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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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.
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?
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.
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.
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.
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
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 @ur momisugly 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.”
DANG! That should have read “The heat input is “240 W/m^2″ to the INNER SPHERE… “ in my last post!
Joseph Postma says:
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.
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.
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.
“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.
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”.
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
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.
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…
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.
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.
“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 🙂
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.
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?
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.
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?
“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).
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.
“ 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?