativistic, learn how to read before you go off on a rant next time.
A laser is not a spontaneous process, I know masers form naturally, but I’ve never seen a CO2 laser randomly appear and function.

April 25, 2013 2:54 am

TomR,Worc,MA,USAs
At April 24, 2013 at 9:32 pm you say

Noelene says:
April 24, 2013 at 3:08 pm

So there is a consensus on the greenhouse effect?

===========================================
I see what you did there.
Not that I agree with you, but well done.
TR

I answered Noelene and you may be interested in that answer. However, you may not have seen it because it was held in moderation for an hour and the thread had moved way past it when it appeared out of moderation.
However, P*stma saw my answer to Noelene because he quoted from it at April 24, 2013 at 4:53 pm and was unjustifiably offended by it.
In case you are interested, my answer to Noelene was at April 24, 2013 at 3:42 pm and this link jumps to it
http://wattsupwiththat.com/2013/04/24/spencer-slays-with-sarcasm/#comment-1286437
Richard

Gary Hladik
April 25, 2013 3:03 am

Max™ says (April 25, 2013 at 1:15 am): “The shell in 1023 does this without the implication that the inner body would increase in temperature.”
Really? OK, let’s assume the inner sphere has a constant heat source, as in the Steel Greenhouse, and is in thermal equilibrium with the shell as in problem 1023. Sphere temp is at T1. We agree that the power radiated by the system to its surroundings is half the power radiated by the inner sphere, as in both problem 1023 and Willis’s Figure 1, yes? (If not, why not?)
Now snatch away the “insulating” shell, real sudden-like. The system is suddenly cooling twice as fast, as per problem 1023, but its power input is unchanged! Oh no! What happens to the temperature of the inner (now alone) sphere? Increase? Decrease? Remain the same?
Let the system come to equilibrium, sphere at temp T2. Now add the shell back. Let the system come to equilibrium. Inner sphere is now at temp T3. What’s the relationship of T1, T2, and T3?
a) T1 = T2 = T3
b) T1 = T3 > T2
c) something else (specify)
and why?

Sergey
April 25, 2013 3:17 am

There are two different mechanisms of radiative cooling of gases: by emitting quants of wavelength corresponding to quantum resonance of moleculae and by emitting at all wavelengthes (so called blackbody radiation). The first mechanism prevails for greenhouse gases, but the second is fairly universal and applies also to nitrogen and oxigen. And even greenhouse gases such as water vapour and carbon dioxide do not necessary emit at their quantum resonance wavelength when excitated by short-range radiation but can transfer excessive energy by collisiom with moleculae of nitrogen and oxigen without emitting anything. How much energy is re-emitted by carbon dioxide and how much simply heat atmosphere without emitting anything is everybody’s guess, because quantum mechanical calculations are too cumbersome to perform even using supercomputers. And heating of atmosphere enhances convection and make convective layer depth higher, which can compensate or even overcompensate backscattering by CO2. So the real contribution of greenhouse effect to surface warming in real atmosphere with convection, evaporation, condensation and radiation cooling can not be calculated, measured or estimated by any other means. It can be arbitrary small or even negative.

steveta_uk
April 25, 2013 3:19 am

Greg House says: April 24, 2013 at 9:13 pm

Now, at the first glance, the “greenhouse gases” are doing exactly that: providing additional energy by back radiating towards the Earth. The problem is, however, that back radiation can not warm the source, this is physically impossible. So we have this situation: back radiation is there, but no effect on the temperature of the source is possible.

(my bold).
Here you are just making an assertion without any backing.
For a simple example of how you are wrong, why are you assuming that the back radiation is from a colder source? Ground frosts at night are very common – they are caused by the surface losing energy by radiation, mostly directly to space, in which case the air close to the surface can clearly be warmer than the surface, esp. when the surface is a good radiator (e.g. a black car).
So even if you don’t believe in the “cooler can warm a warmer thing” idea, you don’t need to. Clearly GHG is providing energy to the surface.

A C Osborn
April 25, 2013 3:28 am

LdB says:
April 25, 2013 at 2:23 am
You are abusing Konrad who has carried out actual experiments which he has published on various blogs, so he does know what he is talking about.
Empirical results which show that theories are wrong are the way that science is supposed to be done.
I suggest that you also take a look at TheFordPrefect’s experiments which show that if Back Radiation exists at all it is a minute amount of energy, nothing like the 50% bandied about by the climatologists.

richard verney
April 25, 2013 3:34 am

Rosco says:
April 25, 2013 at 1:56 am
“…I see people saying stupid things like the back-radiation from the ice of igloos keeps the occupants warm.
I say rubbish.
The igloo traps the warmer air inside, maybe even warming it enough to melt the ice a little. The occupants do not shed all their clothing and romp inside naked.
There is absolute proof about the fact that reducing convective heat loss is the principal means by which we stay warm – radiation trapping has little to nothing to contribute….”
//////////////////////////////////////////////////////
I have lost count the number of times that I have pointed out that Igloos work by reducing convection and thereby by trapping heat.
The same is so of space blankets/insulatuion blankets. To have any substantial effect these need to be wrapped around the user thereby reducing convection and trapping heat. If they worked primarily by re-radiating heat, one could construct a space blanket much like a toilet roll. it could be say 6ft high and 2ft in diameter. The user could stand inside the mirrored roll and enjoy all the back radiated heat. But it would not be effective because of convection; significant heat loss would occur vertically. In fact from the radiative point of view, the diameter could be 3ft or 10ft, and the user would theoretically enjoy the same levels of re-radiated body heat, yet as the diameter size of the roll increases, it becomes less and less effective since convection plays an increasingly important roll.
Where heat can be lost through conduction or through convection or through radiation, then radiation is weak, and is overwhelmed by the other processes.
Today in sunny Spain it is cold (like most of Europe the last 3 months have been cold). I have no heating on in the house. My granite work top (in the kitchen) is cold to the touch. If I place a plastic/poluthene bag on top of the granite and place my hand on top of that bag, it does not feel cold. If I place a sheet of aluminium foil on the granite work top and place my hand on top of the foil, it feels cold.
In short, whatever miniscule radiative effect the aluminium foil has, it does not overcome conduction. The cold surface of the granite is conducted. Whereas the plastic bag is a good insulator and keeps my hand warm (even though plastic is a poor radiator)
If I suspend the aluminium foil horizontally in the air, and place my hand above it, possibly I can feel some reradiated heat when my hand is about 2mm above the foil, but the effect is so slight that I am unsure whether it is real or placebo.
But the point is that in the real world, redatiated heat (energy) is overcome by conduction and convection and this is so in the workings of the Earth’s atmosphere (not also forgetting the phase changes of water and associated latent heat).

April 25, 2013 3:35 am

Joseph E Postma says:
April 24, 2013 at 3:29 pm
“It’s called an incandescent flashlight.”
It is about the resistor (lets use the filament) becoming even hotter still if you shine the flashlight at a mirror. This does not increase the brightness (temperature) of the filament. Trapping the radiation does not increase the temperature of the filament.
Not my prefered item of interest (radiation… a very long time ago) but as far as I remember, if you can concentrate some of the outgoing light (including IR) via a concave mirror back onto the filament, the latter wouldn’t survive for long.
Or with an experiment: make that the outher wall of the bulb is coated with silver and test how long it takes before the filament is burned up, compared to a non-coated bulb… In the case of non-halogen bulbs, only radiation is at work, no conduction or convection.

gbaikie
April 25, 2013 3:37 am

There is no evidence that CO2 in the atmosphere acts like insulation
in a house.
Fiberglass insulation in a house reduces the heat lost from convection.
What fiberglass does prevent the movement of gas molecules- which
inhibits convection of heat. If you inhibit to loss of heat from convection all
gases [including CO2] make good insulator of heat. Fiberglass
also inhibts warmth from entering the house. So insulated house
is easier to be cooler during warmer weather.
Insulation doesn’t make something warmer, it prevent the loss
or transference of heat. A thermos does not make coffee warmer than
the temperature of the coffee which was poured into the thermos.
So the thermos preventing a transfer of heat, so coffee stays warm
for a longer period of time.
To make an analogy of CO2 to insulation is correct in sense
that there is also no way for CO2 to cause the surface or air
at the surface to become hotter. Or an asphalt road in sunlight will
not become hotter because of CO2, What affect the temperature
of this road is the intensity of the sunlight and the loss of heat
from convection of air. If the air is already warm, there will less
heat loss from convection of air. Warmed air and intense sunlight
[sun near zenith and clear skies] will cause it reach it’s max
temperature.
And we have not seen any increases in max surface temperature
from an increase in global CO2 content, nor should this be
expected.
What is expected by some is that CO2 will cause there to less
loss of heat. So generally since it doesn’t cause the day to be
warmer, it could cause the nights to become less cold.
But there is no evidence which clearly indicate that CO2 does
cause less cooling- e.g. there no clear evidence that nights are warmer
due to CO2.
Whether CO2 cause such warming and thereby causes an increase
in average global temperature has not be proven.
But what should rather obvious is that increasing levels of CO2 or
other greenhouse gas are not going make earth unbearable hot
[unless a person presently already thinks it’s unbearable hot].
Or there no reason to assume that greenhouse gases could cause
CAGW like conditions as is commonly described by it’s faithful- or no
reason to assume that Earth could become in anyway similar to
Venus due to greenhouse gases.
No reason to believe greenhouse gases will cause massive widespread
droughts, or numerous other “predicted” drastic changes in climate.
To say the danger from CO2 has been over hyped is an understatement.
Instead it has been series of fabricated stories delivered by mass media
which as had the intention to scare the public. Which has caused many people
a lot of unnecessary stress and concern.
The public has already been conned out hundreds of billions of tax dollars
and has already had numerous coercive and oppression laws designed to
solve a scary problem which is totally fictional.

LdB
April 25, 2013 3:50 am

@max
Then you haven’t looked very hard because the atmosphere and sun form a natural form of a optical pump which is the key component of a laser.
You only need 3 requirements an energy source we call it the sun, for a man made laser we use RF, microwaves, electric discharge anything will generally do its an efficiency issue.
You need an quantum active medium something like CO2, N20 or water vapor will generally will generally do.
Then all you need is a partial mirror which means a different input and output characteristics something like what the atmosphere does and we measure.
All pretty basic and not hard to understand .. see your problem it is that basic.

Sergey
April 25, 2013 3:51 am

A question to Roy: Why we can see the Sun? Why it radiates? Its atmosphere certainly does not contain any greenhouse gases, being almost pure hydrogen with a small traces of helium. What we see is blackbody radiation at 6000 C, and every gas emits according Plank formulae even when its temperature is near 0 K. So it sounds weird that without greenhouse gases Earth atmosphere would not emit anything into space.

April 25, 2013 3:54 am

Now snatch away the “insulating” shell, real sudden-like. The system is suddenly cooling twice as fast, as per problem 1023, but its power input is unchanged! Oh no! What happens to the temperature of the inner (now alone) sphere? Increase? Decrease? Remain the same?” ~Gary
FIrst off, 1023 isn’t about the cooling rate.
To see what would happen though, consider the case where the shell is in contact with the sphere. If the gap was filled with the same material as sphere and shell are made of, then the volume of material goes up while the energy supply remains the same as it was originally, right?
Two spheres with the same power input/composition which only differ in size won’t reach the same temperature, will they?
So now you remove a section from the larger sphere, such that it is the same size as the other sphere, what happens?
Now remove the same amount of material from the larger sphere, except for a thin shell at the original surface, such that the inner sphere is the same size as the smaller sphere, what happens?
Let the system come to equilibrium, sphere at temp T2. Now add the shell back. Let the system come to equilibrium. Inner sphere is now at temp T3. What’s the relationship of T1, T2, and T3?
a) T1 = T2 = T3
b) T1 = T3 > T2
c) something else (specify)
” ~Gary
Interestingly, that is almost the set up for problem 1024, fancy that.
If T_sphere = T_background =T_shell at some point, shouldn’t it remain that way?
If T_sphere > T_background, adding the shell would give:
T_sphere > T_shell > T_background

Myrrh
April 25, 2013 4:05 am

Joseph E Postma says:
April 24, 2013 at 8:07 pm
I did not ever say that the Earth was at +49C. I said this is the equivalent temperature value of the average input. This too is reality. This is very different from the usual assumption of a -18C or 240 W/m2 input.
AGWScienceFiction’s Greenhouse Effect attributes this -18°C to the temperature of the Earth without its variation of “greenhouse gases”, that is, without “longwave infrared imbibing gases”, i.e. “radiated heat imbibing gases mainly water vapour and carbon dioxide”, but, it is crucial to understand this point in order to see the AGW sleight of hand in play, that -18°C figure comes from traditional real world physics and is the temperature of the Earth without any atmosphere at all, that is, also without the bulk of the atmosphere which is nitrogen and oxygen. The comparison in traditional real world physics is with the Moon, which has no atmosphere.
The equivalent temperature of the Moon corresponding to the Earth without any atmosphere at all, is around -23&degC.
In other words, it is the bulk of our atmosphere of around 98% nitrogen and oxygen which is primarily the real world greenhouse gases without which the Earth would be -18°C – the comparison is with the Moon which has no atmosphere.
AGWSF has by sleight of hand science fraud misappropriated this figure and claims it is the temperature of the Earth only without its version of “greenhouse gases, but with the rest of the atmosphere of mainly nitrogen and oxygen in place”.
From this AGWSF claims that “its greenhouse gases bring up the Earth’s temperature 33°C to 15°C. This is an illusion.
Firstly because they have based this on their science fraud of misattribution of the -18°C figure, but also because there is no mechanism they can show for “their greenhouse gases” having the power to raise global temperature 33°C from -18°C to 15°C.
And thirdly, because they have excised the Water Cycle to create their AGWSF Greenhouse Effect Illusion.
Here’s how. In real world traditional science the temperature of the Earth with the atmosphere in place, but without water, think deserts, the temperature of the Earth is given as 67°C.
In other words, in real world traditional physics, it is the bulk atmosphere of mainly voluminous and heavy under gravity nitrogen and oxygen which makes the real thermal blanket around the Earth.
The bulk atmosphere of mainly nitrogen and oxygen which with the inherent properties and process of real gases keeps the Earth from the extreme temperature differences of the Moon, because these real gases not only trap heat through gravity keeping them in place, but also cool the Earth through convection, because in the real world in real physics hot air rises and cold air sinks.
Any meteorologist who doesn’t understand this doesn’t know how we get our winds and weather. Winds are convection currents in our heavy fluid gas atmosphere created by differential heating of volumes (sometimes called packets) of the real gas air. As volumes of heated air rise taking away heat from the Earth, volumes of colder air sink, flowing beneath. Winds are volumes of air on the move. Hot air rises because real gases expand when heated and so become less dense and so lighter than air; these are areas of low pressure, because less dense they weigh less under gravity. Colder heavier more dense air will spontaneously sink flowing down into the low pressure area of less dense hotter air.
Winds blow from high to low. Winds blow from high pressure to low pressure. From denser heavier to less dense lighter. Condensed heavier colder masses of real gas sink, expanded lighter hotter masses of real gas rise. Just as currents are created in the ocean, as volumes of heated ligher water rise and colder heavy volumes sink.
So, with our heavy ocean of real gas atmosphere of mainly nitrogen and oxygen in place, but without water, the extremes of the atmosphere less Moon are avoided and we get a temperature of 67°C.
Putting back water which comprises some 7/10ths of the surface, this 67°C is brought down to 15°C. This is called the Water Cycle.
AGWSF has excised the Water Cycle to create the illusion of “33°C warming by its greenhouse gases”.
The AGWSF Greenhouse Effect is an illusion.
It makes a big difference in what the input can and can not do. The total energy is conserved, the input is equal to the output in terms of total energy, but not in terms of power, not in terms of what the input and output power flux can do. The input can do a lot more work than the output. The input can generate very high temperatures on the surface. The atmosphere is then heated by the surface conductively/convectively and radiatively. That’s all that needs to happen because the input is so hot. Hot heats cold. You can go outside and feel the hot input on a sunny day. Latent heat carries a lot of this input around the planet via water vapor and liquid water.
Water has a very high heat capacity, as heated water becomes even less dense and so even more lighter than air it evaporates more quickly, this fluid gas water vapour taking the heat into the colder heights as it rises where it releases it and condenses back to liquid water or ice, and colder returns to the surface as rain.
The Water Cycle cools the Earth from 52°C from the 67°C temperature the Earth would be without it, but with the bulk of the heavy real gas atmosphere of nitrogen and oxygen still in place.
It is the properties and process of real gases with volume, weight, attraction, which create the lapse rates in conjuction with gravity acting on the real mass; gravity stronger nearer the surface pulling the molecules of gas together, condensing them so they are heavier, allowing them to expand as gravity is weaker away from the surface so they become less dense and so lighter.
AGWSF’s the Greenhouse Effect Illusion is created out of the imaginary “ideal gas”, without mass and therefore nothing for gravity to act on. That’s why they don’t have gravity in their world and don’t understand real world physics. They have no weather at all because their ideal gases with no volume weight attraction cannot expand and condense to create weather. They are climate scientists of a non-existant climate.
All they have is an imaginary world where their ‘atmosphere’ “is empty space with ideal gases zipping through it at great speeds under their own molecular momentum bouncing off each other in elastic collisions and so spontaneously diffusing to mix thoroughly so they can’t be unmixed”. Their massless, weightless “ideal gas molecules” not subject to gravity can only create pressure by bouncing off some imaginary invisible wall of a “container”.
The AGWSF Greenhouse Effect world does not have an atmosphere, they go straight from the surface to empty space. So their “radiation only” world.
Without heat transfer by conduction and convection.
They have nothing to convect.
There is no internal consistency, how could there be, in their fake fisics – how can their “ideal gas carbon dioxide molecules accumulate in the atmosphere for hundreds of years”, when they have no atmosphere and there is no, unknown to real world physics, “invisible container” around the Earth keeping them in? Whatever atmosphere they might have had has long gone into outer space, and so all their gases and weather.
Their “ideal gases travel at great speeds miles apart from each other through empty space under their own molecular momentum” because they don’t have real gases with volume and weight containing them. They have no sound in their world – without gravity acting on real gases with volume they have no medium through which sound can travel.
The only hot air they have is their irrational arguments which they can’t hear because they have empty space around them in their impossible world.
There is no “33°C warming by the AGW greenhouse gases”.
It is an illusion created by science fraud and fake fisics in all its parts.

Kristian
April 25, 2013 4:06 am

Gary Hladik says, April 25, 2013 at 3:03 am:
“Now snatch away the “insulating” shell, real sudden-like. The system is suddenly cooling twice as fast, as per problem 1023, but its power input is unchanged! Oh no! What happens to the temperature of the inner (now alone) sphere? Increase? Decrease? Remain the same?”
The system is radiating out power to its surroundings twice as fast as soon as the surrounding shell is removed, Gary. The sphere isn’t. Q (or J if you like) remains unchanged. The central sphere has been putting out the same power density flux all along, because that is what its emission temperature dictates, the emission temperature in turn being dictated by the heat supplied to the surface of the sphere from its heat source, and only that.
Read my comment
http://wattsupwiththat.com/2013/04/24/spencer-slays-with-sarcasm/#comment-1286834
The flux leaving the surface of the sphere into the vacuum between it and the surrounding shell (Q) will be the same as long as its internal heat source is active and constant. This flux is simply intersected by the shell, making use of some of it (heating the shell and after equilibration, maintaining its temperature) and letting the rest through to the surroundings. In other words, half goes to the shell, half goes to space. The flux from the sphere is the same all along, but the heat transfer between the sphere and the shell (Q’) is only 1/2Q, as is the heat transfer between the shell and space (Q”). Together they make up Q: Q = Q’ + Q” (Q = (Q – Q”) + Q”). The only thing you need to remember is that the entire energy/heat flux ultimately comes from the sphere, the shell simply ‘splits’ it upon absorption, a sort of intermediate reservoir.

DennisA
April 25, 2013 4:23 am

How many angels were there on the head of that pin? My brain hurts. This is my simplistic layman approach to it and I know people will jump all over it, but so what.
Is CO2 increasing?
Year in year out.
Is temperature increasing in tandem?
It seems to be getting colder.
Does it get hot when the sun comes out?
Always seems to.
Does it get colder when the sun is blocked by cloud cover?
Always seems to
Did it get colder in the 60’s and 70’s when CO2 was also increasing.
It really did. I was around then.
Was it warmer than now in the MWP, the Climatic Optimum and the Roman Warm Period?
I wasn’t around then but a lot of credible people say so.
Was CO2 higher then than today?
Apparently it was a lot lower.
Did the LIA happen?
There’s a lot of historical evidence it did.
Were those temperatures normal?
Hope not.
Is it warmer now?
Thankfully
Has there ever been less ice in the Arctic than there is now?
History says yes
Did polar bears die out
History says no.
Did the Industrial Revolution start in 1850-1880?
Ask Sir Clement Clerke and others from 1678, using coal furnaces known as cupolas, or Abraham Darby, who was using coke to fuel his blast furnaces at Coalbrookdale in 1709.
So is CO2 from human emissions warming the planet out of control?
Well, at the end of the day and not to put too fine a point on it, I find it very difficult with the evidence of my own subjective observations, to actually believe that it is.

joeldshore
April 25, 2013 4:40 am

gbaikie says:

What is expected by some is that CO2 will cause there to less loss of heat. So generally since it doesn’t cause the day to be warmer, it could cause the nights to become less cold.

The notion that CO2 can’t cause the days to become hotter is nonsense.
There is “cooling” going on even during the daytime in the sense that the Earth is always radiating heat. It is just that the warming due to the sun is larger than the cooling due to the Earth radiating. If you increase GHGs, you can change that balance and cause it to become warmer during the day too. There are reasons why nighttime temperatures are expected to warm more than daytime temperatures under rising GHGs, but both are expected to rise (and have been rising).

tjfolkerts
April 25, 2013 4:49 am

Kristian says: “What Problem 1026 is telling us is specifically that if R ~ r, the Q = Q’ + Q”.”
More specifically, Problem 1026 is telling us Q’ = Q” = 1/2 Q for R ~ r. In other words, the inner sphere originally emits thermal IR at a rate Q. After adding the heat shield, the inner sphere emits Q/2 IR to the shell. The shell in turn also emits Q/2 to space. So with the aptly named ‘heat shield’, we have reduced the heat loss to 1/2 of what it was, so only 1/2 as much power needs to be supplied to the the sphere to maintain the postulated steady temperature T.
“This is the exact equivalent to the Carnot cycle:”
No, not really. There is no work being done in Problem 1026, and no cycle. Other than that fact that Problem 1026 and the Carnot cycle both have something warm and something cool, there is really not much else that the two have in common.
Even “Q” itself means quite different things. The “Q” for a Carnot Cycle is the total heat transfer. The “Q” in Problem 1026 is the RATE of heat transfer.
“At any point in time, the heat flux leaving the surface of the sphere (Q) corresponds directly to (depends on) its specific emission temperature … “
The heat, ie the net transfer of thermal energy depends both on the temperature of the surface and the temperature of the surroundings. . You are ignoring the “J1” = J/2 that is clearly shown coming back to the inner sphere from the shell due to the temperature of the shell..
The NET loss from the planet is J/2
The NET loss from the shell is J/2.
The power required from a heater to maintain this situation is J/2.
“A star is pretty constantly heated, isn’t it?”
One would think so. But clearly this textbook problem assumes something else — that the TEMPERATURE is constant and the power adjusts to maintain that temperature. They could have instead assumed that the power input was constant, in which case the temperature of the star would have gone up after adding the heat shield.

A C Osborn
April 25, 2013 4:49 am

richard verney says:
April 24, 2013 at 5:10 pm
“I can remember an old design of 3kw electric fire.
So the question is why did the middle element not get noticably hotter and take on a much brighter appearance?”
Richard, come on that is the real world, not the Climatological world.
It should have been about 50% hotter than the 2 outside ones, as they only received half the Radiation (from the other elements) as the middle one.
So all you Physicists please explain, as I notice you didn’t try to when Richard posted it.

April 25, 2013 4:56 am

Tim, you’re mixing two problems, I’ve pointed this out several times now.
1026, regarding the effect adding a shell has on the rate of cooling
1023, regarding the effect a shell has on the power radiated to the surroundings
1023 is the one described as analogous to a star in a dust cloud, with the implication that it has a constant input.
1026 is the one describing a body cooling with a shell added, with the implication that it does not have a constant input.

Bryan
April 25, 2013 5:25 am

A number of posts advocating the greenhouse theory include the idea of a source of constant power.
That is that backradiation plus constant power force the source to a higher temperature.
Now electrical engineers are aware of electrical constant power sources.
But these are specially designed circuits and are not perfect.
I am not aware of any naturally occurring spontaneous constant power sources.
An increase in temperature of the source would imply it was capable of spontaneously increasing the energy QUALITY of its output
If a theory depends on such a violation of the Second Law then it must be false.

April 25, 2013 5:39 am

There’s no link to the thing you were ridiculing either here or on Spencer’s article. That makes it harder to share in the joke. I guess there would be a lot less to talk about on WattsUpWithThat if the greenhouse effect wasn’t true. And there would be a lot less books for Roy Spencer to sell.
In any case, the Tyndall experiment does not prove the existence of an atmospheric greenhouse effect as Monckton claims, merely that a cold gas target will absorb heat from a hot source. In the atmosphere CO2 molecules (and H2O etc) emit as much as they absorb, so there is no greenhouse effect. (My own post on it here.)

Dry desert climes cool faster at night and have a greater diurnal temperature variation than moist climes. Water vapor is the difference. CO2 has a smaller effect, but an effect nonetheless. – Anthony

Regarding the desert at night, it is kept warmer because the water vapour changes the adiabatic lapse rate, not because of IR trapping. Also, humidity makes it feel warmer to a human.
IR trapping cannot make the source warmer than it otherwise was, unless you create energy in violation of the first law of thermodynamics.
Likewise, backradiation cannot make an object warmer than it otherwise was, unless heat moves spontaneously from a cooler to a warmer pool in violation of the second law. Another way of phrasing the second law is: EMR from a cooler object cannot be made to warm a warmer object further still.

TimTheToolMan
April 25, 2013 6:03 am

Mosher writes “Earth radiates to space from the ERL.”
Agreed.
And then writes “When you add GHGs you raise the ERL.”
Maybe.
And then writes “When you raise the ERL the earth radiates to space more slowly.”
Why? Those slides state the following
“Surface and lower atmosphere cool by infrared radiation to space from upper troposphere (ERL= effective radiating level, such that total CO2 above is fixed)”
and
“Infrared radiation leaves earth for space from upper troposphere (ERL). Amount increases with temperature at ERL (immediate). Height of ERL is such that total CO2 above it is constant.”
And the fundamental assumption is that “total CO2 above it is constant” meaning the amount of energy leaving is dependent on opacity. The problem I have with that is that if say the CO2 levels double, then the total number of molecules at the ERL that have sufficient energy to radiate remains the same but the number of CO2 molecules that have sufficient energy to radiate doubles.
Therefore the amount of radiation heading upwards has doubled. So whilst the opacity has increased, so has the number of packets of energy trying to get past so the assumption that “total CO2 above is constant” is intuitively wrong for me.

Yorkshireman
April 25, 2013 6:07 am

A C Osborn says:
April 25, 2013 at 4:49 am…..
My approx. back-of-the-envelope calculations:
(1 cm diameter elements spaced at 5cm)
Assume elements are long enough to disregard effects at ends.
Each element receives ~ 3% of its neighbours radiation. (diameter / 2 * pi * spacing)
Centre element receives ~ 3% more energy input (power) than outer elements or 1.03 times.
Assuming no convection/conduction, using 4th root, centre element is 1.007 times as hot (in Kelvin).
If element temperature is several hundred degrees C, then difference is a few degrees,
This is probably too small to be noticeable by eye.

tjfolkerts
April 25, 2013 6:18 am

Max says: “Tim, you’re mixing two problems … “
BOTH problems have a “star” at a fixed temperature, and then add a “heat shield”. BOTH calculate how much less power is required after adding the shield. The only real difference is that 1023 assumes the shell is “nearby” so the R ≈ r, while 1026 allows the outer shell to be any radius. 1023 is a limiting case of 1026, and both say the total power required is 1/2 as much when the shell is added nearby.

LdB
April 25, 2013 6:40 am

@Sergey
You are sort of getting it and instead of answering your question I will pose two questions which answer you question but with you doing the thinking.
If I looked at the earth in only the infrared section what do I see. Here is a hint shot from Galileo spacecraft at a distance of 1.32 million miles (http://space.about.com/od/pictures/ig/Earth-Pictures-Gallery/Global-View-of-Earth-in-the-Ne.htm) Earth in the far infrared it would be even brighter … why????? You might want to look in contrast at the ultraviolet shot of earth from space shot from rosetta spacecraft (http://www.redorbit.com/news/space/1809886/revealing_earths_ultraviolet_fingerprint/)
So your question why is the infrared image so much brighter than the ultraviolet image when viewed from a distance of space even though the sun emits predominately ultraviolet or visible spectrum.
Second question for you to work on since you are concerned by the suns blackbody temperature. The temperature of the corona of the sun is several million degrees not 6000K so the black body emission temperature discrepancy is actually far far worse than what your poor attempt to portrait the problem does. So how do we resolve a temperature of several million degrees on the corona to a black body temperature of 6000 degree and why isn’t a problem for science.

Greg House
April 25, 2013 7:16 am

steveta_uk says (April 25, 2013 at 3:19 am): “For a simple example of how you are wrong, why are you assuming that the back radiation is from a colder source?”
============================================================
Very simple: per definition. Radiation from a warmer source is not back radiation per definition. You can call it “direct radiation”, if you wish, but a new term is not necessary, because there has been no confusion about it. Well, with the exception of your comment.

beng
April 25, 2013 7:41 am

It’s bad enough to deal w/warmers that inflate known GHG effects to absurd levels. Now add to that slayers that don’t even understand that much (but think they do).

Joel Shore
April 25, 2013 8:07 am

Noelene says:

So there is a consensus on the greenhouse effect?

Actually, the fact that one gets so much argument over something that is such basic physics should totally refute the argument that there is no consensus about AGW because you can find lots of arguments on the web about it.
The mistaken arguments of Monckton and others revered by those at WUWT are a little less silly than the Slayer’s arguments…but not that much.

joeldshore
April 25, 2013 8:17 am

TimTheToolMan says:

And the fundamental assumption is that “total CO2 above it is constant” meaning the amount of energy leaving is dependent on opacity. The problem I have with that is that if say the CO2 levels double, then the total number of molecules at the ERL that have sufficient energy to radiate remains the same but the number of CO2 molecules that have sufficient energy to radiate doubles.

No…You don’t get it. If the total number of CO2 molecules doubles, the atmosphere becomes more opaque and the ERL moves up to a new level where the number of CO2 molecules is less (so the total opaqueness above it is still the same). What you are not understanding is how things work in a gas where you have both absorption and radiation.
Look at plots of the Earth’s emission as observed from space and you will see that the CO2 absorption lines result in suppression of emission at these wavelengths, exactly as theory and models predict. The whole field of remote sensing is based on us having a correct understanding of how this works. So, if you are confused about it, accept that fact that you are confused and need to learn more rather than continuing to believe you are right in the face of overwhelming evidence that you are not.

Stanb999
April 25, 2013 8:29 am

Dry desert climes cool faster at night and have a greater diurnal temperature variation than moist climes. Water vapor is the difference. CO2 has a smaller effect, but an effect nonetheless. – Anthony
——————————————————————————————————————-
Are your sure it isn’t just the H2O being present that changes the dynamic of temperature flow from desert to “moist” areas? Check out the properties of H2O. It is an amazing element. Raise 1 pound of water 1 degree. 1 BTU. Raise that same pound the last degree to steam. 900 BTUs. Free water from it’s solid state… about 140 btu’s..
Of differently. It takes a tremendous amount of energy for the earth to have a temperature different than that of liquid water for a given pressure. The land cools till condensation, Dew/frost. It heats till evaporation caused clouds… This affect has nothing to do with the IR capability of h2o vapor.

tjfolkerts
April 25, 2013 8:32 am

Greg, the “definition” of “back radiation” deals only with direction, not temperature. “Back radiation” is simple radiation directed back toward earth’s surface. By extension we could use your term “direct radiation” or “emitted radiation” or “forward radiation” for the thermal IR leaving any surface. Then “back radiation” or “absorbed radiation” would be the IR absorbed by the surface.
It would be very odd to my thinking to reserve “back radiation” for radiation from a cooler object toward a warmer object. If there is both a warmer and a cooler object around, then a surface would simultaneously be emitting “back radiation” and “direct radiation” from the same atoms. Why use two different names for exactly the same process producing exactly the same spectrum of photons?

Matt in Houston
April 25, 2013 8:38 am

@ Dr. Spencer
” ANYTHING that reduces heat loss from a heated object can increase its temperature.”
This is categorically wrong. It cannot heat anything unless it is at a higher temp than the “heated object”. If there is some insulating object (ie any matter) or gas it REDUCES the RATE of cooling from the “heated object”, under no circumstance can a cooler object HEAT a a HEATED OBJECT.
I don’t think that Dr. Spencer meant his statement exactly the way he stated it though.
Perhaps it can be re-stated in a more thorough fashion Dr. Spencer?
If anyone already addressed this comment, I apologize, I am running out the door and had to post without reading everything in here. Great discussion though and I hope Dr. Spencer and Mr. Watts will try to invite Dr. Pierre Latour to come in and write a response article on this matter. His writings on this subject are articulate, principled and clear.

April 25, 2013 8:50 am

Joel Shore says:
“No…You don’t get it.”
What Joel Shore doesn’t get is the plain fact that the planet is not responding to the rise in CO2 in the way that every climate alarmist was predicting [until the planet showed that it was not cooperating with the endless, incessant, and wrong predictions of runaway global warming].
Who should we believe? Joel Shore? Or Planet Earth?
Because they cannot both be right.

April 25, 2013 8:57 am

Matt in Houston, no it is not wrong. You are being misled about basic thermodynamic principles. I meant my statement exactly as stated: “ANYTHING that reduces heat loss from a heated object can increase its temperature”. Furthermore, that “thing” is almost always at a lower temperature than the heated object. The examples are literally all around you, even the clothes you wear.

Greg House
April 25, 2013 8:57 am

tjfolkerts says (April 25, 2013 at 8:32 am): “Why use two different names for exactly the same process producing exactly the same spectrum of photons?”
=====================================================
a) because it causes confusion and b) because it makes possible to mislead via the trick “the same spectrum of photons”/”warming effect”, although the warming effect is physically impossible.
The notion of “photons” alone is misleading enough, because people fall into the trap and start counting them without any basis in real science. The idea about “all radiation being equal” has no basis in real science either and misleads people as well.

April 25, 2013 9:47 am

Most of science is merely the appearance of confidence. If you make something appear confident, people will follow it, totally independent of any rationality. In fact they’ll create sophistry, and will fully believe themselves, just to defend the apparition of confidence so that they can feel like they’re part of it or that the confidence is part of them. Then, they’ll get really upset about questions which threaten the foundation of the confidence. As we see. Just look at how people call it ridiculous to say that the Earth is heated on one side only by the Sun, and that this is the average solar input. People with PhD’s literally call this ludicrous. They call it ludicrous that the Sun heats the Earth on one side only, and that you can mathematically quantify this and that it physically makes a big difference to the usual assumptions. No one wants to even discuss it calmly, kindly, or openly, they just obfuscate and change goal posts, and all manner of other things. Mostly they get really upset and call people names. They expose themselves for the most part quite uncomfortable with what the scientific method is actually supposed to be about. But they have the appearance confidence, and the masses of number who follow confidence. It is not possible to have a civilized discussion with people who think it is ludicrous to talk about an Earth heated on one side only with a very high forcing temperature. There is no civility to be had between the idea that the Earth is heated on one side, and the idea that it is not. From there, the only thing that happens is that abuse gets thrown about from those who think it is ludicrous to speak of the Earth being heated on only one side. And then the people who say that the Earth is heated on one side, somehow, magically, become blamed for the abuse that other people created. This is all very fascinating.

Reed Coray
April 25, 2013 9:56 am

I want to thank joeldshore (April 24, 2013 at 8:13 pm), LdB (April 24, 2013 at 8:19 pm) and Gary Hladik (April 24, 2013 at 10:34 pm) for responding to my comment. I’ll reply to each response. But before doing that, I want to make it clear– I am not trying to “slay” the “greenhouse effect dragon”!” For one thing, I’m not sure exactly what the “greenhouse effect dragon” is. If the “greenhouse effect” is the statement that some gases in the Earth’s atmosphere absorb/radiate electromagnetic energy in sub-bands of the IR, then I believe in the greenhouse effect. If the “greenhouse effect” is the statement that greenhouse gases in the Earth’s atmosphere will affect the temperature as a function of location on the Earth’s surface, then, again, I agree. If the “greenhouse effect” is the statement that greenhouse gases in the Earth’s atmosphere will increase the “average temperature” (whatever that is) of the Earth’s surface, then I don’t have sufficient knowledge to arrive at an informed opinion. My original comment was a response to Dr. Spencer’s statement: “Greenhouse gases (thermodynamically like insulation in your house) reduce the rate at which heat flows from higher temperatures to lower temperatures, …
Response to joeldshore.
I agree that increasing the amount of CO2 in the vacuum region of a thermos bottle won’t result in improved thermos bottle performance. However, based on the rest of your response I believe you agree with the statement: “adding any amount of CO2 gas to the region between the inner and outer walls of a vacuum thermos bottle will NOT improve thermos bottle performance.” If my belief is valid, then isn’t Dr. Spencer’s immediately above statement incorrect?
Response to LdB.
Who mentioned CLASSIC PHYSICS verses quantum mechanics? The experiment is independent of any particular theory. Injecting a fixed amount of greenhouse gas into the volume between the inner and outer walls of a vacuum thermos bottle will have one and only effect on thermos bottle performance–where thermos bottle performance is defined as the time interval required for the temperature of the coffee to reach the temperature of the heat sink into which the thermos bottle is placed. It will either (a) have no effect (equal time intervals), (b) improve performance (increase the time interval), or (c) degrade performance (decrease the time interval). If either (a) or (c) is true, then Dr. Spencer’s statement is wrong. If (b) then, like all physics’ statements/theories, the experiment doesn’t prove Dr. Spencer’s statement/theory is true under all conditions, but the experiment cannot be used to demonstrate that Dr. Spencer’s statement/theory is wrong.
Next, it seems to me that your introduction of “gain factors” is obfuscating the issue at hand–to wit, will adding CO2 to the vacuum space of a thermos bottle decrease the rate of cooling? And to your comment: “It is that simple and basic and no amount of stupid classic physics with thermos flasks is going to save you … simple question posed is CO2 a quantum gain medium answer yes …. end of story. No possibly way to talk around it unless you want to alternatively explain how a laser works” is both irrelevant and nonsensical. The issue of whether CO2 is a “quantum gain medium” may be relevant to a theoretical understanding of the phenomenon, but has no bearing on the experiment itself. And if the experiment shows that CO2 does NOT reduce the rate of cooling, then I have no need to explain “how a laser works”. The onus, if there is one, would be on you to show why the thermos bottle experiment doesn’t agree with your “quantum gain medium” position.
Response to Gary Hladik.
If I understand what you wrote, you and I agree that adding more CO2 won’t improve thermos bottle performance. Then you mentioned thinking of the top of the atmosphere as “the inner layer of the vacuum thermos bottle”. By “inner layer” did you mean the wall of the chamber that holds the coffee, or the wall of the thermos that is in contact with the heat sink? I agree that with the exception of matter leaving the Earth (e.g., the velocities of many helium gas atoms at room temperature exceed “escape velocity”) radiation is the only mechanism for energy to leave the earth/earth atmosphere system. However, there is a difference between the “inner wall” of the thermos (i.e., the wall of the chamber that holds the coffee) and the top of the atmosphere. In particular, in the case of a vacuum between the walls, radiation escapes to the outer wall only from the inner wall. However, in the case of greenhouse gas in the region between the two walls, in addition to thermal conduction and convection between the inner and outer walls, there is no single “surface” from which radiation to the outer wall takes place. To various degrees, the greenhouse gases in the space between the walls will act like an “inner wall” in that a portion of the radiation from greenhouse gases at all locations between the walls will find its way to the outer wall. As I see it, the issue of an Earth atmosphere greenhouse effect is one of determining the Earth surface temperature for two scenarios: (a) a surface surrounded by a non-greenhouse gas atmosphere such that most if not all radiation that escapes to space originates from the surface, or (b) a surface surrounded by an atmosphere containing greenhouse gases such that radiation escapes to space from both the surface and the greenhouse gases in the atmosphere. As I mentioned at the outset, I have insufficient knowledge to determine these temperatures. I’m open to temperature (a) being both higher and lower than temperature (b).
Finally, if by “inner wall” you mean the thermos wall in contact with the heat sink, then my response is: “let the heat sink be the vacuum of cold space.” The nature of the heat sink will have no effect on whether the CO2 thermos outperforms the vacuum thermos.

A C Osborn
April 25, 2013 10:44 am

Can someone point me to where the actual measured CO2 PPM for the upper troposphere or troposphere are stored?

April 25, 2013 10:45 am

Reed Coray:
I appreciate your challenging statements made. I write to answer a question you put because it comes between two people and – being neither of them – I may be able to help.
In your post at April 25, 2013 at 9:56 am you quote Roy Spencer having said

“Greenhouse gases (thermodynamically like insulation in your house) reduce the rate at which heat flows from higher temperatures to lower temperatures, …”

Spencer’s statement is correct.
(Any statement from him is likely to be correct, and if it were shown to be wrong then he would admit it.)
But it seems you have been confused about that correct statement by – in this case, also correct – statements by joeldshore.
(Your confusion should not dismay you because confusion is a normal outcome of interaction with joeldshore.)
You say

I agree that increasing the amount of CO2 in the vacuum region of a thermos bottle won’t result in improved thermos bottle performance. However, based on the rest of your response I believe you agree with the statement:

“adding any amount of CO2 gas to the region between the inner and outer walls of a vacuum thermos bottle will NOT improve thermos bottle performance.”

If my belief is valid, then isn’t Dr. Spencer’s immediately above statement incorrect?

Your “belief” and Spencer’s comment are both correct. I explain as follows.
The “region between the inner and outer walls of a vacuum thermos bottle” contains a good vacuum. And a good vacuum is a very good insulator. Any gas will not be as effective an insulator as a good vacuum. Therefore, introducing any gas “between the inner and outer walls of a vacuum thermos bottle” reduces its insulation. The gas replaces a good insulator (vacuum) with a less good insulator (gas) and, thus, reduces the performance of the thermos bottle.
The atmosphere is a mixture of gases.
Most of the gases in the air (e.g. nitrogen and oxygen) do not “insulate” the Earth’s surface from flow of radiation from the surface to space. (They are similar to a gas being between the inner and outer walls the thermos bottle).
But greenhouse gases do “insulate” the Earth’s surface from flow of radiation from the surface to space. (They are like a good vacuum being between the inner and outer walls of the thermos bottle).
So, the Earth’s surface is better “insulated” when the gases of the atmosphere include greenhouse gases.
I hope this helps.
Richard

Gary Hladik
April 25, 2013 10:49 am

Reed Coray says (April 25, 2013 at 9:56 am): “By “inner layer” did you mean the wall of the chamber that holds the coffee…”
Yes. This wall touches the coffee on its inside surface and on the outside is surrounded by the vacuum.
“However, in the case of greenhouse gas in the region between the two walls, in addition to thermal conduction and convection between the inner and outer walls, there is no single “surface” from which radiation to the outer wall takes place.”
Correct. The earth’s upper atmosphere, however, does have something like a “surface” from which radiation finally escapes to space, i.e. the “effective radiating layer” which Steve Mosher mentioned above (April 24, 2013 at 3:42 pm).
“As I see it, the issue of an Earth atmosphere greenhouse effect is one of determining the Earth surface temperature for two scenarios: (a) a surface surrounded by a non-greenhouse gas atmosphere [snip] or (b) a surface surrounded by an atmosphere containing greenhouse gases”
Dr. Spencer covers that case. I thought it was a pretty good read:
http://www.drroyspencer.com/2009/12/what-if-there-was-no-greenhouse-effect/

Gary Hladik
April 25, 2013 10:59 am

Matt in Houston says (April 25, 2013 at 8:38 am): “…I hope Dr. Spencer and Mr. Watts will try to invite Dr. Pierre Latour to come in and write a response article on this matter. His writings on this subject are articulate, principled and clear.”
His writings are handwaving. Might I suggest you contact him yourself and ask him why he has not performed the “Yes, Virginia” experiment to prove Dr. Spencer (and innumerable textbooks) wrong and win a Nobel Prize in the process?
http://www.drroyspencer.com/2010/07/yes-virginia-cooler-objects-can-make-warmer-objects-even-warmer-still/

April 25, 2013 11:38 am

BOTH problems have a “star” at a fixed temperature, and then add a “heat shield”. BOTH calculate how much less power is required after adding the shield. The only real difference is that 1023 assumes the shell is “nearby” so the R ≈ r, while 1026 allows the outer shell to be any radius. 1023 is a limiting case of 1026, and both say the total power required is 1/2 as much when the shell is added nearby.” ~Tim
Here, let’s go ahead and see if this is the case, shall we?
“Consider a black sphere of radius R at temperature T which radiates to distant black surroundings at T = 0K.
(a) Surround the sphere with a nearby heat shield in the form of a black shell whose temperature is determined by radiative equilibrium. What is the temperature of the shell and what is the effect of the shell on the total power radiated to the surroundings?
(b) How is the total power radiated affected by additional heat shields?
(Note that this is a crude model of a star surrounded by a dust cloud.)
Solution:
(a) At radiative equilibrium, J – J₁ = J₁ or J₁ = J/2. Therefore T₁⁴ – T⁴/2, or T₁ = ∜T⁴/2 = T/∜2
(b) The heat shield reduces the total power radiated to half of the initial value. This is because the shield radiates a part of the energy it absorbs back to the black sphere.” ~Problem 1023
“A spherical black body of radius r at absolute temperature T is surrounded by a thin spherical and concentric shell of radius R, black on both sides. Show that the factor by which this radiation shield reduces the rate of cooling of the body (consider space between spheres evacuated, with no thermal conduction losses) is given by the following expression: aR²/(R² + br²), and find the numerical coefficients a and b.
Solution:
Let the surrounding temperature be T₀. The rate of energy loss of the black body before being surrounded by the spherical shell is
Q = 4πr²σ(T⁴ – T₀⁴).
The energy loss per unit time by the black body after being surrounded by the shell is
Q’ = 4πr²σ(T⁴ – T₁⁴), where T₁ is the temperature of the shell.
The energy loss per unit time by the shell is
Q” = 4πR²σ(T₁⁴ – T₀⁴).
Since Q” = Q’, we obtain
T₁⁴ = (r²T⁴ + R²T₀⁴)/(R² + r²).
Hence Q’/Q = R²/(R² + r²), i.e., a = 1 and b = 1.” ~Problem 1026

davidmhoffer
April 25, 2013 11:47 am

Anthony
Might I suggest yet one more resource page on WUWT? This topic simply brings out a level of misinformation that I for one find aggravating (which I promptly make worse by getting involved in the discussion).
Might it make sense to have a resource page with a small set of seminal articles on the topic? Willis’ Steel Greenhouse, Ira Glickstein’s series, perhaps one or two more. Putting the credible articles all in one place would be an excellent resource for those new to the discussion and would limit (I’d hope) the confusion and argument by circular reasoning that dominates threads like this one (and I think does more harm than good.)
REPLY: Yes, I’ll consider it. – Anthony

Gary Hladik
April 25, 2013 12:10 pm

Max™ says (April 25, 2013 at 3:54 am): “First off, 1023 isn’t about the cooling rate.”
Ah, I think I see the source of confusion here. Actually problem 1023 is all about the cooling rate. Based on its temp, the sphere radiates (cools) at rate J, but thanks to the shield the system radiates (cools) at rate J1, where J1 = J/2.
In fact problem 1023 is much like problem 1026, which mentions “cooling” explicitly. The only difference is that in 1026 the radius of the shell is variable, but it reduces to problem 1023 when R is very close to r.
I notice you didn’t answer my question, but I’ll be happy to answer yours.
“To see what would happen though, consider the case where the shell is in contact with the sphere. If the gap was filled with the same material as sphere and shell are made of, then the volume of material goes up while the energy supply remains the same as it was originally, right?
Two spheres with the same power input/composition which only differ in size won’t reach the same temperature, will they?”
At equilibrium, the surface of the larger sphere will be cooler than the smaller, because it’s radiating the same power from a larger surface. To calculate temp, use the Stefan-Boltzmann calculator
http://calculator.tutorvista.com/stefan-boltzmann-law-calculator.html
Set emissivity to 1, choose arbitrary constant power input, vary the surface area, and solve for temp.
“So now you remove a section from the larger sphere, such that it is the same size as the other sphere, what happens?”
Surface temp goes up, matches the second identical sphere.
“Now remove the same amount of material from the larger sphere, except for a thin shell at the original surface, such that the inner sphere is the same size as the smaller sphere, what happens?”
Surface temp of the sphere goes up, because according to problem 1026 (your reference), “this radiation shield reduces the rate of cooling of the body”. Same power input, reduced cooling, the sphere’s surface temperature must go up until at equilibrium the shell is radiating as much power as the inner sphere receives.
Congratulation, Max! You’ve used a standard text to derive the Steel Greenhouse! The wizard is slain! Willis would be proud of you! I know I am. Well done.

joeldshore
April 25, 2013 12:29 pm

Gary Hladik says:

Matt in Houston says (April 25, 2013 at 8:38 am): “…I hope Dr. Spencer and Mr. Watts will try to invite Dr. Pierre Latour to come in and write a response article on this matter. His writings on this subject are articulate, principled and clear.”
His writings are handwaving.

“Handwaving” is a rather nice euphemism. “Complete and utter nonsense” is a more accurate summary of what Pierre Latour has to say on the subject.
Joseph Postma says:

Just look at how people call it ridiculous to say that the Earth is heated on one side only by the Sun, and that this is the average solar input. People with PhD’s literally call this ludicrous. They call it ludicrous that the Sun heats the Earth on one side only, and that you can mathematically quantify this and that it physically makes a big difference to the usual assumptions. No one wants to even discuss it calmly, kindly, or openly, they just obfuscate and change goal posts, and all manner of other things. Mostly they get really upset and call people names.

No…We don’t deny that it heats the Earth from one side only. What we deny is that you can ignore half the Earth in your average. And, if this concept is too confusing for you, then don’t work with averages at all…Look at the total energy in and out as I explained in this post: http://wattsupwiththat.com/2013/04/24/spencer-slays-with-sarcasm/#comment-1286615
The reason that we are not so calm and kind is we have been round and round on this before (e.g., a couple years ago on Judith Curry’s blog) and I really think at this point, you should know better. You either have one heck of a monstrous mental block or you really do understand but have, for some bizarre reason, decided that you would rather engage in sophistry. Either way, it is not a pretty sight.

Gary Hladik
April 25, 2013 12:35 pm

Kristian says (April 25, 2013 at 4:06 am): “The system is radiating out power to its surroundings twice as fast as soon as the surrounding shell is removed, Gary. The sphere isn’t.”
When the shell is snatched away (along with its insulating effect), the sphere is the system.
“Q (or J if you like) remains unchanged.”
No, but let’s go on.
“The central sphere has been putting out the same power density flux all along, because that is what its emission temperature dictates, the emission temperature in turn being dictated by the heat supplied to the surface of the sphere from its heat source, and only that.”
No. As Dr. Spencer points out, you need to know both the heating rate and the cooling rate. According to problem 1023, the system was in equilibrium with a cooling rate of J1, which is half of the sphere’s cooling rate J. Snatch away the shell and instantly the cooling rate doubles. Same power input, double cooling rate, the temp must go down. This is basic thermodynamics.
In fact the temp will go down until at equilibrium the sphere’s cooling rate is J1, which was (surprise!) the original (constant) power input.

gbaikie
April 25, 2013 1:05 pm

“joeldshore says:
April 25, 2013 at 4:40 am
gbaikie says:
What is expected by some is that CO2 will cause there to less loss of heat. So generally since it doesn’t cause the day to be warmer, it could cause the nights to become less cold.
The notion that CO2 can’t cause the days to become hotter is nonsense.
There is “cooling” going on even during the daytime in the sense that the Earth is always radiating heat. It is just that the warming due to the sun is larger than the cooling due to the Earth radiating. If you increase GHGs, you can change that balance and cause it to become warmer during the day too. There are reasons why nighttime temperatures are expected to warm more than daytime temperatures under rising GHGs, but both are expected to rise (and have been rising).”
As I said there people who believe this, but there is no evidence which supports it.
The lunar surface during the day is much hotter than Earth is during the day, and the
Moon has no greenhouse effect.
According to greenhouse theory, the greenhouse effect adds 33 C to *average temperature*.
And Earth’s greenhouse effect can’t be making it warmer during the day, as Earth is colder
than the Moon [which has no greenhouse effect] during the day.
During the night Earth is far warmer than the Moon during it’s night. So the only large
effect possible of Earth’s greenhouse effect, or this +33 C which is suppose to added
from the greenhouse effect can only be in keeping the night times warmer.
So there is no evidence of it.
And the theory which believer believe in, does not support the idea that
greenhouse gases increase daytime temperature, yet still many believers
hold this view. [due to being brainwashed].

April 25, 2013 1:08 pm

Again, I just posted the actual text from the problem at 12:10 PM, unaltered, ctrl+F > cooling:
“Consider a black sphere of radius R at temperature T which radiates to distant black surroundings at T = 0K.
(a) Surround the sphere with a nearby heat shield in the form of a black shell whose temperature is determined by radiative equilibrium. What is the temperature of the shell and what is the effect of the shell on the total power radiated to the surroundings?
(b) How is the total power radiated affected by additional heat shields?
(Note that this is a crude model of a star surrounded by a dust cloud.)
Solution:
(a) At radiative equilibrium, J – J₁ = J₁ or J₁ = J/2. Therefore T₁⁴ – T⁴/2, or T₁ = ∜T⁴/2 = T/∜2
(b) The heat shield reduces the total power radiated to half of the initial value. This is because the shield radiates a part of the energy it absorbs back to the black sphere.”
~Problem 1023
Note: I missed your question I’ll check and see what it was real quick… hmmm, not sure which one you meant.
Surface temp of the sphere goes up, because according to problem 1026 (your reference), “this radiation shield reduces the rate of cooling of the body”. Same power input, reduced cooling, the sphere’s surface temperature must go up until at equilibrium the shell is radiating as much power as the inner sphere receives. ~Gary
Ah, so you like 1023 when it agrees with you, but now it isn’t important?
According to 1023 the shell radiates half as much power as the inner sphere receives.
The problem specifically states that it is a simple model of a star in a dust cloud, hence it is roughly analogous to an internally heated sphere and shell.
Perhaps you can find me a Hubble image of a dust cloud glowing as bright as the surface of the stars embedded within, to support this “the sphere’s surface temperature must go up until the shell is radiating as much power as the inner sphere receives” conjecture?

Myrrh
April 25, 2013 1:24 pm

A C Osborn says:
April 25, 2013 at 10:44 am
Can someone point me to where the actual measured CO2 PPM for the upper troposphere or troposphere are stored?
AIRS has it. But it will not release the measurements for the lower or upper troposphere.
However, the scientists working on the project concluded, from all the measurements, that Carbon Dioxide in the atmosphere was lumpy and not at all well-mixed, to their astonishment as they had been brainwashed by this AGW fictional fisics meme, and so insignificant compared with water vapour in the atmosphere.
They released this conclusion at the same time as releasing cherry picked mid troposphere graphics which didn’t show the lumpiness and insignificance of carbon dioxide which they had found to arrive at their conclusion.
Their elementary sleight of hand worked and a flurry of discussions followed as if the conclusion related to the graphics released, which it didn’t. Although to some extent these showed it not well mixed, it didn’t show how lumpy it was.. Instead showing a spread within a few ppm, lumpy suggest areas with zilch.
Carbon dioxide is a real gas, not an “ideal gas” as claimed for the AGW’s Greenhouse Effect.
It has weight because of mass subject to gravity, it is one and a half times heavier than air, this means it will always sink displacing air unless work is being done to change that. Heavier than air gases sink, lighter than air gases rise. It will not readily rise from the ground into the atmosphere, just like the dust on your desk. If it is heated it will as all gases do, expand becoming less dense and so lighter than air will rise, but on releasing this heat it will again condense, real gases condense, and so becoming again heavier than air will sink back to the surface.
Winds, which are volumes of air on the move, that is, convection currents created by differential heating, can move carbon dioxide from one place to another, but, when the wind stops real carbon dioxide will not defy gravity but will sink to the surface. If large amounts of carbon dioxide get into the the great trade winds it can get carried great distances, over the sea to Hawaii for example, but most winds are local and short lived and will move any carbon dioxide around only at a local level. Our great wind systems do not cross hemispheres, so even if carried by these it will not become “well mixed in the atmosphere”. This is what makes carbon dioxide lumpy together with the fact that some areas just don’t produce much and some use all or most of what they produce locally.
And that’s the carbon dioxide which doesn’t come down as rain, all rain is carbonic acid.
AGWSF doesn’t have rain in its “carbon cycle”, firstly because it has eliminated the Water Cycle to create its “Greenhouse Effect”, and secondly because the AGWSF “carbon dioxide” is not real, it is an “ideal gas with no mass, volume, weight, attraction not subject to gravity”, and so not the real carbon dioxide and real water vapour which are greatly attracted to each other in the atmosphere. Real carbon dioxide is in this fully part of the Water Cycle with the residence time of water in the atmosphere of 8-10 days.
There was some Japanese study a short while back which confirmed the lumpiness of carbon dioxide, I think over China, and I recall something about the graphic being pulled and some ambiguous one put in its place.

Peter C
April 25, 2013 1:30 pm

To Gary Hladik,
Why not actually try this for yourself, as an experiment? That way you would know what happens. I have attempted a similar experiment. My result was that the inner black body did not heat up when it’s outgoing. Radiation was reflected back apron it’s self.
My experimental result may be challenged as simplistic and flawed in some respects but in my view it is better than arguing from theory. Maybe some can do a better experiment. Until they do it is my view that cooler objects DO NOT WARM hotter objects, even acting as insulators.

Kristian
April 25, 2013 1:39 pm

tjfolkerts says, April 25, 2013 at 4:49 am:
“You are ignoring the “J1″ = J/2 that is clearly shown coming back to the inner sphere from the shell due to the temperature of the shell..”
In what way am I ignoring it? Did you see the J = (J – J1) + J1?
“The NET loss from the planet is J/2
The NET loss from the shell is J/2.
The power required from a heater to maintain this situation is J/2.”

Nope. The heater will still have to provide J. J = J/2 + J/2. It has to maintain both losses. They both originally trace back to the surface of the sphere, after all … as J.

Kristian
April 25, 2013 1:52 pm

“You are ignoring the “J1″ = J/2 that is clearly shown coming back to the inner sphere from the shell due to the temperature of the shell..”
Ok Tim, so let’s see what this entails …
At first you have the sphere alone in space, powered by an internal heat source providing the surface of the sphere with a constant and evenly distributed power density flux (the heat input) of say 400 W/m^2. This flux brings the surface temperature to a steady 290K and this temperature in turn sets the outgoing flux from the surface of the sphere (the heat loss) to an equal 400 W/m^2. We have energy balance and a steady surface temperature corresponding to the input/output – Stefan-Boltzmann.
Then we surround the sphere concentrically with the shell, a narrow gap of vacuum between the two bodies (R ~ r). What happens?
Well, the internal heat source of the sphere still supplies the surface with its constant flux of 400 W/m^2 which in itself could still only bring the surface temperature to the same 290K as before … and no further. These 290K would in turn by necessity (dictated by the laws of physics) also still produce the same emission flux of 400 W/m^2, output equalling input, as before.
In other words, nothing relating specifically to the sphere and its internal heat source has changed – same input (400 W/m^2), same corresponding emission temperature (290K), same corresponding output (400 W/m^2).
So what has changed with the shell in place? The surface of the sphere now gets 200 W/m^2 in return as ‘back radiation’ from the shell as it’s heated to a temperature of 244K by the outgoing flux from the sphere. That is, it now receives an additional 200 W/m^2 which weren’t there before.
Now, you want to add this flux of 200 W/m^2 to the already supplied flux to the surface of the sphere of 400 W/m^2 from the internal heat source, according to you in the end bringing the new system to a steady state where the shell is at a temperature of 290K and the surface of the sphere is at 345K. At this point the sphere radiates 800 W/m^2 to the shell but gets 400 W/m^2 in return (to a net of 400 W/m^2) and the shell radiates 400 W/m^2 to space, equal to the still constant internal input to the surface of the sphere.
So what’s wrong with this picture?
Raising the temperature of an object or a surface takes absorption of heat. Objects are heated by positive heat transfer. Heat only spontaneously transfers from a warmer to a cooler object.
So how is the 200 W/m^2 flux from the cool shell to the warm surface of the sphere ever going to be an inward positive transfer of heat?
If you want to claim that the 290K sphere warms up to a higher steady-state temperature (by 19%) in the presence of the 244K shell, then you are in effect saying that the shell operates as a second, independent heat source for the sphere.
Why? Because nothing else has changed. The internal heat source still provides but the 400 W/m^2 to the surface of the sphere. This flux will in itself not warm it past the 290K it did before. The sphere also still emits its corresponding 400 W/m^2 flux from its surface based on this temperature.
The 400 W/m^2 from the internal heat source still goes straight out from the surface. It doesn’t in any way linger.
So if the temperature then still rises, then this must somehow be caused by something else. It must be caused by extra absorbed heat from somewhere. A positive transfer of (a gain in) thermal energy.
Well, the only other body in this system is the 244K shell. The only thing that’s different is the additional 200 W/m^2 flux from the cool shell to the warm sphere.
You want to claim that this radiative flux rather somehow slows the cooling rate of the sphere, thus heating it indirectly? Fine, but then you should also explain and show how specifically this is accomplished without the 200 W/m^2 flux of 244K spectrum radiation itself transferring HEAT to it and thereby raising the warmer sphere’s kinetic energy level beyond what it would be otherwise – and hence its temperature? Does it somehow disallow half of the 400 W/m^2 of 290K BB spectrum radiation from leaving the surface of the warm body? No. You’re just adding the 200 to the 400, aren’t you? Effectively as extra transfer of heat to the sphere.

Gary Hladik
April 25, 2013 1:59 pm

Max™ says (April 25, 2013 at 1:08 pm): “Note: I missed your question I’ll check and see what it was real quick… hmmm, not sure which one you meant.”
Not important. You asked essentially the same question in your own way when you derived the Steel Greenhouse.
“Ah, so you like 1023 when it agrees with you, but now it isn’t important?”
1023 and 1026 are essentially the same problem. In fact, as tjfolkerts agrees (April 25, 2013 at 6:18 am) 1026 reduces to 1023 when R is close to r. Now 1023 is cuter, but I like both 1023 and 1026 equally. 🙂
“The problem specifically states that it is a simple model of a star in a dust cloud, hence it is roughly analogous to an internally heated sphere and shell.”
Right, problem 1026. That’s why they made R larger than r, since you don’t see many dust clouds a kilometer or so from a star’s “surface”. In fact in the case of Zeta Ophiuchi
http://www.spitzer.caltech.edu/images/5517-sig12-014-Massive-Star-Makes-Waves
R would be about half a light year, and even then the dust isn’t a perfectly opaque shell.
“Perhaps you can find me a Hubble image of a dust cloud glowing as bright as the surface of the stars embedded within, to support this “the sphere’s surface temperature must go up until the shell is radiating as much power as the inner sphere receives” conjecture?”
Conjecture? Since when are the First Law of Thermodynamics and the Stefan-Boltzmann Law “conjecture”? You don’t believe your own reference? 1023 says the shell cools at a rate J1, half the rate J of the inner sphere. If the system is in equilibrium (no temp change), J1 must be the same as the power input to the sphere. The sphere cools at rate J = 2*J1, but it receives J1 from its power source plus J1 from the shell (look at the diagram), so it’s in equilibrium, too.
Now take away the shell. The sphere (temporarily) radiates/cools at rate J, but now receives only J1 = J/2 from its power source. It must cool until it radiates/cools at rate J1 (plug your own figures for J and J1 into the Stefan-Boltzmann calculator).
Add the shell back, and the sphere’s temp goes back up because, as problem 1023 says, at equilibrium the shell cools/radiates to the surroundings at half the radiation/cooling rate of the sphere, and from Stefan-Boltzmann a surface radiating at J = 2*J1 is hotter than a surface radiating at J1. Easy peasy.

April 25, 2013 2:02 pm

Myrrh says:
April 25, 2013 at 1:24 pm
Myrrh,
The language used by the AIRS people is quite confusing: calling a variation of a few ppm’s (+/- 2% of the full range) over the seasons as “not well mixed”, while about 20% of all CO2 in the atmosphere is exchanged with CO2 from other reservoirs over the same seasons. That only shows that CO2 in the atmosphere is well mixed and quite rapidely. Any huge change in CO2 at ground level is mixed within days to weeks for the same altitude and latitude band, it takes weeks to months to mix it over a full hemisphere and 1-2 years between the hemispheres.
Anyway, over the full globe, except for the first few hundred meters over land, the measured CO2 levels are within +/- 2% of the full range. That is for over 95% of all air…

Reed Coray
April 25, 2013 2:07 pm

richardscourtney says: April 25, 2013 at 10:45 am
Thanks for trying to remove any confusion that might exist. Maybe I can’t see the forest for the trees; but Dr. Spencer did not qualify his statement by caveating that it applies to the Earth and its atmosphere. In fact, he parenthetically included a reference to a house. Taken at face value, Dr. Spencer’s statement:
Greenhouse gases (thermodynamically like insulation in your house) reduce the rate at which heat flows from higher temperatures to lower temperatures, …
imply that everything else being equal, if two objects exist at different temperatures the presence of a greenhouse gas between the objects will, relative to the absence of that greenhouse gas, reduce the rate heat flows from the high-temperature object to the low-temperature wall. If this statement is true, then a CO2 thermos should outperform a vacuum thermos. I haven’t conducted the experiment; and it may turn out that, yes, a CO2 thermos outperforms a vacuum thermos. I don’t believe that is the case–i.e., I believe the vacuum thermos will outperform the CO2 thermos. If the vacuum thermos outperforms the CO2 thermos, then Dr. Spencer’s statement without caveats is not correct. The presence of a greenhouse gas does NOT reduce the rate heat flows from the thermos hotter inner wall to the thermos cooler outer wall.
I’m probably nitpicking here, but from my perspective many people on both sides of the AGW fence arrive at conclusions based on statements made by “experts”. I agree that Dr. Spencer is an expert. But if Dr. Spencer makes a statement that applies only under certain conditions but when making that statement does not identify those conditions, then I’m concerned that people will inappropriately apply that statement to conditions not covered by the caveats. Occasionally the process snowballs out of control leading to incorrect conclusions with societal impact.
Put into yes/no questions, in your opinion
(a) For hot coffee inside the chamber of a vacuum thermos bottle located in a room at a lower temperature than the coffee, is there one or more hot objects and one or more cold objects?
(b) Relative to the outer wall, is the thermos bottle chamber wall a hot object?
(c) Relative to the chamber wall, is the thermos bottle outer wall a cool object?
(d) Will heat flow from the chamber wall to the outside wall?
(e) Will the rate of heat flow from the chamber wall to the outside wall be reduced if CO2 gas is injected into the space (a vacuum prior to CO2 gas injection) between the inside wall and the outside wall?
If the answers to these questions are (a) yes, (b) yes, (c) yes, (d) yes, and (e) no, then do those answers constitute a counterexample to the statement: “Greenhouse gases reduce the rate at which heat flows from higher temperatures to lower temperatures?”
If your answer to the last questions is “No”, then I’d appreciate hearing your rationale.
Thank you for your time.

joeldshore
April 25, 2013 2:24 pm

gbaikie says:

As I said there people who believe this, but there is no evidence which supports it.
The lunar surface during the day is much hotter than Earth is during the day, and the
Moon has no greenhouse effect.
According to greenhouse theory, the greenhouse effect adds 33 C to *average temperature*.
And Earth’s greenhouse effect can’t be making it warmer during the day, as Earth is colder
than the Moon [which has no greenhouse effect] during the day.

And the theory which believer believe in, does not support the idea that
greenhouse gases increase daytime temperature, yet still many believers
hold this view. [due to being brainwashed].

No…They believe it because they understand physics.
Look, you can only get so far by comparing the Earth and the moon. The fact that one has a greenhouse effect and the other doesn’t is only one difference. There are other important differences: The moon has very little atmosphere at all and does not have water. This means that the effective heat capacity or “thermal inertia” is much smaller so temperatures vary more dramatically between day and night. Furthermore, the moon has a much longer day than the Earth…about 30 times longer, so this also favors larger temperature variations. Neither of these differences should affect the average temperature (or, more precisely, the average of T^4), which is determined by radiative balance, but they do affect the range of temperatures.
For heaven’s sake, think things through before making statements about people being brainwashed. You might find that you can actually learn from people who have thought about these things much more than you apparently have.

April 25, 2013 2:25 pm

Myrrh says:
April 25, 2013 at 1:24 pm
It has weight because of mass subject to gravity, it is one and a half times heavier than air, this means it will always sink displacing air unless work is being done to change that. Heavier than air gases sink, lighter than air gases rise. It will not readily rise from the ground into the atmosphere, just like the dust on your desk.
Think one moment further: why should the dust on your desk or the sand of the Sahara travel thousands of km through the air, with a density some 100 times that of air and CO2 at only 1.5 times heavier would sink out? Lookup “Brownian motion” works as good for dust and feathers or sand as for CO2 molecules (or even for far heavier CFC molecules reaching the stratosphere…).