Guest Post by Willis Eschenbach
Short answer? Of course not, that would violate the Second Law of Thermodynamics … BUT it can leave the hot object warmer than it would be if the cold object weren’t there. Let me explain why this is so.
Let me start by introducing the ideas of individual flows and net flows. Suppose I owe you twenty-five dollars. I run into you, but all I have is a hundred dollar bill. You say no problem, you have seventy-five in cash. I give you the hundred, you give me the seventy-five, and the debt is paid.
Now, there are two equally valid ways to describe that transaction. One way looks at both of the individual flows, and the other way just looks at the net flow. Here they are:
Figure 1. Net flows and individual flows. The individual flows are from me to you, $100, and from you to me, $75. The net flow is from me to you, $25.
What does this have to do with cold and warm objects? It points out a very important distinction, that of the difference between individual flows of energy and the net flow of energy, and it relates to the definition of heat.
Looking at Figure 1, instead of exchanging dollars, think of it as two bodies exchanging energy by means of radiation. This is what happens in the world around us all the time. Every solid object gives off its own individual flow of thermal radiation, just as in the upper half of Figure 1. We constantly radiate energy that is then being absorbed by everything around us, and in turn, we constantly absorb energy that is being radiated by the individual objects around us.
“Heat”, on the other hand, is not those individual flows of energy. Heat is the net flow of energy, as represented in the bottom half of Figure 1. Specifically, a heat flux is the net flow of energy that occurs spontaneously as a result of temperature differences.
Now, the Second Law of Thermodynamics is only about net flows. It states that the net flow of thermal energy which we call “heat” goes from hot to cold each and every time without exception. However, the Second Law says nothing about the individual flows of energy, only the net flow. Heat can’t flow from cold to hot, but radiated energy absolutely can.
When an object emits radiation, that radiation goes on until it hits something that absorbs it, whereupon it is converted to thermal energy. The individual temperatures of the emitting and absorbing objects are not significant because these are individual energy flows, and not the net energy flow called “heat”. So there is no violation of the Second Law.
Here’s the thing that keeps it all in balance. If I can see you, you can see me, so there are no one-way energy flows.
Which means that if I am absorbing radiation from you, then you are absorbing radiation from me. If you are warmer than me, then the net flow of energy will always be from you to me. But that says nothing about the individual flows of energy. Those individual flows only have to do with the temperature of the object that is radiating.
So how do we calculate this net energy flow that we call “heat”? Simple. Gains minus losses. Energy is conserved, which means we can add and subtract flows of energy in exactly the same way that we can add and subtract flows of dollars. So to figure out the net flow of energy, it’s the same as in Figure 1. It’s the larger flow minus the smaller flow.
With all of that as prologue, let me return to the question that involves thermal radiation. Can a cold object leave a warm object warmer than it would be without the cold object?
While the answer is generally no, it can do so in the special case when the cold object is hiding an even colder object from view.
For example, if a person walks between you and a small campfire, they hide the fire from you. As soon as the fire is hidden, you can feel the immediate loss of the radiated energy. At that moment, you are no longer absorbing the radiated energy of the fire. Instead, you are absorbing the radiated energy of the person between you and the fire.
And the same thing can happen with a cold object. If there is a block of wood between you and a block of ice, if you remove the wood, you’ll get colder because you will be absorbing less radiation from the ice than you were from the wood. You no longer have the wood to shield you from the ice.
Why is all of this important? Let me offer up another graphic, which shows a simple global energy budget.
Figure 2. Greatly simplified global energy budget, patterned after the Kiehl/Trenberth budget. Unlike the Kiehl/Trenberth budget, this one is balanced, with the same amount of energy entering and leaving the surface and each of the atmospheric layers. Note that the arrows show ENERGY flows and not HEAT flows.
These ideas of individual flows, net flows, and being shielded from radiation are important because people keep repeating over and over that a cold atmosphere cannot warm the earth … and they are right. The temperature and the radiation are related to each other by the Stefan-Boltzmann equation. When we apply the S-B equation to the 321 W/m2 of downwelling “back radiation” shown in the graphic above, it tells us that the effective radiating level is somewhere around freezing, much colder than the surface.
BUT a cold atmosphere can leave the earth warmer than it would be without the atmosphere because it is hiding something even colder from view, the cosmic microwave background radiation that is only a paltry 3 W/m2 …
And as a result, with the cold atmosphere shielding us from the nearly infinite heat sink of outer space, the earth ends up much warmer than it would be without the cold atmosphere.
To summarize …
• Heat cannot flow from cold to hot, but radiated energy sure can.
• A cold atmosphere radiates about 300-plus W/m2 of downwelling radiation measured at the surface. This 300-plus W/m2 of radiated energy leaves the surface warmer than it would be if we were exposed to the 3 W/m2 of outer space.
My best regards to all,
w.
My Usual Request: When you comment, please QUOTE THE EXACT WORDS THAT YOU ARE DISCUSSING, so that we can all understand the nature of your objections.
My Second Request: Please keep it civil. Speculation about the other person’s motives and cranial horsepower are greatly discouraged.
Further Reading: My post entitled “The Steel Greenhouse” looks at how the poorly-named “greenhouse effect” work, based on the principles discussed above.
Math Notes: There’s an excellent online calculator for net energy flow between two radiating bodies here. It also has the general equation used by the calculator, viz:
with the following variables:
and Q-dot (left-hand side of the equation) being the net flow.
Now, when the first object is totally enclosed by the second object, then area A2 is set to a very large number (I used a million) and the view factor F12 is set to 1. This is the condition of the earth completely surrounded by the atmosphere. For the general case, I’ve set area A1 to 1 square metre. Finally, I’ve made the usual simplifying assumption that thermal IR emissivity is 1.0 for the surface and the atmosphere. The emissivity values are greater than 0.9 in both cases, so the error is small. With those usual assumptions, the equation above simplifies as follows, courtesy of Mathematica:
But sigma T ^ 4 is simply the Stefan-Boltzmann radiation for the given temperature. That is why, in the energy budget above, we can simply add and subtract the energy flows to produce the budget and check to see if it is balanced.
So, let’s say that a jar of hot water, first, is sitting ALONE. The jar of hot water, thus, has a quantity of “warmth” that is different without an ice cube present.
Now we introduce an ice cube into close proximity to (but not touching) the jar of hot water. The ice cube, obviously, has a SMALLER quantity of “warmth” than the jar of hot water.
But the claim is that … (1) the ice cube can leave the jar of hot water warmer than the jar of hot water would be if the ice cube weren’t there.
It this is so, then the “warmth” that the jar of hot water has WITHOUT the ice cube will INCREASE, when the ice cube enters. The ice cube, thus, heats the jar of hot water. A colder object heats a warmer object.
How is Claim #1 not a blatant violation of the Second Law? … merely worded differently?
Robert Kernodle,
It this is so, then the “warmth” that the jar of hot water has WITHOUT the ice cube will INCREASE, when the ice cube enters.
That is not what Willis said.
First, the jar of hot water will cool in both cases (as there is no heat source to keep it warm).
Second, the cooling in space (at 3 K background temperature) is faster without the ice in the neighborhood.
So, there is no increase of warmth in both cases, only less loss of warmth with the ice…
If there is a constant (electrical) heat source under the jar, there will be an increase of warmth of the jar if you bring the ice near it, for the simple reason that with the same input, you reduce the heat loss to space from the jar, thus the temperature must go up to get rid of the same incoming amount of energy…
See the temperature curve of my calculation sheet…
“thus the temperature must go up to get rid of the same incoming amount of energy“
Quick question: if the heated object were sealed inside a perfectly internally reflective casing (along with its source of power), would that object keep heating itself up, by its own emitted power, indefinitely?
Bear in mind that if your answer is no, it would seem you disagree with the logic in your final paragraph.
Tony,
If every outgoing photon is reflected back, then of course the heated object would heat up further until something gets wrong. As there is only supply and no removal, the temperature would go up until eternity…
I think that this is the fundamental point where one side of the argument diverges from the other. I would say that the slayers think the frequency spectrum of the radiation the object emits can’t be increased by itself, or the fact that the object receives its own reflected energy. Only with an increase of this frequency spectrum would the object rise in temperature.
Tony,
Indeed, the Slayers think that any photon coming from a colder object can’t be absorbed by a warmer one, while every CO2 laser proves that wrong…
As steel absorbs every frequency of IR that is hitting it, the energy contained in the 10 micrometer band sent by the laser is thermalised to increasing temperatures up to the melting point of steel. While getting hotter, the steel is emitting some of that energy as photons with increasing frequency…
Well no, a laser is a completely different matter altogether.
Tony,
Well no, a laser is a completely different matter altogether.
Not at all: the receiving steel doesn’t “know” the origin of the IR photon, which may come from the human next to it (the 10 micrometer wave is at the peak frequency of around a human body temperature) or from a laser. Thus either the steel absorbs it or reject/reflect it, no matter the origin.
As the temperature of the steel increases, a lot (if not all) of the 10 micrometer waves are thermalised, no matter the origin. At the same time, the original low temperature steel emits IR with a peak frequency and intensity directly dependent of its temperature. By thermalising the laser radiation, the steel temperature increases and due to that, the frequency and intensity of its own radiation shifts to shorter – even visible – wavelengths and higher intensity.
Yes, it’s a completely different matter altogether. For the same reasons as are explained every single time a laser is brought up in these discussions. Why you people seem to just want to discuss the same exact things over and over again is beyond me.
Tony November 28, 2017 at 2:30 am
Yes, it’s a completely different matter altogether. For the same reasons as are explained every single time a laser is brought up in these discussions. Why you people seem to just want to discuss the same exact things over and over again is beyond me.
It’s certainly not a different matter, all 10.6 micron photons are the same regardless of their origin, the absorber treats them all the same.
The amount of heating depends on the intensity of the radiation; that would involve the energy in each photon (which you and Ferdinand are discussing) AND the number of photons arriving each second. Lasers used for cutting take wattages in the hundreds and focus the beam until its arriving within a point sometimes less than 0.1 mm in diameter. That translates to an incredibly high irradiance (W/m2) received by the metal at that point.
The heated object trapped inside the reflective casing can neither increase the frequency spectrum of its own emitted radiation, nor increase its own radiant exitance/irradiance (W/m2) to a higher flux.
Tony,
Back to the essence:
I would say that the slayers think the frequency spectrum of the radiation the object emits can’t be increased by itself, or the fact that the object receives its own reflected energy. Only with an increase of this frequency spectrum would the object rise in temperature.
Which is not true. Any addition of EM energy, whatever the frequency, will increase the temperature of the object and thus increase the outgoing frequency spectrum of an object.
That is a little bit for the energy input from cold sources and some more for hot sources and extremely much for a CO2 laser, even if the latter has a very small frequency band, much lower than most emitted frequencies from a heated object…
Of course you are right that the laser provides much more energy than the energy emitted by the temperature of an object. But that is not the point, the point is that steel does absorb and thermolyses low energy IR photons, whatever the source and intensity.
The heated object trapped inside the reflective casing can neither increase the frequency spectrum of its own emitted radiation, nor increase its own radiant exitance/irradiance (W/m2) to a higher flux.
That is only true for a non-heated object, as the amount of energy reflected equals the amount of energy emitted. A heated object still receives extra energy, while emitted and received energy remain equal. As there is conservation of energy, the extra heat provided is used to warm the object further, thus increasing its emission spectrum, which is reflected 100% back,… Until the whole bunch melts down…
Tony:
At a good lighting store, you can purchase special halogen incandescent bulbs that reflect back much of the infrared output of the filament. As a result, the filament runs hotter than in a standard halogen bulb, resulting in higher total radiation output, and a spectrum shifted to higher frequencies (shorter wavelengths) that standard halogen bulbs.
Ferdinand, you and Phil. should patent your perpetual motion machine right away. All you need is a heated object surrounded by internally reflective casing, and according to you, you can get more power out than you put in. The world’s energy problems are saved!
Something is not right with your idea, Ferdinand. You work it out.
Ed, the whole halogen bulb thing has been done already, under this article I think. Either this or the last one at WUWT that ran to a similar number of comments.
You people really need to stop regurgitating the same nonsense all the time. I know you’re getting desperate, now that GHE “denial” is fast becoming more mainstream, but it’s no excuse for laziness. LikeI said before, you guys will be OK.
Tony,
If you still heat (thus add energy) to an object fully enclosed in reflecting mirrors without any loss of energy to the outside world, that will increase the temperature of that object (and as a side effect shift the frequencies to shorter and more energetic).
Never heard of conservation of energy?
Yes, I’ve heard of it, that’s why I know you’re wrong. You also need to look up the basics of heat transfer.
[Remainder snipped for unwarranted nastiness. Keep it civil, Tony. -w.]
Tony:
These bulbs, with the IR-reflective coatings increasing the filament temperature, exist, and they work as advertised. They are not some thought experiment or laboratory measurement error.
They are physical, empirical proof that you have no idea what you are talking about.
Tony November 28, 2017 at 2:07 pm
Ferdinand, you and Phil. should patent your perpetual motion machine right away. All you need is a heated object surrounded by internally reflective casing, and according to you, you can get more power out than you put in. The world’s energy problems are saved!
To late it’s already patented and it works, it doesn’t do what you claim though, you do not get out more power than you put in, no one except you made that claim. What the lamps do do is achieve the desired temperature element and therefore the desired light output for less electrical input and therefore higher efficiency.
Something is not right with your idea, Ferdinand. You work it out.
No need it’s your idea that is faulty.
Ed, the whole halogen bulb thing has been done already, under this article I think. Either this or the last one at WUWT that ran to a similar number of comments.
You people really need to stop regurgitating the same nonsense all the time.
No the nonsense that is being regurgitated is the same false ‘inverse Wien’s Law’ and the bizarre idea that a photon of a wavelength that is able to be absorbed by an object is mysteriously forbidden from being absorbed if it emitted from a source at a lower temperature. Stop regurgitating that nonscientific nonsense every time the subject of radiational heat transfer comes up and we’ll get somewhere.
Tony,
You may know a lot of heat transfer, but you need to recall some elementary calculations: if you add continuously 100 W electrical energy to a lamp inside a lot of mirrors without any loss to the environment, that energy is added to the total energy content of the lamp. The only way that the added energy is conserved is by warming the lamp and its filament up and up, until the filament melts and then it is over and out.
Even if only 1% of the energy reflected by the mirrors hits the filament, that will increase its temperature until melting as long as you add energy from outside the box: all added energy is emitted and 1% of that extra energy again is reflected to the filament. The remaining 99% reflected energy of what is added from outside will heat the rest of the lamp.
If that was not the case, where does the supplied energy resides?
Nothing to do with perpetuum mobile, only conservation of energy…
Oh, Ferdinand asked me if I’d heard of conservation of energy, so I thought we were doing a whole “condescension” bit. But I see it’s only a problem if I do it.
Anyway, all that aside, you really should patent that perpetual motion idea. There is no real scientific check on patents, as is evidenced by the many perpetual motion machines patented throughout history. Of course, these machines never work when they’re built, but that’s another story. You could use this technology:
https://phys.org/news/2016-12-devices-electricity-closer-reality.html
To convert even some of the heat (the infinite temperature Ferdinand mentioned) into electricity, and you would only need the smallest input of electricity to power the object, and the runaway self-heating will take care of the rest. Infinite temperature will equal infinite electricity, so just one of these things should theoretically (according to Ferdinand) power the whole world’s energy requirements.
Ed and Phil, the bulb thing you’re talking about has already been discussed, not sure why you keep bringing it up. See the discussion on it at Rabbett Run that several people were having about it (in enormous detail). Think it was in the “Green Plate Challenge” thread, if you’re not happy with Bret Keane’s earlier comment on this thread at WUWT. It’s no use just going through the same arguments over and over, that have already been had by other people. It won’t change what has already been said.
And just for the record, Phil and Ferdinand:
“the bizarre idea that a photon of a wavelength that is able to be absorbed by an object is mysteriously forbidden from being absorbed if it emitted from a source at a lower temperature“
and
“Indeed, the Slayers think that any photon coming from a colder object can’t be absorbed by a warmer one, while every CO2 laser proves that wrong…“
You really should not put words in people’s mouths; that is something that makes some people around here very angry, and I’m sure they will be along here soon to condemn that sort of thing as I doubt they would want to be seen as one-sided in applying that criticism. I am not aware of any argument made by myself or any one of these “slayer” characters that I’ve observed, to whit that an individual photon from a cooler object CAN’T be absorbed by a warmer one. As far as I’m aware the arguments centre around the idea of LIKELIHOOD of a photon from a cooler object being absorbed and thermalised by a warmer object compared to the likelihood of a photon from a warmer object being absorbed and thermalised by a cooler one. Thinking about individual photons gets a bit silly when there’s just so many of the darn things. Especially with irradiance as high as you can get with a laser beam.
Tony,
If you continuously add 100 W to a lamp inside a set of mirrors, its temperature will go up, as you don’t loose energy to the outside world. All what you can do is using that extra heat to warm water to a maximum of 100 W if there is zero loss to the rest of the world. That is not a perpetuum mobile, that is just conservation of energy.
If there is no cooling at all, the temperature of lamp and filament goes up until the filament melts…
Again my question: where does the 100 W input go if not used to heat up the lamp/filament?
Ferdinand, the temperature the object can get to is already determined by the incoming power. Why is it that you think that the temperature will suddenly increase just because the power supply continues supplying power? Will the mirrors somehow increase the radiant exitance from the object, or the irradiance to it, in reflecting energy all around? Do the mirrors change which molecules of the object are likely to thermalise photons received?
“the temperature the object can get to is already determined by the incoming power”
No, the temperature is determined by the thermal resistance of the pathway of that heat passing to the environment. Think of an electric blanket (without thermostat). Out in the open, it is cool with the power on. Under blankets it is warm. Too many blankets, and it can catch fire.
The electrical analogy is a current source into a resistance. The voltage (temperature) is proportional to the resistance. Here the reflecting mirrors increase the resistance.
If you want in on Ferdinand and Phil’s invention, Nick, please go ahead. You guys could make a fortune from it if it works. Temperatures going up “to eternity” is not something you want to miss out on, especially since you could couple it with that new technology I linked to. But if it does all work out, don’t forget that I gave you the idea to put the two things together, so I will want in on some of that action. If it works.
By the way, does anyone else want to chip in? I’m not sure if me arguing with four other commenters at once is enough.
Tony,
Ferdinand, the temperature the object can get to is already determined by the incoming power.
Tony, you are completely mistaken on this: the temperature of an object is determined by the incoming power and the outgoing power. If these are in equilibrium, then the temperature of the object is stable. If there is a difference between the two, either the temperature of the object goes up or goes down.
In the case of the mirrors, no/less energy is leaving the system to outside the mirrors than is supplied by the electrical energy, so the whole internals of the system get hotter, or you are destroying energy…
Yes, Ferdinand, the incoming power to the object will be the (presumably electrical) power supplied to the object. The outcoming power from the object will be the thermal radiation that the object emits.
Tony:
Yes, Ferdinand, the incoming power to the object will be the (presumably electrical) power supplied to the object. The outcoming power from the object will be the thermal radiation that the object emits.
For the lamp, the energy input is not only the 100 W electric power, but also the 100 W EM emitted by the lamp which is reflected by the mirrors. That is 200 W incoming in total, 100 W outgoing, 100 W used to heat up the total lamp.
Even if only 1% of the reflected energy is absorbed by the filament, that will heat it further, thus increasing temperature and radiation, including a shift to shorter waves.
I get that you think that reflected output from an object can be double-counted as input to the object. That is necessarily the condition which leads to your idea of runaway self-heating of the object to eternity. Restating your initial position is not going to advance this conversation.
I suggest that if you believe that objects can warm themselves through their own reflected energy, you patent and build your machine forthwith. Utilising the technology I linked to as part of the design will mean that you can get an electrical output from the machine without having to compromise the internally reflective casing. If you believe that what you say is true, why would you not be doing this?
Us talking about it, back and forth, until the end of time, is not going to settle anything.
Tony, you say: “Ed and Phil, the bulb thing you’re talking about has already been discussed, not sure why you keep bringing it up.”
We keep bringing it up because these things that you say could not possibly exist do in fact exist. They work, they do what we claim, and you can buy them in stores.
They are an absolute refutation of your position.
“they do what we claim”
The temperature goes up until eternity (Ferdinand’s words)? Astonishing.
Ed, your bulbs are a distraction, do not represent what we’re talking about (energy still leaves the bulbs, for a start – that’s why they’re bulbs) and they have been discussed to death elsewhere. I get that you’re desperate, but cooler heads have prevailed. Sorry.
Tony:
I get that you think that reflected output from an object can be double-counted as input to the object.
I don’t only think that, it is proven by the MIT lamp with IR mirrors, proven bij high yield halogen lamps, proven by the lamp test by Anthony, etc…
There is a constant input of energy to the lamp. All of that energy must be conserved.
If the lamp is free standing in vacuum, all energy supplied is emitted at the temperature of the filament at 3,000 K and nothing back radiated. Per second: 100 W.s supply, 100 W.s emitted to space, no further heating of the filament and lamp.
If the lamp is covered with mirrors with 100% reflection and no ecape to space and no absorption of the mirrored energy, then in time:
Second 1: 100 W.s energy in, 100 W.s energy radiated out, bouncing around within the mirrors.
Second 2: 100 W.s energy in, 100 W.s radiated out, 200 W.s energy bouncing around.
Second 3: 100 W.s energy in, 100 W.s radiated out, 300 W.s energy bouncing around.
….
Even if only 1% of the bouncing energy is trapped again, the temperature of the lamp and/or filament goes up.
There is no place to hide for the energy supplied other than heating the lamp and its filament.
Except if you have another explanation where the supplied energy resides…
Lol, back to the lamps again…
These lamps where, we’re conclusively told that you don’t get more power out than you put in. It’s just that you get the same power out for less power in. Hilarious:
“To late it’s already patented and it works, it doesn’t do what you claim though, you do not get out more power than you put in, no one except you made that claim. What the lamps do do is achieve the desired temperature element and therefore the desired light output for less electrical input and therefore higher efficiency.“
Either you are all effectively saying that you DO get more power out than you put in (I’m afraid saying that you get the same power out for less power in is the same thing), in which case you are DISAGREEING with me, and the laws of physics, or the bulbs do NOT get more power out than you put in (you just get more in the visible light spectrum and less IR, at both lower power in AND lower total power out), in which case you AGREE with me, and Brett Keane, and the person commenting at Rabett Run, and the laws of physics, etc etc etc.
Tony,
It’s just that you get the same power out for less power in.
Please… Who said that? You get more visible light for less power in. And you get the same temperature of the filament for less power in by recycling IR back to the filament. That is what is said.
Something which is impossible by your theory: IR at a lower average energy level increases the temperature of the filament to much higher temperatures than for the power suplied…
https://wattsupwiththat.com/2017/11/24/can-a-cold-object-warm-a-hot-object/comment-page-1/#comment-2679985
Tony,
We agree on:
you just get more in the visible light spectrum and less IR, at both lower power in AND lower total power out
That is what the lamps do because they reflect IR back to the filament by mirrors. So we may agree on the result, we do disagree on the reason why: the hot filament receives extra energy from the reflected IR which is at a lower frequency, still heats the filament for 2/3 of the total energy needed to reach the desired temperature.
BTW, where gets the energy input if you enclose everything in mirrors without escape to space?
So, Ferdinand, the lamps do NOT give you a higher power output from the same power input so they do NOT prove that
“reflected output from an object can be double-counted as input to the object.”
because the reflected IR is NOT shown to increase the temperature of the system. In the bulbs case the system includes that which is emitted outwards by the bulb (so it includes the room the bulb is in) and in our case (the original, frequently overlooked case which we are actually supposed to be talking about) the system does not include such output and space since everything is contained within the reflective casing. Which is the crucial difference you continue to miss.
In answer to your question, the power output from the heated object doesn’t go anywhere. It bounces around indefinitely inside the reflective casing. Unless you are making the mistake that energy is like matter, and is therefore going to keep building up and building up until something “bursts”, then you shouldn’t have a problem with that.
Tony:
they do NOT prove that
“reflected output from an object can be double-counted as input to the object.”
because the reflected IR is NOT shown to increase the temperature of the system.
Sorry? If you do reduce the power input to 1/3 of the original and the filament still shows the same temperature of 3,000 K, only by reflecting a lot of IR back to the filament, that is NOT increasing the temperature of the filament???
Is there any physical reason why reflected IR is NOT absorbed by the object that emitted it?
The reflected IR which is at a lower frequency than average what is emitted, still heats the filament for 2/3 of the total energy needed to reach the desired temperature.
The same for the mirrored case: every bit of input, even over weeks, is bounced back and forth without ever reaching – and heating – the lamp or its filament, even if it hits it directly?
Any reason why that energy is not absorbed and thus simply reflected in your opinion?
You either agree with me that you’re not getting more power out than you’re putting in, or you don’t.
If you don’t understand, keep reading through my responses until you do.
Tony,
We, that is Ed Bo, Nick Stokes, Phil. and me, do agree that you can’t get more energy out of a system than you put into that system.
Where we do disagree is that within a system, you can have accumulation of energy (and thus temperature) that is retained before it reaches the outside of the system.
That is the case for reflected IR which heats an object within the system higher than from the external input alone.
That is the case for mirrored EM which heats a lamp until the filament melts if there is only energy supply to the system and no energy loss out of the system.
If you don’t agree, give me the reasons why an already warm filament can’t thermalyse (part of) its own reflected EM energy…
Ferdinand. I’ve already made the point that you are not seeing the difference between what constitutes “the system” in the two DIFFERENT cases of the heated object within the internally reflective casing and the light bulb. You insist on referring to the heated object within the internally reflective casing as “a filament”, for instance. The filament is a part of the bulb system, which includes the outer casing and the output from that outer casing. The heated object in the internally reflective casing, on the other hand, is not a filament. It’s just an object.
With the heated object inside the internally reflective casing, that outer casing (the other side, not the reflective side) and the output from that side, are NOT parts of “the system”.
Now take this in. Walk away from the conversation for a day or so. Mull over the implications of it. Read through all the posts again. Mull it over some more. And then, and only then, bother to respond. If you keep replying before then I’m just going to refer you to previous posts.
Tony,
You are just diverting the attention from the essence…
The essence of the whole discussion is that a hot object can and does receive and thermalise radiated energy from a colder object, even if the “colder” object is its own reflected IR radiation.
It doesn’t matter if the energy is transmitted by a filament inside a bulb or from a heated globe. If a part (or all) of the outgoing radiation is reflected back, the temperature of either the filament or the globe will go up for the same initial energy input.
Except if you have proof that for any reason the objects can’t thermalise any reflected radiation…
https://wattsupwiththat.com/2017/11/24/can-a-cold-object-warm-a-hot-object/comment-page-1/#comment-2680265
Plus, not to forget:
https://wattsupwiththat.com/2017/11/24/can-a-cold-object-warm-a-hot-object/comment-page-1/#comment-2679660
And finally, back to the beginning…
https://wattsupwiththat.com/2017/11/24/can-a-cold-object-warm-a-hot-object/comment-page-1/#comment-2678573
Then just keep reading through until you get it. Give it a couple of days. I have faith in you.
Sorry, first put on the wrong place:
Tony,
No need for a few days, I am learning fast…
Back to your answer to the last reference:
I would say that the slayers think the frequency spectrum of the radiation the object emits can’t be increased by itself, or the fact that the object receives its own reflected energy.
I don’t see why that is impossible: if even only 1% of the energy contained in the reflected EM is absorbed, the temperature will go up and the total received energy increases and thus the object’s temperature and the outgoing spectrum gets more shortwave and energy intensive.
Even if, as you say, the possibility of absorbance for low energy IR is less than for high energy IR, if the full outgoing spectrum is reflected, half of it is higher in energy than the peak frequency, thus with a high possibility of absorbance.
In realistic figures:
Start conditions: object (sphere, plate,…) with external heating 100 W in thermal equilibrium with its own emissions spectrum of 100 W in vacuum, no mirrors.
Inserted within a mirrored case, no external losses; 10% of the reflected waves absorbed by the object; time delay between object and mirrors and back: 1 second (needs a box of 150,000 km diameter, but is only used to show the energetic balances).
Second 1: 100 W.s in; 100 W.s out.
Second 2: 100 W.s in; 100 W.s out; 10 W.s absorbed; 90 W.s bouncing around.
Second 3: 100 W.s in; 5 W.s heating the object; 105 W.s out; 19 W.s absorbed; 171 W.s bouncing around.
Second 4: 100 W.s in; 9 W.s heating the object; 115 W.s out; 28 W.s absorbed; 248 W.s bouncing around.
etc…
These are not exact figures, but when I have some spare time, I can make a sheet like the above where one can play with energy inputs, % reflected, % absorbance, etc…
No matter that only 10% of the reflected energy is absorbed and thermalised, the temperature of the object goes up until something goes wrong.
There is no violation of any physical law, as the extra warming is caused by energy lost from the object in the previous (nano)second and (in part) directly absorbed again. Even if that would be 100%, the absorbed energy is from its own previous emissions.
The emission spectrum plays little role, as in every case (near) half of the spectrum is higher energetic than half of the emitter’s own spectrum, thus anyway a substantial part of the bounced energy is absorbed, even if you don’t believe that every photon, whatever its energy level, is absorbed…
There’s just no point in this conversation continuing. You are not adding anything new with what you’re saying.
Perhaps that’s unfair. You are, at least, listening to what I’m saying and taking on board new information. That’s more than we see from a lot of people…I apologise.
Tony November 29, 2017 at 1:15 am
And just for the record, Phil and Ferdinand:
“the bizarre idea that a photon of a wavelength that is able to be absorbed by an object is mysteriously forbidden from being absorbed if it emitted from a source at a lower temperature“
You really should not put words in people’s mouths; that is something that makes some people around here very angry, and I’m sure they will be along here soon to condemn that sort of thing as I doubt they would want to be seen as one-sided in applying that criticism. I am not aware of any argument made by myself or any one of these “slayer” characters that I’ve observed, to whit that an individual photon from a cooler object CAN’T be absorbed by a warmer one.
I’m not putting anything in anyone’s mouth, that comment has frequently been made here and I will continue to post rebuttals to it. Yes I have on occasion referred to the dichroic bulbs as an example of recycling of IR allowing more efficient generation of visible light and will continue to do so where necessary.
Do whatever you want, Phil.
Tony:
You are, at least, listening to what I’m saying and taking on board new information. That’s more than we see from a lot of people…I apologise.
No need to apologise… I always try to understand what another’s opinion means and try to refute it (or accept it) with arguments.
The difference in opinion still seems to be that you and others in this discussion don’t accept that reflected waves of the same and less intensive wavelengths can be thermalised by a hot object. With nuances between to a lesser extent – depending of the probability – and not at all.
The other side, myself included, are of the opinion that whatever wavelength, for a black body every radiation is absorbed and thermalised, no matter the temperature of the origin and the destination.
The latter opinion is supported by lots of real life applications: CO2 and other lasers, light bulb internal reflection,…
Some heavy, but interesting literature about lasers used in metallurgy:
https://www.diva-portal.org/smash/get/diva2:999341/FULLTEXT01.pdf
Fig. 3 at page 27 shows the graph of absorbance: Iron absorbes about 6% of the energy of a CO2 laser at 10.6 micrometer. According to that study, the absorbance increases with temperature and even gets a boost in the molten metal…
Ferdinand, I can see it’s important to you that you have the last word. That’s OK with me.
But it may (or may not) be a revelation to you that not all absorbed radiation is thermalised. That’s another thing to process.
Tony:
But it may (or may not) be a revelation to you that not all absorbed radiation is thermalised. That’s another thing to process.
Radiation that is absorbed must be thermalised (or you are destroying energy). The reason why and how is in the first book of that dissertation. But a lot of the IR beam is reflected, depending of the smoothness of the surface… The best results seems to be with a small layer of rust…
It’s not up for discussion. Absorbed radiation isn’t always thermalised. Look it up.
Your model is a straw man. The claim is that a cool atmosphere that envelops a warmer planet, reduces the planet’s cooling rate over what it would be in its absence.
Your hot water is enveloped by ambient air and the ice cube is a fart in a windstorm
You are a sophist, you cannot delay cooling without creating ”heat” you cannot create heat with a weaker resonance making a higher resonating frequency resonate even higher, to keep ”heat” in the equation, it needs to be created,
Robert:
If you took an actual formal thermodynamics course, you would learn that the first step in analysis is to define the system completely, INCLUDING the ambient conditions. Otherwise you are lost before you start.
You say: “a jar of hot water, first, is sitting ALONE.” First mistake! There is no ALONE. What is the ambient? If the ambient is the walls of your room at 23C (296K), then adding the ice nearby will result in the water being cooler than without it.
However, if the ambient is the 3K effective radiative temperature of deep space, adding the ice nearby will result in the water being warmer than without it.
In both of these cases, if there is no separate power supply for the water, the presence of the ice nearby just changes the rate at which the water cools to ambient. In the first case, it would accelerate it; in the second, it would slow it.
But if there is a constant power input to the water, the presence of the ice nearby will change the resulting steady state temperature of the water. In the first case, it will lower the SS temperature; in the second case, it will raise the SS temperature.
The situation we are talking about with the earth is this final case, where there is a separate power input (the sun), and a very cold ambient (the 3K of deep space). In this case, the presence of nearby ice, being hotter than ambient, will lead to a higher SS temperature for the water.
Ed, I have to thank you for your clear and dispassionate answers to people throughout this thread. Your description above is exactly what would happen—if you put an ice cube next to a warm object in space, the object will be warmer than it would be without the ice cube.
You are also quite clear about the steady state case—with an external power input, the object will end up at a HIGHER steady-state temperature.
w.
Ok, let’s examine the “ambient” claim.
If the water is warmer than both the walls and the ice, will either object increase the temperature of the water?
If both the wall and the ice cube had no starting energy of their own other than the background radiation received from space, would the water cool more slowly with the presence of the ice cube?
Does oxygen, nitrogen or carbon dioxide have any starting energy of their own? Or are all three dependent on what is received from the sun and earth’s surface (which is itself dependent on what it receives from the sun) to be warmed? Has there been any change in the energy received in the global warming/Greenhouse Effect claims by changing the atmospheric composition? NO! All the claims are based on the energy emitted. The amount received doesn’t change. The sun warms the earth. The earth cools to space. The earth is warmed until the energy it emits is equal to the energy it absorbs. The amount being absorbed hasn’t changed, by the conversion of some Carbon and Oxygen into Carbon Dioxide, therefore the amount being emitted hasn’t been changed by it either.
The speed at which something cools has no relevance to what something will eventually cool to. An warmer object will ALWAYS cool to the temperature of its surrounding environment if that environment is cooler. It will ALWAYS warm to the temperature of its surrounding environment if that environment is warmer. So “cooling more slowly thereby making it warmer” is just nonsense. You may create temporary changes of the objects in a system by moving those objects around, but the total energy of that system won’t change as long as the energy absorbed from the heat source remains constant.
wickedwenchfan,
You need to make a differentiation between what happens underway and the ultimate balance.
Once the whole system is in equilibrium, whatever happens in intermediate steps, the amounts of incoming and outgoing energy are the same. So far so good.
Where then is the difference? If you introduce a hindrance to the outgoing emissions, no matter if that is some insulation or something that reflects/re-emits IR, that reduces the outgoing radiation, thus introducing an energy imbalance. That energy imbalance increases the temperature of the emittor (the earth’s surface), until the amount of outgoing energy after the hindrance is the same as the incoming energy at the emittor, thus before the hindrance.
In both cases, the ultimate energy balance is zero. In the second case the earth’s surface is warmer.
In the case that there is no incoming energy, the cooling would be slower for the second case with extra insulation/backradiation.
The ice cube, obviously, has a SMALLER quantity of “warmth” than the jar of hot water.
Remember, though, both are radiating.
But the claim is that … (1) the ice cube can leave the jar of hot water warmer than the jar of hot water would be if the ice cube weren’t there.
If the hot water exist in a infinite, empty vacuum with nothing else that radiates, and you add an ice cube which does radiate, it is trivially obviously that the ice cube must warms the hot water, since an object receiving radiation must be warmer than if it does not receive said radiation. This only seems strange to you because you’re generally surrounded by radiating objects which are warmer than ice cubes, not an infinite zero-radiation vacuum.
“How is Claim #1 not a blatant violation of the Second Law? … merely worded differently?”
The 2nd Law governs the average net flow, it doesn’t speak to the elements of the flow.
Tony,
No need for a few days, I am learning fast…
Back to your answer to the last reference:
I would say that the slayers think the frequency spectrum of the radiation the object emits can’t be increased by itself, or the fact that the object receives its own reflected energy.
I don’t see why that is impossible: if even only 1% of the energy contained in the reflected EM is absorbed, the temperature will go up and the total received energy increases and thus the object’s temperature and the outgoing spectrum gets more shortwave and energy intensive.
Even if, as you say, the possibility of absorbance for low energy IR is less than for high energy IR, if the full outgoing spectrum is reflected, half of it is higher in energy than the peak frequency, thus with a high possibility of absorbance.
In realistic figures:
Start conditions: object (sphere, plate,…) with external heating 100 W in thermal equilibrium with its own emissions spectrum of 100 W in vacuum, no mirrors.
Inserted within a mirrored case, no external losses; 10% of the reflected waves absorbed by the object; time delay between object and mirrors and back: 1 second (needs a box of 150,000 km diameter, but is only used to show the energetic balances).
Second 1: 100 W.s in; 100 W.s out.
Second 2: 100 W.s in; 100 W.s out; 10 W.s absorbed; 90 W.s bouncing around.
Second 3: 100 W.s in; 5 W.s heating the object; 105 W.s out; 19 W.s absorbed; 171 W.s bouncing around.
Second 4: 100 W.s in; 9 W.s heating the object; 115 W.s out; 28 W.s absorbed; 248 W.s bouncing around.
etc…
These are not exact figures, but when I have some spare time, I can make a sheet like the above where one can play with energy inputs, % reflected, % absorbance, etc…
No matter that only 10% of the reflected energy is absorbed and thermalised, the temperature of the object goes up until something goes wrong.
There is no violation of any physical law, as the extra warming is caused by energy lost from the object in the previous (nano)second and (in part) directly absorbed again. Even if that would be 100%, the absorbed energy is from its own previous emissions.
The emission spectrum plays little role, as in every case (near) half of the spectrum is higher energetic than half of the emitter’s own spectrum, thus anyway a substantial part of the bounced energy is absorbed, even if you don’t believe that every photon, whatever its energy level, is absorbed…
Sorry, put on the wrong place…
Paul you only ever have in-flight photons in the chamber they never stop, a minute bit of free flux yet to impinge and replace, photons containing a fraction of the radiation’s resonance from the warmer block.
The thermal resonance of both blocks are near identical if you make the temperature differential 1 kelvin only, all outward emitted fluxes replace all outward fluxes it really is that simple, the blocks are virtually identical a 0.3% temperature differential the chamber back ground is 0 kelvin., [magically].
A is replacing B’s emission, and B is replacing A’s emission, at the same time, virtually instant, however B’s emission and Only B’s s radiating resonance contains some slightly more curried up photons that thermalise.
Gary Ashe,
You haven’t looked at Wien’s displacement law: every object above 0 K emits IR at a range of frequencies, as every atom/molecule in mass has its own “temperature” which may be 0 K to 500 K as kinetic energy. All what we measure as “temperature” is the average kinetic energy of a lot of atoms/molecules.
The difference between 274 K and 275 K is only a small shift in peak wavelength and both objects have largely overlapping emissions at both sides of the peak wavelength.
So, even in your -wrong- reasoning that only more energetic waves are absorbed: near half of the photons emitted by the “colder” object have frequencies which are high enough to be absorbed by the average atoms/molecules in the “warm” object and thus are thermalised…
That even very hot objects do thermalise IR waves in bands that are in the “cold” range is proven every day by CO2 lasers: these emit all their energy around the 10 micrometer band, that is appr. the peak temperature of a human’s body. Despite that, steel at 800ºC still is warmed further up to its melting point by these “cold” waves.
If the blocks are at equilibrium all outward emissions are just replacing all outward emissions from both blocks, there is no variance, none of the continued resonance exchange creates ”heat”. and they do not slow each others cooling, they simply swap the exact same resonance for eternity. at 275 kelvin….. never again will either blocks emissions or absorption resonance act thermally………….it is redundant energy, trapped for eternity.
Unless you stick your finger in.
Not if the object was full of water Ferdinand, you could plug the object directly into the sun and shield it perfectly all you like, you still wont get the h20 above 100c.
Gary,
We were discussing radiation, not boiling. Still once all water evaporated, the jar would melt in the sun…
OK, Ferdinand, let me ask you this, if it works for an Ice Cube wouldn’t it work better for 2 objects at the same Temperature, (like the reflected Energy)?
Both objects are at 60C with a very small Air Gap, will either of them get warmer when exposed to the others Radiation Energy?
If this doesn’t happen why would it happen for and object with lower temperature Radiation?
Willis, re the statement made by both yourself and Dr Spencer on his measuring DWLIR experiment post, which I have reread yet again, including all the comments.
The statement in response to my “concentrating DWLIR” is that diffuse radiation cannot be concentrated.
His very carefully designed experiment proved the opposite.
So please take off your “scientific consensus” head and put on your usual common sense logical head that gave us your thunderstorm regulator posts.
Here is the data I would like you to process and explain to me and anybody else still reading this post where My conclusion is wrong.
The very carefully designed scientific experiment created by Dr Spencer managed to obtain a reduction of 4 degrees F below ambient.
A complete layman produced a reduction of approximately 20 degrees F below ambient, (confirmed by Universities), with no special equipment and no special insulation.
This reduction of 5 times Roy’s was achieved using a Solar Oven as I explained to you before.
What I should have explained is that it is actually a Solar Collector used as an oven.
This Solar Collector, based on those results appears to also be a DWLIR collector also.
To confirm this without realizing it’s significance Roy added a Solar (Radiation) collector to his experiment and it made a difference, however Roy did not appear to update his original graph with the new data so we do not know by how much.
So in your opinion does the Solar Collector “Concentrate” DWLIR or not based on that data?
The comments mentioned quite a few times the Solar Oven’s design and that is why Roy incorporated it to improve the reduction of temperature in his experiment.
Dr Spencer also made an incorrect statement in response to someone who suggested that increased CO2 at TOA would increase Cooling and the good Dr said it would not and would decrease cooling.
This statement by Dr Roy is completely wrong, as NASA’s own satellite data has shown.
Refer to my link to the 2013 AGU meeting for confirmation.
A C:
I’m afraid you are completely missing the point. Some people observe that you can concentrate the energy in solar radiation to get a very high temperature (above ambient) at the point of concentration, and wonder why you cannot concentrate the longwave infrared “back radiation” in the same manner to obtain temperatures above ambient.
Willis (and others) correctly answer that while solar radiation is highly parallel (with only 1/2-degree spread) and so can be highly concentrated with parabolic reflectors, the LWIR radiation is completely diffuse, so cannot be concentrated.
You are talking about cooling effects producing temperatures below ambient — completely different.
Ed, without the “Collector” only -4 degrees below zero was achievable.
So how did the Collector increase the rate of cooling to achieve -20 degrees?
AC:
You don’t provide a link to the experiment you cite, so I can’t comment on the specifics. But, I repeat, Willis was answering questions about possiblly concentrating radiation to get higher temperatures. You are talking about achieving lower temperatures.
You and Willis are both good at quoting each other.
Here is the link.
http://www.drroyspencer.com/2010/07/first-results-from-the-box-investigating-the-effects-of-infrared-sky-radiation-on-air-temperature/
Please read all the comments and then get back to me and explain just how it works.
I also have the same question for you that I asked Ferdinand.
If an Ice Cubes Radiation makes an object warmer wouldn’t it work better for 2 objects at the same Temperature, (like the reflected Energy)?
Both objects are at 60C with a very small Air Gap, will either of them get warmer when exposed to the others Radiation Energy?
If this doesn’t happen why would it happen for an object with lower temperature Radiation?
A C Osborn,
If both objects are not heated from an outside source, in both cases bringing them together in space will reduce the speed of cooling (never get one of them warmer!).
If they are both equal in temperature and size, the speed of cooling would be exactly half the speed without the second one. If the second one is cooler (or smaller) the speed of cooling would be faster in ratio to the difference in mass and temperature.
Why?
Isn’t either of the Objects being bombarded and absorbing Extra Photon Energy?
If it doesn’t work for them how does it work at night for CO2, when the Surface has No Direct Source of Energy and is in the same cooling state as the 2 objects at 60C.
Of course it would work for the CO2 because it is still getting energy from the Surface.
A C Osborn,
If there was no CO2 or other GHGs, the earth at night would cool with all the IR emitted per unit of time for its temperature.
With CO2 (even much more with clouds) part of that emitted IR is sent back to the surface. That is always less than what is emitted by the surface, as not all IR is sent back. Thus all what happens is that the speed of cooling is reduced.
As clouds and water vapor are much stronger reflectors, under cloudy or wet conditions, the nights are hardly cooling, but even the small contribution of CO2 in open skies is measurable…
The message of this thread to me is that the rate of cooling and warming can be reduced by an insulating agent which in the case of the earth is the atmosphere. During the day it reduces the warming, during the night it reduces the cooling (compare to the moon). The clouds increase both effects. What role CO2 has in this? If prof Will Happer is right (and I have not seen it debunked, even Eli Rabett admits it) how much or little LWIR the CO2 molecules in the atmosphere are absorbing, they remain at almost exactly the same temperature as the other air molecules.To me this means that CO2 has no specific insulating capacity. It acts like the other gases of the atmosphere.
a radiant gas instantly (speed of light) distributes the radiation around the globe and out to space.
the word for that is dispersion, not trapping.
“they remain at almost exactly the same temperature as the other air molecules”
Yes. When they absorb energy, they rapidly transfer it to neighboring molecules. Everything gets a little bit warmer.
Because of their warmth (shared with the air) the GHG molecules also emit photons. Then everything gets a little bit cooler again. This mostly works out to a steady state.
Nick: so you agree that CO2 has no specific insulating capacity?
Well if one has two objects without atmospheres near to one another in space both being subjected to the same external radiation source and being at different temperatures then due to the net interchanges of radiation between them the warmer one would heat up the cooler one until both were at the same temperature. It is therefore the net flow between them that matters even with bi directional radiation fluxes.
How could the radiation from the cooler one to the warmer one make the warmer object even warmer when there is no net flow from the cooler to the warmer?
The issue here is whether that final temperature OF BOTH would become elevated above that which the S-B equation predicts.
I suggest not because both having equalised at the same temperature one cannot be cooler than the other and no additional reduction in the cooling rate of either can be caused by the proximity of one to the other.
Both will stabilise at the S-B predicted temperature.
So why is it proposed that a steel shell or an atmosphere around a planet would behave any differently?
Now add an atmosphere to both of the objects.
If the atmospheres were inert they would acquire the same S-B temperature as the surfaces and both surfaces and both atmospheres would be isothermal at S-B. Either via conduction or radiation fluxes the S-B equation would be satisfied.
An atmosphere is not inert, it convects. That is the critical different because calculations based on radiation fluxes alone no longer work.
Suddenly one has to consider non radiative energy transmission processes AND the fact that KE becomes PE in rising air or PE becomes KE in falling air such that an isothermal atmosphere cannot develop.
In that situation the surface temperature DOES rise above S-B and it is nothing whatever to do with radiation.
If each object in space had a different mass of atmosphere then the surface temperatures would be different but the radiation fluxes between them would both be the same as if there were no atmospheres because both objects would be radiating to space and to each other at S-B just as the Earth with its surface temperature enhancement radiates to space at S-B.
Well if one has two objects without atmospheres near to one another in space both being subjected to the same external radiation source and being at different temperatures then due to the net interchanges of radiation between them the warmer one would heat up the cooler one until both were at the same
temperature. It is therefore the net flow between them that matters even with bi directional radiation fluxes.
How could the radiation from the cooler one to the warmer one make the warmer object even warmer when there is no net flow from the cooler to the warmer?
The issue here is whether that final temperature OF BOTH would become elevated above that which the S-B equation predicts.
I suggest not because both having equalised at the same temperature one cannot be cooler than the other and no additional reduction in the cooling rate of either can be caused by the proximity of one to the other.
Both will stabilise at the S-B predicted temperature.
So why is it proposed that a steel shell or an atmosphere around a planet would behave any differently?
Now add an atmosphere to both of the objects.
If the atmospheres were inert they would acquire the same S-B temperature as the surfaces and both surfaces and both atmospheres would be isothermal at S-B. Either via conduction or radiation fluxes the S-B equation would be satisfied.
An atmosphere is not inert, it convects. That is the critical different because calculations based on radiation fluxes alone no longer work.
Suddenly one has to consider non radiative energy transmission processes AND the fact that KE becomes PE in rising air or PE becomes KE in falling air such that an isothermal atmosphere cannot develop.
In that situation the surface temperature DOES rise above S-B and it is nothing whatever to do with radiation.
If each object in space had a different mass of atmosphere then the surface temperatures would be different but the radiation fluxes between them would both be the same as if there were no atmospheres because both objects would be radiating to space and to each other at S-B just as the Earth with its surface temperature enhancement radiates to space at S-B.
The most fundamental point being that the temperature value predicted by the S-B equation is an absolute MAXIMUM that can be achieved via radiation alone from a given external energy source.
It cannot ever be increased by INTERNAL radiation that is not INDEPENDENTLY generated.
In order to exceed S-B one must add non radiative energy transmission processes and since they are inevitably slower than radiation they must raise temperatures of surfaces affected by them.
Thereafter, to maintain hydrostatic equilibrium within those atmospheres, the non radiative processes must net out to zero which is exactly what happens in adiabatic uplift and descent.
Well, I guess I am unsurprised that one more emphatic posting on this topic has failed to resolve all disagreements and put the issue in the rearview mirror once and for all.
It sure does seem to me that at least some of the highly contentious disagreements we find in this article and comment thread ought to be resolvable by one or a series of carefully designed experiments.
It seems very unlikely that these sorts of disputes can ever be resolved by hashing it out at length.
Radiant-heat trapping by freely convective gases has never been demonstrated experimentally.
Convective gases trap conductive heat which depletes the amount of radiant energy able to escape to space,
However, the effect only occurs during the initial uplift of the atmospheric mass off the surface. Once hydrostatic equilibrium is achieved there is no further transfer of radiant energy to convective energy via conduction.
Instead the energy required for hydrostatic equilibrium is simply recycled up and down within convective overturning indefinitely in a net zero loop.
Many years ago I described it as the adiabatic loop as opposed to the diabatic loop.
Overall this is true, but this is still a bad framing: “While the answer is generally no, it can do so in the special case when the cold object is hiding an even colder object from view.”
The vacuum of space is not an object, it is the absence of objects. It’s more accurate to just say “all objects radiate and warm other objects, even objects that are warmer than they are.”
talldave2
Objects in space that are cooler than an adjacent object can only use their own radiation to slow the rate at which they are warmed by radiation from the adjacent object.
They can’t make that warmer object any warmer than the S-B prediction.
In due course the cooler object warms up to a point where both objects satisfy the S-B equation.
AGW alarmists and Willis are trying to tell us that the colder object can make the warmer object reach a temperature higher than the S-B prediction from radiation alone. That must be wrong because if it were possible there would be a perpetual warming process/ loop between the two objects as the colder one continually warms the warmer one and the warmer one then continually warms the colder one a bit more.
AGW radiation theory is the true perpetuum mobile and an abuse of the S-B equation.
The S-B temperature is the maximum temperature one can get for either object at a given level of external radiation reaching both objects simultaneously and in due course both objects will reach that maximum temperature because the net flow is from the warmer object to the colder object.
Unless one adds convecting atmospheres that is but in that event the radiative only S-B equation ceases to apply.
Oh dear oh dear oh dear, I cannot believe anybody can be so slow on the uptake. This is very categorically NOT what Willis is saying. Go and read the article again.
The warmer body benefits from a constant heat source. In the Earth’s case that is short wave radiation, mainly visible and UV, from the sun, arriving at the Earth’s surface, being absorbed and warming it. That warm surface then emits that energy in the form of LWIR. However, it cannot just radiate directly into space – it has that darn pesky atmosphere in its way. That atmosphere, despite being colder than the surface, slows down the rate at which the constantly arriving energy absorbed by the earth’s surface can be dissipated back out to space.
If the rate of energy dissipation is slower than the rate of energy absorption then the amount of energy at the surface starts to build up, in which case the temperature has to rise. A body at a higher temperature emits radiation at a greater rate than a cooler body. Therefore as the temperature rises the rate of heat loss increases until a new equilibrium is reached where the rate at which energy escapes from the earth’s surface is the same as the rate as it arrives from the Sun. That new equilibrium is at a higher temperature.
It is total common effing sense! All the S-B equation does is give you the constant that allows you to calculate the temperatures of the various bodies involved. You don’t need it to understand the principle.
Both you and Willis are talking about a cooler body reducing the rate of cooling of the warmer body such as to raise its ‘equilibrium’ temperature. You both refer exclusively to radiation and so must take into account the S-B equation which is a solely radiative calculation.
In doing so you seek to avoid the implications of the S-B equation. That equation gives a temperature that inevitably follows from external radiation impacting a blackbody that absorbs all that radiation of whatever wavelength and emits it all as sensible heat in the form of the narrower IR wavelengths.
I am pointing out that in the absence of a convecting atmosphere that equilibrium temperature cannot rise above the S-B prediction because the S-B equation defines the perfectly efficient scenario for the generation of IR from multiple other wavelengths.
Simply adding a non-convecting atmosphere can only result in the atmosphere AND THE SURFACE reaching the S-B temperature. Effectively, the atmosphere would have to behave as a solid for that to be possible.
Since the atmosphere is actually a partially radiative gas and not a solid it follows that if the surface goes above the S-B prediction then the total of energy radiated directly to space BY THE SURFACE and energy radiated to space from WITHIN THE ATMOSPHERE will be more than radiation coming in from outside the system which is not possible if the atmosphere is to be held indefinitely in hydrostatic equilibrium.
To be able to go beyond the S-B temperature AT THE SURFACE when the atmosphere also has radiative capability you need the atmosphere to be convecting so that the surplus radiation that would otherwise be coming from the hotter surface cannot be radiated to space IN ADDITION TO the radiation going to space from within the atmosphere. Instead, it is going to fuel ongoing convective turnover via further conduction.
Surface kinetic energy (sensible heat) cannot BOTH radiate and conduct simultaneously.
Thus the surface temperature enhancement is a consequence of mass convecting up and down in a gravity field and not radiation fluxes going back and forth.
Only that way can conservation of energy be complied with.
You are right to say:
“That atmosphere, despite being colder than the surface, slows down the rate at which the constantly arriving energy absorbed by the earth’s surface can be dissipated back out to space.”
But it must be done by conduction and convection and not by radiation for the reasons I stated.
TLM, “The warmer body benefits from a constant heat source. In the Earth’s case that is short wave radiation, mainly visible and UV, from the sun, arriving at the Earth’s surface, being absorbed and warming it.”
No!
It is NOT CONSTANT.
Do you not Understand the difference between Day Time and Night Time.
It is total common effing sense!
So night and day is actually the sun blinking on and off then? Who knew!
Objects in space that are cooler than an adjacent object can only use their own radiation to slow the rate at which they are warmed by radiation from the adjacent object.
Again, that’s sort of true but a poor framing. It’s more accurate to simply say both objects radiate, and some of that radiation reaches the other, ergo a cold object must warm a warmer one — indeed, all object not at absolute zero will radiate and warm all other objects, irrespective of temperature.
S-B just tells you what an object radiates at a given thermodynamic temperature. It doesn’t change any of the above.
That must be wrong because if it were possible there would be a perpetual warming process/ loop between the two objects as the colder one continually warms the warmer one and the warmer one then continually warms the colder one a bit more.
No, that’s only true when the received radiation is greater than the emitted. You’re confusing “getting warmer” with “warming” If I light a fire in a forest at -40F, it will certainly warm me, but I may very well get colder anyway.
You are forgetting that both objects are being continuously irradiated by a single external source.
S-B tells us that at a given input of external irradiation falling onto an ideal blackbody a given temperature will be achieved. It is a two way equation. EITHER a given amount of irradiation will achieve a given temperature OR a given temperature will radiate at a given level.
My point is that whatever radiation is being swapped between those two bodies you cannot go higher than the temperature predicted by the S-B equation because S-B defines the thermal outcome from the fastest and most complete conversion of incoming shortwave to outgoing longwave. All that will happen is that both bodies will come to match the S-B prediction PROVIDED that there is no additional energy source.
If you suggest that for ANY reason the radiation from either body drops below the S-B prediction whilst external irradiation continues at the same rate then you have a potentially catastrophic positive feedback loop.
That is why I have proposed two separate energy loops, one radiative matching energy in from space with energy out to space and one- non radiative matching convective energy in ascending air with convective energy in descending air.
Both loops then remain in steady state as long as external irradiation continues and any internal radiative imbalances are neutralised by convective adjustments.
That is exactly what we observe.
TLM,
If the rate of energy dissipation is slower than the rate of energy absorption then the amount of energy at the surface starts to build up, in which case the temperature has to rise. A body at a higher temperature emits radiation at a greater rate than a cooler body. Therefore as the temperature rises the rate of heat loss increases until a new equilibrium is reached where the rate at which energy escapes from the earth’s surface is the same as the rate as it arrives from the Sun. That new equilibrium is at a higher temperature.
Exacfly! I don’t know why this is so hard to understand.
However, it cannot just radiate directly into space – it has that darn pesky atmosphere in its way. That atmosphere, despite being colder than the surface, slows down the rate at which the constantly arriving energy absorbed by the earth’s surface can be dissipated back out to space.
This is another instance of being basically correct but with framing problems — objects radiate exactly the same way no matter where they are. The atmosphere isn’t “in the way” so much as it is radiating back at the warmer objects (even though it’s colder than they are!), especially when there are big lumps of water vapor in it.
I propose a Skeptic mantra: It’s the water vapor, stupid.
Paul Bahlin
November 29, 2017 at 8:45 am
So night and day is actually the sun blinking on and off then? Who knew!
This hardly deserves a response.
I am sitting here in the UK at midnight and the Sun has been effectively “turned off” for 8 hours now and the temperature is around Zero C, we have another 8 hours of cooling before the Sun effectively turns back on again tomorrow.
The “Constant Average Solar Energy at TOA” is not real it is a construct, the Sun will be varying from Zero Energy as now to around 1200W/m2 tomorrow at around midday at the equator and quite a bit less at this Latitude.
So the Sky and CO2 above me is not supplying energy to my surface with a constant solar energy supply, it is currently zero on my surface, how about yours?
Talldave2 says: “The vacuum of space is not an object”
Well, you could always look at the BIG picture and say the atmosphere is hiding radiation from the ubiquitous glowing gas from when the universe was ~ 380,000 years old. 🙂
Yeah, here in the actual universe, those ancient, red-shifted photons are still getting here from our lightcone-horizon, due to inflation and expansion. But it too warms whatever it hit 🙂
Willis E says: Further Reading: My post entitled “The Steel Greenhouse” looks at how the poorly-named “greenhouse effect” work, based on the principles discussed above.
Math Notes: There’s an excellent online calculator for net energy flow between two radiating bodies here. It also has the general equation used by the calculator,
The attached link should demonstrate why your Steel Greenhouse is wrong and leads to why this post is wrong. You neglect all view factors except F1-1 and thus make bad assumptions about the what the temperature of the inner shell will be and its effect on the original surface.
https://www.scribd.com/document/15665685/Concentric-Spheres-with-Radiation
mkelly:
You cite a very different problem. In your linked problem, the inner sphere is maintained at 1000K, regardless of what is happening around it.The power to maintain the sphere at this temperature could vary depending on conditions.
Willis’ steel greenhouse problem has a fixed power input to the sphere, with the temperature varying depending on the conditions around it.
Thermodynamics professors love to throw variations like these at students to catch the unwary, in the hopes of making sure those students who eventually pass the course can spot these distinctions.
Ed Bo
November 28, 2017 at 1:01 pm: So, how do you explain the cessation of statistically-significant warming, Ed?
“Willis’ steel greenhouse problem has a fixed power input to the sphere, with the temperature varying depending on the conditions around it.”
The earth does not have a fixed power input to the sphere. Clouds ozone, dust, aerosols, and CO2 itself changes the input power.
I’m still looking for a steel greenhouse.
Heck , not even any glass up there. !!
It is FANTASY !!
Ed,
That is total horse manure. This is the same problem. You just don’t like the outcome
SGW:
You don’t understand the difference between a problem where the power input is held constant and one where the temperature is held constant.
Seriously???
Now I understand why so many universities use thermodynamics as a “washout” course to eliminate people without the technical chops from rigorous courses of study. There are those who just cannot get the basic concepts.
Brett:
There is a huge difference between the existence of the atmospheric greenhouse effect — you cannot bring the surface power balance within 200 W/m2 without it, and possible subtle changes to that effect.
No one thinks the earth’s surface is out of balance more than about 1 W/m2 averaged over all the surface and a year. The interesting arguments are whether it is closer to 1.0 W/m2 (alarmists) or 0.0 W/m2 (skeptics).
There are many factors other than increasing CO2 levels that create effects of that magnitude. There are no other factors other than downwelling longwave infrared (“back”) radiation that can close the 200+ W/m2 “power gap” at the surface.
Ed,
You must have washed out in your thermo course, since you cannot grasp the concept of the First and Second Laws of thermodynamics. The shell in Willis’s thought experiment cannot transfer heat to the sphere (Second Law), so it will not rise in temperature. If the sphere did rise in temperature, it would violate the First Law as well since it would be at a higher temperature and energy state (violation of the First Law by creating energy. The sphere being the ONLY source of energy)
The whole GHE scenario is just an example of super-bad physics. The sun is the ONLY source of energy available to heat the earth. The earth/atmosphere system is purely passive. Using Trenberth energy balance diagram data, the GHE takes 341 W/m2 of the solar radiation at the TOA, and magically multiplies it to 494 W/m2 at the earth’s surface. The GHE creates energy out of nothing..
SkepticGoneWild November 28, 2017 at 11:55 pm
When a man starts throwing mud as you are doing, it is a sure sign that he has no ammunition. Ed is correct, as any thermo text will explain in great detail.
ENERGY is not HEAT!!! Nobody said the shell was “transferring heat to the sphere”. It transfers ENERGY to the sphere, which leaves the sphere warmer than it would be without that energy.
Heat cannot flow from cold to hot … but radiant energy certainly can. If it could not, then our eyes could only see objects warmer than body temperature …
w.
“When a man starts throwing mud as you are doing, it is a sure sign that he has no ammunition“
Ah, so Ed must have been out of ammo here:
“Thermodynamics professors love to throw variations like these at students to catch the unwary, in the hopes of making sure those students who eventually pass the course can spot these distinctions.“
and especially here:
“Now I understand why so many universities use thermodynamics as a “washout” course to eliminate people without the technical chops from rigorous courses of study. There are those who just cannot get the basic concepts.“
Ed,
Well said!
Why is that so many unschooled people in these endless debates do not understand that the whole scenario under discussion hinges around a ‘steady state’ model involving, in particular, two important fixed values:
(1) constant flow of energy input from the Sun
(2) a consequent steady-state global mean surface temperature
In that model, the Sun’s input is fixed, and (consequently) so is the earth’s surface temperature. Such a system necessarily involves a constant and equal energy flow in from the Sun and out to space. So there is no warming! And no cooling!
Only if the atmospheric composition and/or surface albedo is changed does the surface temperature begin to transition to a new fixed value. During this process of transition, the system is not in steady-state and it is legitimate to describe the surface as ‘warming’ or ‘cooling’.
Perhaps Anthony should consider asking the moderators to reject the weasel words ‘warming’ and ‘cooling’ when a contributor is quite clearly addressing a steady-state condition!
SGW:
Clausius, “father of the 2nd Law”, anticipated confusions like yours, and specificially dealt with this issue. He even had a specific term for what you say cannot exist because it would be in violation of the 2nd Law. He called it the “ascending transmission of heat” (ascending in the temperature sense).
He said this transmission could occur as long as there was “compensation”, and that the ascending and descending transmissions “compensate each other”.
These days, also to try to prevent confusions like yours, we generally only refer to the net transmission as heat, as Willis does here, with the gross transmissions referred to as some form of energy transfer.
I’ve looked at dozens of college textbooks, both physics and engineering, that discuss radiative heat transfer. Every single one discusses it in terms of radiative exchange of energy. And by the way, they all discuss the two bodies simply having temperatures of T1 and T2, or Ta and Tb. None talks about “Thot” and “Tcold”.
The discussion, figures, and equations in these texts show T1 radiating power to T2 as a function of its own temperature and emissivity, and T2 radiating power to T1 as a functiona of its own temperature and emissivity. This is true whether T1 is greater than T2, they are equal, or T2 is greater than T1.
Scientists and engineers have been educated this way for over a century (nothing to do with climate), and miraculously somehow, their thermal designs have worked.
Tony:
I have taught advanced technical subjects at the university level, and I am completely serious when I say that if I had a student that could not understand a basic issue like the difference between an object with a constant power input and an object maintained at a constant temperature, I would be telling that student that he does not belong in a technical field.
SkepticGoneWild November 28, 2017 at 11:55 pm
Ed,
You must have washed out in your thermo course, since you cannot grasp the concept of the First and Second Laws of thermodynamics. The shell in Willis’s thought experiment cannot transfer heat to the sphere (Second Law), so it will not rise in temperature. If the sphere did rise in temperature, it would violate the First Law as well since it would be at a higher temperature and energy state (violation of the First Law by creating energy. The sphere being the ONLY source of energy)
It doesn’t create energy it recycles it.
The whole GHE scenario is just an example of super-bad physics. The sun is the ONLY source of energy available to heat the earth. The earth/atmosphere system is purely passive. Using Trenberth energy balance diagram data, the GHE takes 341 W/m2 of the solar radiation at the TOA, and magically multiplies it to 494 W/m2 at the earth’s surface. The GHE creates energy out of nothing..
I guess you didn’t take a heat transfer class?
Radiative heat transfer from an object: P=k(T^4-Tc^4) where T is the temperature of the object and Tc is the temperature of the surroundings.
In Steady State:
For an object with a transparent atmosphere illuminated by a star like our sun we get:
P=k(T1^4-3^4)
For an atmosphere like ours which illuminates the planet with ~300 W/m^2, T~270K we get:
P=k(T2^4-270^4)
So for the same solar irradiance T1^4-3^4=T2^4-270^4
Therefore T2^4=T1^4+270^4-3^4
So T2 is hotter than T1, if T1 was 300 then T2 would be about 340K
No violation of thermodynamic laws, energy in equals energy out.
LMAO. More hand waving from Ed.
Phil,
I guess you’ve never heard of the First Law of Thermodynamics. Energy cannot be created. Please calculate the energy output of the sphere before the shell was placed, and then after. Then think about the First Law.
skepticgonewild November 30, 2017 at 1:02 pm
Phil,
I guess you’ve never heard of the First Law of Thermodynamics. Energy cannot be created. Please calculate the energy output of the sphere before the shell was placed, and then after. Then think about the First Law.
No, explicitly same solar input and same output only the recycle back to the surface has changed.
mkelly November 28, 2017 at 11:19 am Edit
mkelly, you have missed the essential difference between your cited analysis and mine. In your analysis, the inner sphere is maintained at a certain constant temperature.
In mine, it is NOT maintained at a certain temperature. Instead, it has a certain constant energy input. This is a totally different situation, with (predictably) a totally different steady-state. As a result, your claim that
is simply not true. It is looking at a completely different thought experiment than mine.
Regards,
w.
Willis,
I am going to remove the insulation from my water heater and place multiple steel shells around it. How many shells do I need until the water in the water heater starts to boil?
Ed,
You’re so smart. How many steel shells do I need?
I have a 50 gallon cylindrical steel walled water heater maintained at 50 degrees C, How many steel shells until the water starts to boil? Come on daveburton. Help me out.
SkepticGoneWild,
False question…
You can’t reflect more than 100% of all emitted radiation from the water heater, thus at maximum you can maintain the water temperature, not heat it up. Alle what you can is reducing the heat loss, no matter if that is by insulation or reflection. The latter is used in space with multiple layers of metal sheets. You see, it works that way…
OMG Willis. I am not flinging anything but the laws of thermodynamics. Your sphere increased in temperature. That’s what happens when you transfer heat to it. You are just calling it “energy transfer”. But the result is the same. A passive object heated by the sphere, yet colder than the sphere, magically increases the energy and temperature of the sphere. And then you conveniently stop at one iteration. For the sphere at the new temperature will warm the shell more.
This type of pseudoscience begets all types of nonsense, such as the idea that the earth actually heats the sun a minute amount. Half the people at Spencer’s blog subscribe to that fiction.
Ferdinand. No. This is the same example as the Willis Steel Greenhouse. Not a false question. The temperature is not fixed. There is nothing to prevent it from heating up.
SkepticGoneWild November 29, 2017 at 12:16 am
Skeptic, you seem to have missed the part about the vacuum between the planet and the shell … if you put a steel shell around your water heater and remove the air from between the two, you’ve made a very effective thermos bottle. If you think that won’t affect your water heater, well, I fear you are beyond my poor power to add or detract.
However, the water in your tank is thermostatically controlled so that it never boils … as a result, all that putting your water heater in a thermos bottle will do is to reduce the amount of gas needed to heat it.
So the answer is, as long as it is thermostatically controlled it will NEVER boil … but then you knew that.
Or maybe you didn’t know that. But in any case, if the thermostat fails, you see that funny little valve up there at the top of the heater? That is a pressure relief valve. Why is it needed? Because if the thermostat fails, the water will boil with exactly zero steel shells around it …
Best regards, pay attention to the law of holes,
w.
“ENERGY is not HEAT!!! Nobody said the shell was “transferring heat to the sphere”. It transfers ENERGY to the sphere, which leaves the sphere warmer than it would be without that energy.”
I’m glad that you’re starting to recognise that energy is not heat. That being the case, you should now be able to see that in saying a transfer of energy from the shell leaving the sphere warmer than it would be without the energy, is exactly the same thing as saying that the shell has transferred heat to the sphere. Different semantics won’t change the meaning of what you’re saying.
Your next problem is that heat doesn’t transfer from the shell to the sphere, it only transfers in the reverse direction, up until equilibrium, when the transfer stops.
Let’s write that sentence better:
That being the case, you should now be able to see that saying a transfer of energy from the shell leaves the sphere warmer than it would be without the energy, is exactly the same thing as saying the shell has transferred heat to the sphere.
SkepticGoneWild November 29, 2017 at 1:26 am
Ferdinand. No. This is the same example as the Willis Steel Greenhouse. Not a false question. The temperature is not fixed. There is nothing to prevent it from heating up.
There certainly is, it’s called a thermostat! It is not the same as Williis’s Steel Greenhouse which had a constant energy input your example had a variable energy input and is maintained at constant temperature.
If it had constant energy input then improving the insulation would increase the temperature reached, usually what would be done then would be to reduce the energy input to maintain the desired temperature, i.e. improve the efficiency.
Willis,
Heat is defined as energy transferred due to temperature differences Your steel greenhouse is transferring energy from the shell to the sphere and raising its temp. That is heat transfer, but from a cold object to a warm one. This is where your BA in psychology is not helping.
SkepticGoneWild November 29, 2017 at 5:54 am
Willis,
Heat is defined as energy transferred due to temperature differences Your steel greenhouse is transferring energy from the shell to the sphere and raising its temp. That is heat transfer, but from a cold object to a warm one. This is where your BA in psychology is not helping.
On the contrary it’s really good engineering, see for example radiational heat transfer text by Hottel, so it would appear that the BA in psychology is no impediment, whatever your degree is in appears to be causing problems however.
SGW starts off OK with “Your steel greenhouse is transferring energy from the shell to the sphere and raising its temp. ”
But then errs with “That is heat transfer, but from a cold object to a warm one.”
No, the HEAT transfer is still from warm to cold — from the sphere to the surroundings. Its just a smaller heat transfer than before.
* If the sphere radiates to the extremely cold regions of outer space (3K), the energy out from the sphere to space(call it E(1,2) is large and the energy back from space to the sphere (E(2,1)is basically zero. The net exchange (which is called “heat”) is Q(1,2) = E(1,2) – E(2,1) is large.
* If the sphere radiates to the mildly cold shell, the energy out from the sphere to the shell (call it E(1,3) is still as large as E(1,2). However, the energy back from shell to the sphere, E(3,1) is now much larger than E(2,1). The net exchange is now Q(1,3) = E(1,3) – E(3,1) is much smaller than Q(1,2).
If the sphere had been in a steadystate condition when radiating to space, it will no longer be in steadystate if the shell is added. It will be losing less but still gains the same, meaning it must warm.
SGW:
You need to acquaint yourself with the “multi-layer insulation” commonly used to surround industrial furnaces. It has layers of reflective metal interspersed with non-conductive layers. This allows the radiated energy from the hot furnace to be reflected back, while minimizing the conductive transfer out. This permits the furnace to achieve much higher temperatures than could otherwise be realized.
According to your analysis, this is all nonsense. But people keep building furnaces this way, and it actually works!
Tim,
Replace the sphere with the actual sun. Take away the shell and substitute it with planet earth, What your fantasy physics tells us is that that reflected EM from the earth travels to the sun and heats it up a minute amount:
“If the sphere[SUN] had been in a steadystate condition when radiating to space, it will no longer be in steadystate if the shell [EARTH] is added. It will be losing less but still gains the same, meaning it must warm.”
You can find that silliness with a boatload of people over at Spencer’s site as well. I don’t think you will find such nonsense in your standard physics textbook.
The steel greenhouse is pure fantasy. A thermodynamic nightmare of a thought experiment. It is up to Willis to confirm that it might actually work with a real experiment, conducted by people who have actually studied physics, and published in a reputable science journal. That is how real science happens.
Ed,
By all means please provide a link that provides a thermodynamic analysis of these furnaces.
SGW:
Consult an engineering heat transfer textbook. I suspect you’ve never read one.
Ed,
I am still waiting for the analysis of your fantasy furnace.
When stumped, Ed punts.
SGW:
One minute (4:48pm to 4:49pm) and I’m stumped. Sheesh!
I am completely serious when I say this is the stuff of textbooks, and introductory ones at that. But if you can’t understand the difference between an object with a constant power input and one maintained at a constant temperature, it is completely obvious that you have never even cracked such a textbook.
This no more belongs in a scientific paper than does the argument that if an apple breaks free of a tree, it will fall to the earth.
The argument is very simple for anyone who understands the very basics of thermodynamics and heat transfer.
If you apply a constant thermal power to an object (such as a furnace), it will reach a steady-state temperature when it outputs as much power to ambient as it receives from the source (1st Law). Of course, the higher the object’s temperature the more power it will output to ambient.
Anything that you do to increase the thermal resistance between the object and ambient will lessen the power it transfers to ambient for a given object temperature. This means the object must increase in temperature to restore the power balance, outputting as much power to ambient as it receives from the source.
One of the modes of heat transfer is radiative (in a vacuum it is the only one). If you can inhibit radiative transfer from the object to ambient, either by reflecting some of it back, or by absorbing some of it and re-radiating from a temperature in between that of the object and ambient, increases the radiative thermal resistance, reducing the power transfer from object to ambient. This means that the object is receiving more power than it outputs, so it increases in temperature until the power balance is restored.
This is such a basic concept that I would expect students to understand it completely within a couple of weeks of starting an introductory course.
Don’t your arms get tired with all the hand waving?
SGW:
Qualitative analysis is all it takes to reject your silly argument that adding radiative insulation cannot result in a higher temperature of a separately powered object. If you had any experience whatsoever in dealing with these kinds of problems — which obviously you don’t — you would understand that you always do this type of analysis before getting to the equations, finite element modeling, etc.
One of the root-level errors I see over and over again on “Slayer” websites and comments is a rush to plug numbers into equations before doing this kind of qualitative analysis. As a result, they use the equations in completely wrong ways, leave out important effects, etc. And they have no clue where they went wrong.
In this case, I don’t even need to get to the equations to demonstrate the wrongness of your arguments.
If you want to start with some quantitative analysis, use Willis’ “steel greenhouse” example. I had many problems like this when I was studying thermo and heat transfer. I encounter many systems like this in thermal systems in my professional work. Willis’ analysis is correct, and until you understand his very simple system, you will never even begin to start to understand the issues at play.
Skeptic,
“Heat is defined as energy transferred due to temperature differences Your steel greenhouse is transferring energy from the shell to the sphere and raising its temp. That is heat transfer, but from a cold object to a warm one. This is where your BA in psychology is not helping.”
Your whole premise is wrong. No heat is being transferred from the shell to the sphere. The internal heat source of the sphere is supplying heat to it. That internal heat is what causes the sphere to rise in temperature when its outflow of heat is reduced by shell. No 2LOT violations.
Skeptic,
“Heat is defined as energy transferred due to temperature differences Your steel greenhouse is transferring energy from the shell to the sphere and raising its temp. That is heat transfer, but from a cold object to a warm one.”
Not at all. No heat is being transferred from the shell to the sphere. The internal heat source of the sphere is supplying heat to it. That internal heat is what causes the sphere to rise in temperature when its outflow of heat is reduced by shell. No 2LOT violations are involved.
Nate,
You are ignoring First Law violations as well. The sphere has doubled in energy output. Where did that extra energy come from since the sphere is the only source? The First Law says NO ENERGY CREATION. You have energy output greater than energy input. That does not happen in the real world. Only in a fantasy world of made-up physics.
And then you stop the iteration at one cycle. Why? The new higher temperature of the sphere has to warm the shell more.
Ed,
You are seriously confused.
See the following publication:
https://tinyurl.com/ybmbxwxr
Notice that it states:
“An alternative method is to use radiation shields between the heat exchange surfaces (Holman, 2009). These shields do not deliver or remove any heat from the overall system; they only place another resistance in the heat-flow path so that the overall heat transfer is retarded”
Skeptic,
“You are ignoring First Law violations as well. The sphere has doubled in energy output. Where did that extra energy come from since the sphere is the only source? The First Law says NO ENERGY CREATION.”
You have to very careful to keep track of all energy flows before claiming a 1LOT violation. First of all, there IS a source of energy-there is input to the sphere from the source. The source CAN store extra energy in the sphere if less has gone out for a time. (e.g. my salary was the same but my bills were lower this month, so I put $$ into savings)
“The sphere has doubled in energy output.” The shell has warmed up. Therefore the net flow of heat, sphere to shell, which is proportional to (Tsp^4-Tsh^4) does not double. In fact it reaches its original value in steady state.
Nate,
So the First Law is not important to you, Just your adherence to the GHE fiction. I get it. However, the First Law violation is staring you right in the face.
Furthermore the idea the that the sphere radiating at 235 W/m2 causes the shell to radiate at 470 W/m2 is absurd as well.
You never did answer the question regarding why the process was stopped at one iteration or heating cycle?
.
Let me edit the second paragraph above which had an extra “the”:
“Furthermore, the idea that the sphere radiating at 235 W/m2 causes the shell to radiate at 470 W/m2 is absurd as well.
SGW:
You simply continue to display the fact that you don’t understand the most basic concepts of thermodynamics and heat transfer.
Look at the 2nd sentence of the paper you cite: “In this study, a simplifying approach for calculating the radiant energy is achieved using the concept of net radiation heat transfer and provides an easy way for solving a variety of situations.”
Note the word “NET” in “net radiation heat transfer”. This is the whole point of Willis’ post — distinguishing between gross and net transfers — but you can’t under stand it. You can look at the transfers either way, AS LONG AS YOU ARE CONSISTENT (which you are not).
Let’s look at Willis’ steel greenhouse example in terms of the paper you cite approvingly.
The sphere, which has an input power of (235 * Area) watts, has a “net radiation heat transfer” of (235 * Area) watts to the shell. So it is in energy balance by the First Law, therefore in steady state conditions.
The shell, which receives a “net radiation heat transfer” of (235 * Area) watts from the sphere, has a “net radiation heat transfer” of (235 * Area) watts to space. So it also is in energy balance by the First Law, therefore in steady state conditions.
Note particularly that the shell is a “shield [that does] not deliver or remove any heat from the overall system; [it] only place[s] another resistance in the heat-flow path so that the overall heat transfer is retarded.”
SGW:
You say: “Furthermore, the idea that the sphere radiating at 235 W/m2 causes the shell to radiate at 470 W/m2 is absurd as well.”
False premise! In the steady state condition, the sphere is radiating at 470 W/m2 (gross). THIS causes the shell to radiate at 470 W/m2 (235 from the inside surface, 235 from the outside surface).
The 1st Law is satisfied for the sphere, as it is receiving 235 from its power source, it is outputting 470 (gross) towards the shell, and receiving 235 (gross) from the shell. If you prefer, you could say the sphere is receiving 235 from its power source, and outputting 235 (net) towards the shell.
The 1st Law is satisfied for the shell, as it is receiving 470 (gross) from the sphere, outputting 235 (gross) from its inner surface, and outputting 235 (gross) from its outer surface. Or if you prefer, it is receiving 235 (net) from the sphere, and output 235 (net) towards space.
Ed,
Your poor, poor students. How is the First Law satisfied when the sphere’s 235 W/m2 becomes 470 out of thin air? That is energy creation. The shell has NO energy of its own. OMG! Your explanation is pure theatrics, handwaving, just like your magical furnace.
I am DONE with your nonsensical fantasy physics. It’s time Willis (and you) put up or shut up. Willis came up with this monstrosity of a thought experiment, now it’s time to perform some REAL science and follow it up with a real experiment per the scientific method. Of course we know that will never happen.
Don’t respond with more theatrical handwaving.
SGW:
This very simple but definitive experimental demonstration of the effect you consider impossible was put up on WUWT years ago:
https://wattsupwiththat.com/2013/05/28/slaying-the-slayers-with-watts-part-2/
SkepticGoneWild on November 29, 2017 at 12:55 am asked, “I have a 50 gallon cylindrical steel walled water heater maintained at 50 degrees C, How many steel shells until the water starts to boil? Come on daveburton. Help me out.”
It will never boil. Water heaters have thermostats to prevent them from boiling, regardless of how much insulation you wrap around them. (Steel shells are not the best insulators, BTW.) But if you insulate them better the energy they consume to maintain water temperature goes down.
However, I posed and answered a similar question here.
Ed,
More hand waving I see.
“Water heaters have thermostats to prevent them from boiling,”
OMG Dave. A thermostat in a water heater prevents the water from cooling when set at a certain temperature. Not the opposite.
“A thermostat in a water heater prevents the water from cooling when set at a certain temperature. Not the opposite.”
Breathtakingly ignorant, SGW. It does both.
Daveburton writes,
“It will never boil. Water heaters have thermostats to prevent them from boiling, regardless of how much insulation you wrap around them. (Steel shells are not the best insulators, BTW.) But if you insulate them better the energy they consume to maintain water temperature goes down.”
Thermostats doesn’t prevent boiling since it is never set to boil the water in the first place,settings are commonly around 110-115 F.
The concern is the Pressure build up in the tank.
You need to think a little more before you start banging on those ivories.
SkepticGoneWild December 2, 2017 at 6:06 pm
Remember my rule of thumb, that whoever uses OMG in a scientific conversation is wrong? Here is another example.
Skeptic, a thermostat in a water heater does two things. When the water gets too cool it turns on the gas. And when the water gets to the desired temperature, it turns off the gas.
Now, THINK ABOUT WHAT KEEPS IT FROM BOILING. Or as Paul Bahlin said in a more genteel manner:
You’re not doing your reputation any good with these kinds of OMG!!! incorrect but passionate assertions.
w.
SGW:
So direct experimental evidence is “handwaving” to you.
We’ve already seen that standard thermodynamic analysis that any scientist or engineer learns as an undergraduate is “handwaving” to you.
It’s now obvious that anything beyond your incredibly limited capabilities you write off as “handwaving” because you cannot understand it. Truly pathetic.
OMG Willis. I was correcting Dave’s false statement:
“Water heaters have thermostats to prevent them from boiling”
If you cannot understand the error of this statement, then you should not be posting on a science blog. Sunsettommy also corrected Dave’s error as well.
My reputation? Worry about your own.
Here’s my suggestion. Go take a general physics course at your local university. Not the dumbed-down version for climatologists. The rigorous courses for physics and engineering majors usually have a prerequisite of a year of calculus, and a maybe a semester of differential equations, so you have your work cut out for you. Then maybe people will take you seriously.
Like I state earlier, it is your responsibility to attempt to confirm this bizarre thought experiment with a real experiment. But it’s been like 8 years since the original post, so we know that’s not going to happen.
I went back and read some of the original steel greenhouse post. Some of the comments were brutal:
“I’m sorry but I thought the article was tremendous nonsense, and will cause immense damage to the reputation of this website”
“This article should be removed. The author obviously knows nothing about radiative heat transfer”
“This article is utter nonsense, and a simple logical analysis reveals this without even resorting to physics (and I am a physicist).As requested by others above… please remove this article”
“Thank you Anton. It wasn’t just me who spotted that the logic is simply busted. Willis, next time contact a physicist before doing something like this.”
“I agree withn John A. (20:11:22) this load of [snip] has no place on Wuwt. Wuwt’s reputation has now been tarnished. This is not science but fantasy. The thing is so sorely messed up that it doesn’t even qualify for print Has Wuwt fallen prey to a belief in pseudoscience? I am truly disappointed!”
“Really sad to see this kind of rubbish published on this great site. The physics is completely wrong. Anthony, please do not let this kind stuff published on this site, it will take the value of the whole site down.”
I could go on, but you get the point.
“OMG Willis. I was correcting Dave’s false statement: “Water heaters have thermostats to prevent them from boiling””
Dave’s statement was not false, SGW; it just pointed out one of the modes of a thermostat. It did not assert that there was no other mode.
In thinking about your various comments and rebuttals, I’ve come to the conclusion you like to dominate others through pettifogging and personal attacks. It’s just an opinion but, like a$$holes, everyone has one.
Wait a minute Ed. I’m going to test my 1800 W hairdryer, and direct it into my concave bathroom mirror. See if I can get 3600 watts out of it. If I’m not back in a few minutes, call 911.
Without changing the subject, and without any need for models or esoteric physics nobody is talking about, answer this simple question…….
Here’s the shot:
Your claim is that IR from a 90 k source incident on a 91 k target, does nothing to the target. If true it has to then either pass through it (albedo of 0) or be reflected from it (albedo of 1). You further claim if that very same source hits the very same target but now the target has a temperature of 89 k it is absorbed, at least partially.
Here’s the chaser:
If you would grant me the liberty of putting that in an equation, it would be albedo= f(T). Do you agree that my equation correctly supports your claim?
SkepticGoneWild December 3, 2017 at 1:00 am
OK, I’ll bite … why doesn’t the water in a water heater boil? I mean, it’s got a gas flame under it … so what keeps it from boiling? Good looks? Mystical incantations? What keeps it from boiling?
Your move, SGW …
w.
Ed,
OK. I am back. The experiment went well. I only have second degree burns, and I tripped the circuit breaker. Upon further inspection, the power is off in our neighborhood as well, and I hear sirens outside my door. So yes. I was wrong. Power output CAN exceed input.
OMG Willis!!!!
Seriously, though. The question was not about why water heaters don’t boil. The issue is with how a thermostat operates. Say for example, your steel sphere was a water heater. A thermostat does not prevent the water from heating up. A thermostat only fixes the lower temperature limit. Any external energy source (like a shell) can cause the temperature to rise (per your theory). The thermostat will not cool it down. Trivial point, really.
So why not go back to school? Take a physics class. You love the stuff, so what’s stopping you?
I love this sight because every time I come here I learn something I was formerly ignorant of.
All my life I have had the good fortune to recognize the old saying; Ignorance can be fixed but stupid is forever. I also hate to speak I’ll of anyone.
In your case I must break the latter and invoke the former of these two credos.
You are stupid.
SGW, doubling down [“A thermostat does not prevent the water from heating up. A thermostat only fixes the lower temperature limit.”] is not helping with your image.
Again, a thermostat works both ways. I trust my tender body every day to the operation of my water heater thermostat, both ways.
Although, you may prefer one that allows your water heater to get as hot as its energy supply allows. Who knows or cares.
I couldn’t have asked for a better placement for that comment from Paul. Thanks, Paul.
Welcome.sorry to break your thread
Oh my goodness. Well then, SkepticGoneWild & Sunsettommy, I confess to imprecise language. I wrote:
“Water heaters have thermostats to prevent them from boiling…”
I should have said:
“Water heaters have thermostats which prevent them from boiling…”
Or:
“Water heaters have thermostats to prevent them from getting too hot or cold…”
Better?
And Sunsettommy, water heaters are typically set to 120-150 °F, not 110-115 °F. 110-115 °F is tepid. Baby bottles are commonly heated nearly that warm. When your coffee gets that cool you probably put it in the microwave oven for 20 seconds or so, to reheat it.
To prevent the growth of dangerous bacteria in your water heater, the thermostat should be set to at least 140 °F. But you need to be careful because 140°F water can scald a person in about 5-6 seconds:
http://www.accuratebuilding.com/images/services/charts/hot_water_burn_scalding_lrg.gif
SGW:
You challenged me to cite experimental evidence for my views. I did exactly that, linking to a set of experiments posted on WUWT several years back that clearly show with quantitative evidence that my views are correct.
https://wattsupwiththat.com/2013/05/28/slaying-the-slayers-with-watts-part-2/
You simply responded by saying: “Ed, More hand waving I see.”
And you have completely refused to engage with any of the results of those experiments, desperately trying to change the subject. I wonder why?
I guess I shouldn’t expect much from a guy who can’t understand the difference between a problem with an object held at constant temperature one with an object provided with a constant power input.
Thanks, Ed. You’ve made up for Paul’s comment moving upthread. Now we’ve got perfect placement again!
Skeptic,
Put my answer in wrong thread. https://wattsupwiththat.com/2017/11/24/can-a-cold-object-warm-a-hot-object/comment-page-1/#comment-2683050
Skeptic,
Sh*t, I mean here: https://wattsupwiththat.com/2017/11/24/can-a-cold-object-warm-a-hot-object/comment-page-1/#comment-2684013
Earlier in the discussion, I suggested a shell be placed around a water heater in similar fashion to the steel greenhouse thought experiment. After all, even in the absence of a vacuum, the wall of the water heater emits IR which will warm the shell, which in turn will emit IR back to the water heater just like the steel greenhouse. But Willis said:
“However, the water in your tank is thermostatically controlled so that it never boils”
The premise was that the shell could heat the water in the water heater up. Willis and the rest of you apparently don’t understand how a thermostat operates. Just because a thermostat turns off the flame when the water heater reaches a certain temperature does not mean the water does not have the ability to heat up more (from the shell in my example). If you set the thermostat at 140 F, it will not go any lower, but there is nothing to prevent it from going higher,
What the hell? Do you think the thermostat provides cooling? NO.
So let’s give the McFly award to the following people:
Dave Burton
Dave Fair (you get the dumb***, dumb****, dumb****,and **** for brains award as well)
Phil
Paul
Willis is exempt since he may not have understood the premise.
SkepticGoneWild December 3, 2017 at 8:46 pm Edit
Skeptic, do you have to work at being that unpleasant, or does it come naturally to you?
In any case, to the example. I put a shell around a water heater. As the water warms, so does the shell. It radiates some of the heat back to the water heater. The water keeps heating.
At some point the water reaches the thermostat’s set point, call it 140°F, and the gas turns off.
What has happened so far is exactly the same as if the shell were replaced by insulation. We’re slowing heat loss from the tank. We’ve taken the water to the same temperature, but it has required less gas to do so.
But contrary to your claim, when the thermostat shuts off, the entire system, water, water heater, and shell begins to cool.
Of course, the system will cool more slowly because of the shell. Again, it is slowing heat loss from the system by intercepting the radiation and returning half of it to the heater.
So while you are correct that “If you set the thermostat at 140 F, it will not go any lower, but there is nothing to prevent it from going higher”, there is also nothing making it go any higher. The shell won’t, it just slows the rate of heat loss. The gas is off. So it is unclear just what you are fantasizing will make the water become hotter once the gas is turned off.
Note that this is different from the steel greenhouse. There, unlike in the water heater, there is no thermostat so the energy supplied by the internal radioactive decay continues unabated … which is why it can get hotter than it would with no shell. But of course, that doesn’t happen when the gas shuts off in the water heater. It just cools from that point onward, there is no energy being added to the system AS THERE IS IN THE STEEL GREENHOUSE.
Do you realize how ugly, immature, and unpleasant it makes you look when you pontificate, call a bunch of people stupid and shit for brains … and then it turns out you are wrong, wrong, wrong?
I guess you must not, or you wouldn’t do it. My advice is to emulate the rooster, and wait until it is actually dawn before you start crowing …
w.
Willis,
So you are allowed to break the laws of thermodynamics, but I am not. I need a rule book to keep things straight.
Your steel greenhouse sphere magically doubles in energy output when there is no other source of energy in your system. That is a violation of the First Law. You cannot create energy out of nothing.
You are the one transferring heat from the cold shell to the warmer sphere, so don’t give me some highbrow lecture, when you routinely break the Second Law as well..
And furthermore, it was Mr. Dave Fair who called me an asshole, but you never got on his case.
Willis,
And I don’t need a science lecture from someone with a BA in Psychology and a [pruned] certificate.
You will just be moaning on and on in another 8 years about the steel greenhouse, since your thought experiment will continue to be just an unconfirmed thought experiment.
SkepticGoneWild wrote, “…wall of the water heater emits IR which will warm the shell, which in turn will emit IR back to the water heater… the shell could heat the water in the water heater up…. Just because a thermostat turns off the flame when the water heater reaches a certain temperature does not mean the water does not have the ability to heat up more (from the shell in my example). If you set the thermostat at 140 F, it will not go any lower, but there is nothing to prevent it from going higher…” and then, ironically, “you are allowed to break the laws of thermodynamics, but I am not.”
SGW, is your name Paul W. McDonald, or perhaps Justin Reese Chrivia?
SkepticGoneWild wrote, “OMG Dave. A thermostat in a water heater prevents the water from cooling when set at a certain temperature. Not the opposite.”
And Willis replied, “You’re not doing your reputation any good with these kinds of OMG!!! incorrect but passionate assertions.”
He’s not doing his reputation any harm, either, Willis, because he hides behind an anonymous alias. My guess is that, were that not the case, he might be less careless with his words.
(In contrast, if I counted correctly, five of the seven people he’s arguing with, who he thinks are all dunces, use their own names. )
SGW: the rules are, at the point you clearly show they’re in error, they:
1) Change the subject, or
2) Put their fingers in their ears and stop responding, then further down-thread, carry on making the exact same mistakes as if nothing was ever said, or
3) Respond with a comment which completely (and sometimes quite creatively) misses your point, whilst being as condescending as possible, or
4) Respond with a comment which focuses on chastising you (usually for insults or something like that) whilst ignoring the identical (or worse) behaviour from themselves and those in agreement with them, or
5) Simply lie about what has been said so far in the discussion, even though the comments are there to be read by anyone.
Or, sometimes it’s a mixture. Being pathologically incapable of conceding even the slightest point when it comes to this subject, what else can you expect from them?
Daveburton, your response at December 4,12:36 am, was a mind-blowing example of number 3), creatively missing the point. You quote SGW explaining how (relating it to the boiler example) the steel greenhouse violates the laws of thermodynamics, and in your response indicate that you agree that this is a violation of the laws of thermodynamics. So, you agree with SGW, whilst acting like you disagree. Now this has been pointed out to you, I reckon there could be a number 2) on the horizon (never a pleasant thought).
There is only one energy source (and thus only one heat source), the sphere. There would be no reason for this to double in energy output on addition of the shell unless the shell were an additional source of heat (which it isn’t, it’s a passive object). Given that the steel greenhouse example DOES regard the shell as an additional source of heat, when relating it to the boiler example, this additional heat source will be able to continue to heat the warm water further, even when the gas is switched off by the thermostat. In reality, it wouldn’t, since the passive shell is not a heat source. It provides energy, but not heat.
Yup, the steel greenhouse is THAT easy to debunk. The only reason the debunking hasn’t been accepted yet is due to endless confusion over the definitions of heat vs energy, and due to the consistent application of the aforementioned rules 1-5 by teams of dedicated professional obfuscators.
Willis, on the other hand, argues that when the gas is off, the shell around the boiler no longer functions as an additional source of heat. It miraculously changes to being just a passive object. If it were a treated as a passive object whilst the sphere was providing heat (as it does continuously – no thermostat – in the steel greenhouse thought diversion) then the sphere would never undergo that first miraculous rise in temperature in the first place. It would (and should) carry on at the same temperature it was before the shell was added, and the shell will just warm until equilibrium (which will be at a lower temperature than the sphere, due to the larger surface area of the shell).
Skeptic,
“Furthermore, the idea that the sphere radiating at 235 W/m2 causes the shell to radiate at 470 W/m2 is absurd as well.”
Nobody is saying that. Where do you get the idea shell is radiating 470?
Again, if you desire sphere and shell to be at the same temp (why I dont know), it aint gonna work. That would a violation of 0th and 1st laws of thermo. You ok with that?
In order for temperature of sphere to not explode, it MUST be outputting all of the input heat (235) to the shell (1ST LAW my friend!).
In order for heat to pass from sphere to shell, there must be a temperature difference between sphere and shell. (0TH LAW my friend!). The amount of flow is sigma(Tsph^4-Tsh^4). No way around this.
“Nobody is saying that. Where do you get the idea shell is radiating 470?“
Ed Bo says that, during this discussion. Though if you asked him he’d probably say that he didn’t say that.
Nate:
The sphere must radiate 235 W/m2 NET outward to balance the internally generated 235 W/m2 for it to be in steady state conditions (1st Law conservation of energy).
Since the sphere is receiving 235 W/m2 GROSS from the inner surface of the shell (which must radiate 235 W/m2 outward to space — this is both gross and net, because there is nothing coming back — to keep the whole sphere+shell system in 1st Law balance), the sphere must radiate a total of 470 W/m2 GROSS to output (470 – 235) = 235 NET.
Remember that this whole post is about keeping straight the difference between GROSS and NET transfers. Willis understands it completely, but there are several others here who just cannot get it…
Tony:
Were you talking about the moon as your alternate imaginary earth or not? You have been very emphatic both ways about it — have you made up your mind yet?
You’ll get over it, Ed.
Ed Bo,
Agree. Folks keep telling me that you said the SHELL is radiating 470. I did not find you saying this anywhere. As usual people see what they want to see.
Ok 470 radiated from shell in both directions. Yes. That was what Ed said. My bad thinking outward only.
Still, as he makes clear, that is not NET flow. Net flow outward from shell = net flow out from sphere = net flow from source.
So there is no ‘energy creation’ or problem with 1LOT or 2LOT, as claimed by Tony, SGW.
It’ll be OK, Nate.
Ed, your comments were no more nor less offensive to SGW than what you received in return, yet Ol’ W only chastises those on one side of this argument. My criticism was of Ol’ W, not yourself. I couldn’t care less about insults, either received or given.
Tony November 29, 2017 at 10:40 am
A couple of comments.
First, when someone starts playing grade-school games with my name, I know that they know that they are losing the debate.
Second, I snipped what you said because it made you sound both vicious and vindictive. Maybe I’m an incurable optimist, but I don’t think you are either one. I figured you were just having a bad hair day.
However, if those nasty words actually reflect your inner self, I’m happy to let you rave on … you might note, though, that such personal attacks don’t do your reputation any good.
Sadly,
w.
Your double standards are obvious to anyone rational.
Tony, “Ed Bo says that”. Nope, not that I can see.
https://wattsupwiththat.com/2017/11/24/can-a-cold-object-warm-a-hot-object/comment-page-1/#comment-2683256
Phil,
The First Law states energy cannot be created or destroyed. The Willis steel greenhouse is creating energy. The sphere is the ONLY energy source. Calculate the energy leaving the sphere before the shell arrives. Now calculate the energy leaving the sphere after the shell is put in place. The energy output of the sphere has doubled! OMG the stupidity. You need to crack open a physics or thermo text and review the FIrst Law again.
Skeptic, measure the energy leaving the sphere and going to outer space before the shell arrives. For convenience, let’s say it is 240 W/m2, the amount of energy making it past the clouds in the earth system.
Now measure the energy going to outer space after the shell is put in place and steady-state is re-established … again, it is 240 watts. It has to be.
In both cases, 240 watts is entering the system, and 240 watts is leaving the system … OMG, as you say, no energy has been created or destroyed. No First Law violation.
Let me try another example.
Suppose you have a hot light bulb at steady state. Energy in = energy out. Now put an asbestos blanket around the light bulb. Will the light bulb get hotter? Sure. It will heat up until the same amount of energy is leaving the total system as before the blanket was put on. At that point energy in once again equals energy out. … but the light bulb getting hotter doesn’t mean that the First Law is violated as you incorrectly assert. It just means that the energy loss has slowed, so the bulb heats up until the steady-state is re-established.
w.
PS—another of my rules of thumb is that when someone says OMG in a scientific discussion, they’re most likely wrong … just saying …
Skeptic,
“So the First Law is not important to you, Just your adherence to the GHE fiction. I get it. However, the First Law violation is staring you right in the face.”
1LOT is important to me. Just dont see the violation that you see. You have not made it very clear where the violation is.
If you want to obey 0th and 1st law of thermodynamics, the sphere and shell cannot be at the same temp. The sphere MUST be warmer than shell in order for it to be able transfer heat to the shell (by ordinary radiative laws). It must output heat to shell at same rate as input (1LOT)
In turn, the shell must radiate at same rate as input to it (1LOT).
I have no idea what the ‘iteration’ discussion is about.
Borrow an IR camera and look at a partially cloudy night sky. You will see that the clouds are warmer than the sky. You know this because IR from the clouds is being detected by the camera sensor. Detection involves absorbing the IR, so the detector must be infinitesimally warmer due to the IR even though the cloud is -70 degrees and the detector is at room temperature..
So Stephen, you do grant that the atmosphere is a “partially radiative gas”? Yes?
If yes than do you grant that the atmosphere acts as a partially effective interceptor and reradiator of LW?
Yes but:
Since the entire atmosphere and the oceans have a similar effect via conduction and convection any effect from ghgs counts for nothing in comparison.
Ghgs simply change the lapse rate slope as I explained before so that they radiate to space more effectively from a lower warmer height instead of warming the surface.
Conduction and convection move lots of energy around, inside the earth-atmosphere system, yes. But they only effect the system’s radiation budget if they manage to move radiating gas to more favorable position for escape. Anything else is just internal churn (weather).
So without the handwaving internal porridge, show the math that makes radiation 1st law balance inside and outside the system: atmosphere, surface, and system balance.
I’m particularly interested in where to find the insignificant bits.
Radiation in from space 255k
Radiation out to space 255k
KE to PE in atmospheric uplift 33k
PE to KE in atmospheric descent 33k
Surface temperature 288k
As per observations
Odd that you would choose 255k as a radiation figure, but I assume you mean 240 w/m^2 where you use it. Same for the 33k?
If you’ll permit me to restate ….
You have 240 w/m^2 in and out. Good. The surface though, I don’t follow. You have a ke-pe exchange, equal and opposite in energy that is raising the temperature of the surface without increasing outbound radiation. How does that work?
Simply because the same unit of kinetic energy at the surface cannot both conduct away and radiate away simultaneously. That is why S-B should not be applied to a surface beneath a convecting gaseous atmosphere.
Of the 288k at the surface only 255k gets past the conductive barrier presented by the convecting mass of the atmosphere.
As regards the units I was assuming a reader would realise that the radiation intensities were ‘worth’ a 255k contribution to the 288k surface temperature.
So instead of quibbling over numbers and units let’s say we just stick to letters and energy flow (w/m^2). Also I’m just an engineer so I like to visualize things.
If I understand what you are saying we have a surface and a box. The box contains your pe-ke thingy. The surface has A incident SW and leaving the surface we have B AND C longwave. B LEAVES the system and C goes into the box. Lastly we have A=B.
Do I have that right?
OOPS. C is not longwave it is conduction. Correct?
“If I understand what you are saying we have a surface and a box. The box contains your pe-ke thingy. The surface has A incident SW and leaving the surface we have B AND C longwave. B LEAVES the system and C goes into the box. Lastly we have A=B.
Do I have that right?”
Not quite.
The surface has A (incident SW) plus C (conduction) from the box and leaving the surface we have B (IR to space) plus C (conduction) into the box.
The two Cs net out to zero leaving incident SW (A) having the same thermal energy value as the outgoing IR to space (B).
You then have radiation in from space equal to radiation out to space and conduction up the same as conduction down.
In effect the radiation from space gets a free pass through the system once the two Cs reach the same figure which occurs when the system stabilises at hydrostatic equilibrium for the atmosphere as a whole.
You can also envisage it as two distinct energy loops.
Radiation in and out (A and B) comprises a diabatic energy loop which is solely radiative whereas Conduction up and down comprise an adiabatic energy loop which is solely non radiative.
AGW theory and all proponents of it omit the downward C from the non radiative adiabatic heating provided by descending columns which puts the energy budget out of balance so to make it balance they have to invent a proposed surface heating effect from DWIR which is physically untenable because convection adjusts to neutralise it.
I think I’ve got it now. A is SW into the surface. B is LW out of the surface (free pass to space) and A=B. Simultaneously we have atmosphere and box exchanging equal amount of C at the surface and the whole thing is in steady state.
From this I would deduce that the atmosphere and surface are exactly equal in temperature and this exchange of C contributes no net energy to the surface therefore the atmosphere is not increasing energy of surface and the surface temp would be given by Stefan-Boltzmann, the same as for an atmosphere free planet.
Correct?
No,the atmosphere does warm the surface above S-B because you have to add the kinetic energy that is being cycled up and down to the continuing incoming insolation hence 288k instead of 255k.
The thing is that kinetic energy at the surface must EITHER radiate OR conduct.The same parcel of KE cannot do both simultaneously.
So, if you look at the system as a whole from space it is at 255k as per S-B because that represents the amount of radiative energy passing straight through.
However, the surface beneath the atmosphere must be at 288k in order to sustain hydrostatic equilibrium as well as match radiation out with radiation in.
AGW theory says that a surface at 288k MUST radiate AT 288K but it cannot do so if 33k is being conducted and convected instead.
That is how confusion arises if you try to fit a radiation only equation to a mixed radiative and non radiative scenario.
In your previous reply, you had the energy exchange by conduction between surface and box net 0, correct? There is no difference to energy flow be it radiation, conduction, or convection.
So here is the energy flow into the surface…..
A+C
Out is….
B+C
Net is….
(A+C)-(B+C)=0
There is no energy available to raise surface temp above airless planet. Not for steady state flow.
I can see your conceptual difficulty but how best to address it ?
In short, the energy tied up in conduction and convection must warm the surface above S-B and maintain that warmth otherwise there would be insufficient KE remaining at the surface after radiation out to space to keep convective turnover going.
If you have 255k radiating in and 255k radiating out and the surface is at 255k how to you fuel the constant upward pressure gradient force that opposes the downward gravitational force in order to keep atmospheric mass off the surface?
Do you say that the atmosphere can be held off the surface at zero energy cost?
If so, how?
My conceptual difficulty is the 1st law of thermodynamics. You are creating energy!
Well before you get to the handwaving inside the box, you need to resolve the 1st. Every bit of pe created in the box is returned as ke, C in = C out.
Everything inside that box is transport. The only time that’s not true is the very first time you feed it some joules.
The very first time it was fed some joules was during the very first convective overturning cycle when all the energy in the atmosphere was drawn by conduction from energy that would otherwise have radiated to space but which was in fact conducted from surface to atmospheric gas molecules to be convected upwards and which created all that PE that now acts as a permanent reservoir for continuing convection and which in the process heats the surface by 33k above S-B.
So, you see, you agree with me once you realise the significance of that very first cycle of convection.
I’m not creating energy at all. I am pointing out that the system is storing energy internally and thereby creating the so called surface temperature enhancement.
Actually, I don’t agree with you at all! You are creating energy. You agreed previously to what the energy flows were at steady state. You continue to add temperatures which is totally bogus.
With the model you previously agreed to, the surface temp could never rise above 255k. It can dip while energy is used to bring the box to steady state. But this model has to approach 255k at the surface over any reasonable integration interval.
At steady state the box is all transport. Sure it can move gobs of energy away from the surface in one place but it will deliver it all back someplace else.
Read again, carefully:
“The 33 in rising columns has no effect on surface temperature because it is taken from surface KE that would otherwise have gone to space via radiation so that the surface radiates to space at 255k and not 288k.
The 33 in falling columns does affect surface temperatures because it is arriving in addition to continuing insolation at 255k
You get a surface temperature of 288k despite the S-B prediction of 255k and it is observed that only 255k is going out to space.
That squares off the whole debate perfectly.”
You need to stop with the 33 up add 33 down. You need to stop with the 255. They are nowhere in the model under discussion.
They are obfuscations to divert attention from the facts of 1st law energy accounting that you can not balance.
Stick to the A,B,C that we stipulated upfront and show me the increase in surface energy content that makes the surface temp rise.
You Can’t heat a room by lifting weights.
You can heat a room by delaying heat loss and that is what convection achieves since it works more slowly than radiation.
Even when the up and down flows reach zero balance between themselves that is only hydrostatic equilibrium and not a removal of the energy stored as PE which is constantly being returned to the surface as KE in addition to continuing insolation.
To try and keep it really simple:
The 33k in rising air is taken from the outgoing radiation rather than directly from the surface itself.
The 33k in falling air is added directly to the surface in addition to continuing insolation.
Due to the out of phase timing that developed during the first convective overturning cycle one is borrowing energy from outgoing IR and lending it to incoming shortwave for a period of time and that forces up the equilibrium temperature of the surface to 33k above S-B.
No energy being created. Just a delay in the throughput of 33k due to conduction and convection being one convective overturning cycle slower than radiation.
That raises surface temperature because one is delaying the radiative loss of 33k by the length of time it takes the first convective cycle to complete.
That delay is then repeated ad infinitum.
Have you got it yet?
In an accounting of STEADY STATE energy balance there is no phase delay allowed. That’s why it is a steady state accounting.
You are handwaving. Your atmosphere box creates energy if, as you claim, it has raised surface temp. You have failed to show an energy imbalance that would increase surface energy content
Then you cannot or will not see the obvious.
You can not increase the temperature of something unless you increase its energy content. That means you have to put more into it than what leaves it.
You can have a gazillion joules flowing in and out of the surface from lapse rate hand waving but as long as those flows are equal, the effect on the surface is nada.
Should have added that the effect is nada over a reasonable integration interval.
And I didn’t say that the value of C is zero. I said it was net zero for the up and down processes taken together. C is still a sizeable number comprised of all the PE and KE in all the molecules that are not in contact with the surface.
It is that interplay between the proportions of KE and PE at different heights and latitudes that gives us climate zones and all forms of weather.
Do you say that a convecting atmosphere has no energy content?
Take any figure for the value of C as representing all the KE and PE in the atmosphere, say 66 for simplicity
Thus A and B cancel out but you have 33 of C in rising columns and 33 of C in descending columns.
The 33 in rising columns has no effect on surface temperature because it is taken from surface KE that would otherwise have gone to space via radiation so that the surface radiates to space at 255k and not 288k.
The 33 in falling columns does affect surface temperatures because it is arriving in addition to continuing insolation at 255k
You get a surface temperature of 288k despite the S-B prediction of 255k and it is observed that only 255k is going out to space.
That squares off the whole debate perfectly.
Note that when the atmosphere first lifted off the surface outgoing radiation dropped by 33k but the surface stayed at 255k as per S-B. To drop the surface below S-B conduction and convection would have to work faster than the speed of radiation and we know that is not possible.
When upward convection reached its maximum height and the subsequent descent then reached the ground then the adiabatic loop closed and since then the system has been in steady state with the surface temperature enhancement caused by conduction and convection storing all that energy in the atmosphere and recycling it up and down indefinitely as long as external insolation continues.
Stephen — you ARE violating conservation of energy.
“when the atmosphere first lifted off the surface outgoing radiation dropped by 33k but the surface stayed at 255k as per S-B.”
No! If 240 W/m^2 are arriving as sunlight and 240 W/m^2 are leaving as thermal IR @ur momisugly 255K, then there is simply no extra power available to warm the air.
If, for example, 20 W/m^2 are being siphoned off to warm the air and initiate convection, then there are only 220 W/m^2 left to radiate and the surface will only be 250K. As long as 20 W/m^2 are being stolen, the surface stays at 250K. If that circulating air eventually returns and delivers the 20 W/m^2 back, then the net lost from convection becomes zero, and the net radiation could rise to 240 W/m^2.
Tim said:
““when the atmosphere first lifted off the surface outgoing radiation dropped by 33k but the surface stayed at 255k as per S-B.”
No! If 240 W/m^2 are arriving as sunlight and 240 W/m^2 are leaving as thermal IR @ur momisugly 255K, then there is simply no extra power available to warm the air.
If, for example, 20 W/m^2 are being siphoned off to warm the air and initiate convection, then there are only 220 W/m^2 left to radiate and the surface will only be 250K. As long as 20 W/m^2 are being stolen, the surface stays at 250K. If that circulating air eventually returns and delivers the 20 W/m^2 back, then the net lost from convection becomes zero, and the net radiation could rise to 240 W/m^2.”
Tim, your error is assuming that the ‘stolen’ 20 W/m2 reduces surface temperature below S-B. It doesn’t. The surface stays at 255k as per S-B and the ‘theft’ is instead from the departing radiation some of which is diverted by conduction into convection. From space one sees a drop in the entire system temperature to 250k during convective cycle one because less radiation is escaping past atmospheric mass but the ground remains at 255k.
Then, when the first convective overturning cycle completes the surface temperature rises to 260k in your example (would have been better if you had stuck with the real world number of 288k that I used) and viewed from space the system returns to 255k as per S-B.
You see, the energy held in the atmosphere after completion of the first cycle doesn’t disappear just because energy up equals energy down. That just signifies hydrostatic equilibrium has been achieved. The additional energy accumulated during cycle one. is still there in the system forever so there is extra power to warm the surface.
You have to add the KE arriving at the surface in descending columns to the continuing 240 W/m2 from insolation and the combination raises Earth’s surface to 288k instead of it settling at the S-B level of 255k.
if conduction is ‘stealing’ radiation (an entirely new form of energy transfer) then you need to show that as an additional flow, let’s call it D, with an arrow depicting a fork to B going into the box. This makes outbound LW = B-D.
You also say that the stored energy in the atmosphere is the “power” that raises surface energy. Power is not energy flux but I assume your Intent was energy. If so, then you need to show that as an additional output from the box, we’ll call this one E.
So now we have C exchanging equal and opposite energy between box and surface ( this is the ke-pe pumping) and stolen D going into the box equal to E going back to the surface by, unstated, transfer mechanism.
The only problems i see here are the radiation imbalance at the system boundary, the new radiation-Conduction transfer science, and the new science where stored energy becomes power and gets used twice to raise the energy level of something with no net energy exchange.
Oops. Guess I’ve still got a lot to learn.
Paul, that is not an accurate or coherent summary of what I am saying. Others can make their own judgments.
In particular I resent being accused by you in another comment of presenting a slayer model when I agree with Willis that a cool atmosphere heats a warmer surface. My only difference with Willis is as to the mechanism.
It is either DWIR or convecting mass that does the job and the debate is wide open on that.
A lot of commentators are now focusing on that very issue so to block debate of it here would be very odd.
”Others can make their own judgments.”
Stephen, as I judged previously, your imagination of surface uneven warming with neat little packets of rising and descending air parcels is not found in the real world. For example, clouds are simply not popping up and down all the time & everywhere, they have bases (ceilings in aviation).
Settlement of a debate on physics is in the lab not the imagination or prose. Here is a short experimental video showing you how surface convection is much messier than you imagine.
That is not an accurate summary.
I am considering the effects of convection averaged globally including the large Hadley Cells and others.
The surface stays at 255k as per S-B”
Why in the world would you think this? The surface cannot simultaneously 1) receive 240 W/m^2 and 2) radiate 240 W/m^2 and 3) transfer some, additional power to warm the atmosphere directly by conduction and 4) maintain a steady temperature of 255K.
Any three of those can happen simultaneously, bit not all four!
S-B gives the temperature achievable by a blackbody at a given level of irradiation.
That assumes all incoming and outgoing energy runs at the speed of light.
In order to make the surface temperature fall below S-B as proposed by you a method of energy removal faster than light would have to be applied.
Conduction and convection operate at a slower rate and so cannot reduce the temperature of the irradiated surface below the S-B temperature.
Instead, conduction and convection take their energy from the outgoing radiation AFTER the S-B temperature has been achieved and cannot reduce that pre existing surface temperature.
The same parcel of kinetic energy at the surface cannot both radiate and conduct simultaneously so whatever goes into conduction reduces the number of photons emitted to space. Observed from space the planet would appear to drop below S-B but at the surface it would not have done so.
That being the case the surface temperature must then rise above S-B when KE is released from PE beneath descending columns of air and observed from space the planet would return to S-B despite the surface actually being warmer.
That is exactly what we observe in the real world.
”That is not an accurate summary.”
Seems it is when Stephen writes:
“The 33 in rising columns…The 33 in falling columns..”
Not observed in tests. While this test does show a sort of rising convective column as observed with clouds and gliders, there is no falling column as the test shows much spreading laterally and does not get back to the surface as cleanly if at all.
Simple testing shows Stephen’s “The 33 in falling columns does affect surface temperatures..” is imaginative at best. Whereas sky DWLIR is a much more constant glow measured incident on the surface which is affected by atm. opacity and T.
The surface energy balances across the globe show as much up convection as down convection, no net energy added or removed from the surface so no effect on median T at the surface or TOA. As opacity changes, each convective component will change but overall the convective net will remain zero modulation of surface T observed over 4-10 years or more.
It is non zero during the process of the first convective overturning cycle and that is where the energy for the surface temperature enhancement comes from.
The rest of your post is inaccurate and irrelevant.
“KE is released from PE beneath descending columns of air..”
As the simple test shows, and observations of clouds at various ceilings, descending columns of air are simply imaginary. Although severe downdrafts are observed locally, by and large as Stephen often writes the atm. is in hydrostatic equilibrium most of the time. Sure, rising convention significantly disturbs it some days but not generally. The atm. glow is much more constant up and down.
Darned, Trick, thanks for the information; all along I thought there were deserts at certain locations on the surface of the earth. Downwelling dry air, anyone?
Wow, that’s a crazy trick.
All this air goes up in convections and updrafts……. but NONE comes back down…..
Some sort of weird mass balancing happening, wouldn’t you say ;-).
“but NONE comes back down…..”
Only in Andy’s imaginations. Not observed coming back in columns Andy, look at the test: as in the atm. the blue spreads in from lower z height higher pressure system replacing the lower pressure system created by the rising convective ~columns as shown. Stephen’s descending columns are simply imaginative.
Yes Dave deserts are regions of descending dry air on the lee side of mountain ranges and certain atm. constant circulations. Note not descending columns which are imaginary as shown in the test. Convection goes up in reasonabe columns, descent is not columnar. As the test shows. There should be more testing around here less imagination.
Most deserts occur beneath the subtropical high pressure cells in both hemispheres and they are large scale regions of primarily descending air. There are other such cells at the poles
Stephen, yes, Atacama, Death Valley: lee side. Sahara, poles: circulation.
Lee side of mountains is not sufficient on its own for deserts. Also need to be under or near the subtropical high pressure cells.
Every region of higher than average pressure is a column of descending air. That constitutes half the bulk atmosphere.
ROFLMAO.
So the air DOES come back down.
Thanks for that, 😉
”It is non zero during the process of the first convective overturning cycle..”
Simply more imagination from Stephen, not reality.
Of course, testing and observations are inaccurate if they do not agree with Stephen’s imagination. I prefer to go with testing and observations not Stephen’s imagination. Imagination does work in movies and blogs for entertainment but that’s about it. Though some science based imagination does eventually get confirmed by test.
Please specify how any uncompleted initial convective overturning cycle could fail to have a non zero energy effect at the surface
That’s easy peasy EIN=EOUT. No net change.
I don’t think anyone would dispute localized imbalances both geographic and temporal but you need to think about integrating over the planet and over reasonable time periods to get one vector in and one vector out. They MUST be equal and opposite at steady state or energy is either being created or destroyed.
That is for a completed cycle only.
Whilst incomplete it is a non zero process.
We are talking about equilibrium states, not steady states.
Right. No net energy exchange.
You have built a fantasy with no net energy exchange except for that pesky one that has ‘equal’ exchange at the surface-atmosphere boundary. Except in one direction it acts like energy transfer and in the other it moves energy without moving energy.
Answer my question.
Why does a developing atmosphere need to reduce surface temperature below that determined by insolation when it gets all the energy it needs from energy that would otherwise have radiated to space?
If it fails to reduce surface temperature during initial uplift then it must raise the temperature above S-B when KE is released in the subsequent descent.
Not going there! Discussion is about 4 billion year old system in thermal equilibrium….
No net energy flows.
A developing atmosphere is not in thermal equilibrium
“Every region of higher than average pressure is a column of descending air.”
Not in tests and observations only in Stephen’s imagination. Watch the blue dye test Stephen, it spreads in laterally underneath at the surface just like in the windy atm. as the moderator notes. Lower pressure system being replaced by higher pressure.
Not a rotating sphere, not a gas, no density gradient with height. A very silly attempt at diversion.
“Please specify how any uncompleted initial convective overturning cycle could…” do something.
Please show a physical example of a test/actual observed uncompleted initial convective overturning cycle and we can discuss what it could do. I can’t possibly determine what such a process could do in Stephen’s imagination.
Simple logic should do it.
A rising molecule takes energy away from the surface having received energy by conduction.
Is it imagination to see that as a non zero effect on surface energy until such time as the molecule falls back to the surface?
Yes
That implies that you think a developing atmosphere requires no energy to lift it off the surface.
Bizarre.
“A rising molecule takes energy away from the surface having received energy by conduction.”
Ok, then my answer is as shown in the test video, by simple logic. Your further comments should be in line with the test not just in your imagination.
What Trickie is saying is that more energy is moved in the upward direction than downwards.
ie, the atmosphere has a net COOLING effect.
Thanks Trickie. We KNEW that. 🙂
“S-B gives the temperature achievable by a blackbody at a given level of irradiation.”
More specifically, the maximum temperature achievable if the only energy input is a given level of EM radiation. As it happens, we are discussing a situation where the only input is EM radiation, so in this setting this statement works fine
“That assumes all incoming and outgoing energy runs at the speed of light.”
No — that assumes that all incoming energy leaves as thermal IR. “Speed” has nothing to do with this. It is conservation of energy at its simplest. If some energy ‘leaks’ away via conduction or convection, then there is simply less left fly off as thermal radiation.
No energy leaks away in conduction and convection.
It later returns to warm the surface beneath descending air which constitutes half the atmosphere.
“We KNEW that.”
Sure, good for you Andy. The atm. is radiating LW energy to the sink of deep space; if not the world would be a very different place.
“It later returns to warm the surface beneath descending air which constitutes half the atmosphere.”
Not as shown by test Stephen, only in your imagination. And anyway you many times write the atm. is mostly in hydrostatic equilibrium so not much descending or rising air is the norm. There is lateral fluid movement also at the same T as shown by the blue dye at the surface and red dye at the top.
Don’t you think you have discredited yourself enough yet ?
Try answering the question I put to Paul.
Why does the inception of an atmosphere need to reduce surface temperature when it gets all the energy it needs from energy that would otherwise have radiated out?
Please show a physical example of a test/actual observed “inception of an atmosphere” and we can discuss what it could do to previous energy balances & if any different than the results of the test in the video. I can’t possibly determine what such a process could do in Stephen’s imagination.
4:00pm ”Why does a developing atmosphere need to reduce surface temperature below that determined by insolation when it gets all the energy it needs from energy that would otherwise have radiated to space?”
Stephen now proposes a developing atm. at 4:00pm not inception of an atm. as at 3:43pm.
I am not really sure what you are imagining here Stephen, which is why I ask for physical example to answer any question you may have, like in the test video. I am guessing you are looking for an answer you already have imagined and any answer not in accord with your imagination will be called inaccurate, irrelevant and discredit the author.
So please, if you really want a serious question answered in comments on a post about can a cold object warm a hot object, ask the question in a way that can be tested for accuracy, relevancy, and to credit the answerer.
A C Osborn November 28, 2017 at 3:37 am
I have no clue what you are on about. Let me once again say, since you and others seem determined to ignore it:
I don’t know which of my words you are babbling about. I have no idea which expeiment of Dr. Roys you think proves something, or what it proves. Your post is meaningless to me.
Come back when you are willing to specify your objection. This is tiresome.
… added …
Also, read what Ed Bo says above. If that is what you are talking about, Ed is right and you are wrong. The use of night-time radiative cooling has a long history. People in the Southwest US used to use it to freeze water when the air and ground temperature was above freezing … but that has absolutely nothing to do with concentrating diffuse IR.
w.
PS—Your comment about “take off your ‘scientific consensus’ head” is plain nasty. I don’t have two heads, I have one. My respect for you has taken a big hit with this comment of yours …
And your response has reduced my respect for you also.
We have been having this discussion about How a Solar Oven makes things colder, you originally denied all knowledge of what I was talking about and suddenly you state “The use of night-time radiative cooling has a long history. ”
So you appear to have known about it all about it all along.
Your exact words
Willis Eschenbach November 27, 2017 at 11:06 am
1. Nobody knows any way to “concentrate” environmental thermal IR since it is diffuse and coming in from every angle. You cannot focus it with a mirror or a lens … how are you planning to “concentrate it?
In his first experiment here
http://www.drroyspencer.com/2010/07/first-results-from-the-box-investigating-the-effects-of-infrared-sky-radiation-on-air-temperature/
Roy W. Spencer, Ph. D. says:
July 29, 2010 at 2:38 PM
you cannot focus an extended source of thermal radiation.
And reference actually INCREASING the affect of DWLIR here as suggested by others because a Solar Oven does so
Roy W. Spencer, Ph. D. says:
July 29, 2010 at 3:42 PM
OK, I added a funnel made of aluminum sheeting, and I think I can see enhanced cooling right away. We have cirrus clouds right now from surrounding thunderstorms, but it typically clears during the night. If it does, we’ll see whether we get greater cooling below ambient tonight.
So I ask you yet again If you cannot Collect, Concentrate or Focus DWLIR from CO2 or Space
How does it increase the cooling affect?
As to your taking offense at a figure of speech about “putting on another head” I appologise.
Just clarify my question
How does adding the Aluminum Funnel increase the cooling affect?
Or how does a Solar Collector oven do the same thing?
AC:
I repeat: there is nothing in that thread about concentrating thermal infrared radiation. The “funnel” that was added was simply to block radiation from the side angles, which would come from warmer sources. No concentration, and especially no heating.
A C Osborn November 29, 2017 at 8:29 am
Ed is right. Adding shielding to your setup such as the aluminum funnel is common when you want maximum radiative cooling. It shields what you are cooling from the effects of the ground, the trees, the building, and the nearly horizontal radiation of the lowest, warmest layer of air. This allows the object to reach a cooler temperature than without the shielding.
However, it is NOT concentrating diffuse IR, as you seem to think. As an example of the problem, try using a magnifying glass to concentrate the light on a day which is heavily overcast. A magnifying glass only works with essentially parallel rays like those of the sun.
If you think can truly concentrate IR, I encourage you to either find a reputable source that shows how it can be done, or to do an experiment yourself. As far as I and many other knowledgeable people on this blog know, it can’t be done … but then as Feynmann said, “science is the belief in the ignorance of the experts”.
Let us know what you find.
w.
A good answer, but again not good enough.
Why would you use a Funnel of Shiny Material, why not just use any old material if it “just a mask”?
The Solar oven is Shallow and allows any and all radiation to enter it and still works.
A C Osborn November 29, 2017 at 4:22 pm
Good question. You use aluminum, not because it is shiny, but because the emissivity of aluminum is low. It typically ranges from 0.05 to 0.4, depending on the surface treatment.
This is an advantage, because at a given temperature an aluminum shield radiates less than high emissivity materials. At say 15°C, a body with an emissivity of 0.9 emits about 350 W/m2. With an emissivity of 0.4 it would emit about 155 W/m2, and with an emissivity of 0.1 it would emit only about 40 W/m2. Since the point is cooling, we want to use a low-emissivity shield to block out the low-angle radiation from trees, buildings, and the lowest (and warmest) atmosphere.
As to the “Solar oven”, I have no clue what you mean by that. As mentioned before, you can concentrate collimated radiation, but not diffuse radiation. Solar radiation is collimated, so it’s not applicable to the current question of Dr. Roy’s experiment.
Regards,
w.
A small addition to Willis’ description. If the sheet of material at 15 C with emissivity = 0.1 was in a room at 15 C, you would measure approximately … 390 W/m^2 coming from it. It will only EMIT ~ 40 W/m^2 itself (10% of 390 W/m^2), but it will REFLECT about 350 W/m^2 from the room (90% of 390 W/m^2).
If the reflective sheet is pointed at the sky that might only be producing 200 W/m^2, then the total IR from the funnel would be something like (0.1 * 390 + 0.9 * 200) = 220 W/m^2. A lot more than 40 W/m^2, but still WAY less that ~ 390 W/m^2, so it would make for an effective cooling system.
Also, I think the ‘solar oven’ is something like this: https://www.scientificsonline.com/product/solar-oven
Basically, it would serve as the ‘funnel’.
AC Osborn: The correct answer to all these mis–directors and takers of offense is to channel the Gypper – “There y” go again”. Never mind, you are doing well, all here who won’t wear the tin hat.
Well if you had accorded me the courtesy of actually reading the Dr Spencer post which I referred to, I know you are a busy person but what is the point of asking for “references” if you are not going to follow them up?
If you had read it and the comments you would have seen multiple references to Solar Ovens (Collectors) including links to photos.
Due to the suggestions of the commentors to add a similar kind of “collector feature” Dr Spencer did.
In his quote that I supplied he called it a Funnel, not a mask or shield or block and it worked.
The definition of a funnel is “A tube or pipe that is wide at the top and narrow at the bottom, used for guiding liquid or powder into a small opening.” So replace water or powder with Radiation and what do you get?
You disparigingly said in your previous post “If you think can truly concentrate IR, I encourage you to either find a reputable source that shows how it can be done, or to do an experiment yourself. As far as I and many other knowledgeable people on this blog know, it can’t be done”.
Well I don’t need go anywhere to find a “reputable Source” , that post was in 2013 and many of those knowledgeable posters obviously believed that a “collector” would have some kind of an Affect and Dr Spencer proved it.
Universities have tested and found it can reduce temps to 20C below Ambient.
With that evidence I think it is up to you to provide an alternative reason for it working which proves it is not a Funnel (DR Spencer’s description”) or some kind of Collector.
And just to avoid confusion this is what a Solar Oven of Collector types looks like.
https://www.bing.com/images/search?view=detailV2&ccid=wzMzM1Cy&id=4579BB3763543FE29CFE7C54B595DADBE1354769&thid=OIP.wzMzM1Cyms39KnwuoU9HmwEgDY&q=photos+of+solar+ovens&simid=608005888856884813&selectedIndex=31&qpvt=photos+of+solar+ovens&ajaxhist=0
https://www.bing.com/images/search?view=detailV2&ccid=WCYY9Nc2&id=1E1249F3E1D932EFF39E477B8A95AD69776DEB09&thid=OIP.WCYY9Nc2twrnJt2-eBPIpAGoCT&q=photos+of+solar+ovens&simid=608044397524550803&selectedIndex=33&qpvt=photos+of+solar+ovens&ajaxhist=0
https://www.bing.com/images/search?view=detailV2&ccid=WCYY9Nc2&id=1E1249F3E1D932EFF39E477B8A95AD69776DEB09&thid=OIP.WCYY9Nc2twrnJt2-eBPIpAGoCT&q=photos+of+solar+ovens&simid=608044397524550803&selectedIndex=33&qpvt=photos+of+solar+ovens&ajaxhist=0
https://www.bing.com/images/search?view=detailV2&ccid=422xT3Bc&id=FB7164846187C39D4B8322CB4576C9933A7DAE7C&thid=OIP.422xT3BcmBGi0d8gZhGEUQEsDh&q=photos+of+solar+ovens&simid=608029012954909515&selectedIndex=3&qpvt=photos+of+solar+ovens&ajaxhist=0
There is also this
https://johnosullivan.livejournal.com/18334.html
A C Osborn November 30, 2017 at 2:09 am Edit
If you were citing a comment you should have actually cited it. When you cite a post, I assume you are citing the post. Foolish me.
I’m sorry, but I seem to have wandered into an alternative universe. People were asking why you couldn’t concentrate IR radiation to make it hotter, so that it could do work, in the same way that solar can be concentrated. The question was, can we use IR to make things hotter, not make them cooler.
You’ve given as an example a night-sky radiator, which attempts to shield its contents from IR in order to cool them down. But as Ed Bo noted:
So no, the shielding on Dr. Roy’s experiment doesn’t “concentrate” IR. And calling it a “funnel” just means that it is funnel-shaped, not that it is actually functioning as a funnel. Nor does someone calling it a “collector” mean that it actually collects anything. It reminds me of the old joke: “How many legs does a cow have if you call a tail a leg?”
…
The answer is “Four … because calling a tail a leg doesn’t make it one”. And calling something a “funnel” doesn’t make it one either.
But the biggest problem is that you are answering a question that wasn’t asked—is it possible to cool something by shielding it from warm IR? The answer to that question, as you noted, is assuredly “yes”. Build an insulated box, point the open side at the cold sky, shield it from warm ground radiation, and it will cool … so?
The problem is, that says nothing about the original question, which was whether you can concentrate and focus IR to get it hot enough to do useful work.
Finally, can you stop with the solar ovens? I know how they work, I used to teach Peace Corp Volunteers how to build them, yes, they have a reflective funnel to focus the COLLIMATED solar rays, and I know that they have nothing, repeat nothing, to do with whether we can focus diffuse IR to bring it to higher temperatures. As far as I know we can’t, and you haven’t provided one thing that says that we can.
As to O’Sullivan’s risible claims, he says that if you point a solar oven at the sky you can freeze water when the air temp is above freezing. This is true. We used to note that they could cool as well as heat when we built our solar ovens, although we never got anything to actually freeze because Peace Corps trainings are generally in the summer.
However, this does not, as he claims, “prove” that there is no downwelling radiation from the atmosphere. He says:
Please tell me that you know that downwelling radiation is real. It is measured routinely around the planet by dozens of scientists and has been for years. And as to whether it is absorbed at the surface, if it were not absorbed it couldn’t be measured as the scientists are doing. Nor does the global energy budget balance without downwelling IR, it is way, way off without it.
The problem is that we only get about 170 W/m2 at the surface from the sun … which equates to an S-B temperature of minus forty degrees (C or F, take your pick, they’re the same). So if there is no downwelling IR as O’Sullivan foolishly claims … what is providing the energy to keep us from going into the icebox? It’s not gravity, gravity can’t provide ongoing energy unless the earth is shrinking, so what is it keeping us from freezing if not downwelling IR?
And if you’re going to claim that all those scientists measuring downwelling IR all around the planet are either lying or mistaken, if you believe O’Sullivans fantasy that downwelling IR “doesn’t exist in the real world”, then I fear we have little to discuss.
Finally, back to the original question, O’Sullivan certainly doesn’t show that you can use a solar oven to concentrate IR. Like I said, I don’t know of anyone who knows how to concentrate diffuse radiation.
w.
One of my favorite examples of IR exploitation is smudge pots in orchards. They create copious amounts of a gas (that can not be named) that converts upwelling LW energy to downwelling LW energy that keeps fruit at a toasty above freezing temp.
The gas can be we’ll below freezing when it does this.
Stupid farmers don’t know that this Can’t work
AndyG55 November 28, 2017 at 1:43 pm
No, it is what is called a THOUGHT EXPERIMENT. Einstein used them extensively. Go accuse him of FANTASY and stop exposing your ignorance here. Everyone is entitled to an opinion … but not every opinion is worth blurting out in public.
w.
More like thought bubble.
Irrelevant to anything to do with the atmosphere.
But hey…. stick to your thought bubbles if that’s what it takes to convince yourself.
Are you really saying that your steel greenhouse actually exists?
If not, then it is a fantasy. It is make-believe.
You can call it a “thought experiment” if you like, but so is me planning a trip to Tahiti.
Not based on reality.
This makes me sad. Complex systems are ALWAYS analyzed by picking away at the components in simple, constrained ‘thought’ working from known and we’ll understood fundamentals towards what we don’t know in little bitty steps. It’s called science. It’s called engineering.
Anybody who criticizes these exercises has no business here because it demonstrates complete incomprehension of what is going on.
Indeed, most of these experiment threads go down ratholes invented to obfuscate and divert attention from the unpleasant truths that conflict with the favorite meme of the day.
If you think It’s fun and snarky to point out there isn’t a steel greenhouse, duh, then you certainly have missed important learning in that classic post and likely Can’t comprehend what it intends to teach.
I know exactly what is going on. It makes me sad that you obviously don’t.
This is an imaginary thought bubble, irrelevant to anything to do with the atmosphere.
And conflating it with Einstein? Seriously !!!
So Andy … would you say that Schrodinger’s Cat was just a useless “fantasy”? How about Maxwell’s Demon? What about Einstein’s five thought experiments that reshaped physics? Just “fantasies”.
Your callous dismissal of valuable scientific thought experiments as “fantasy” is hilarious. You truly believe that? Really?
w.
Again,
conflating with Einstein.. You are getting WAY ahead of yourself., as you often do.
Some hypothetical thought processes are “useful”..
others..
just meaningless thought bubbles.
Yes Andy, we are talking to pygmies here, in terms of understanding. Good thing they are in a sidestream while the real deconstruction of the false physics continues. Particularly by defunding, you might notice. Deasperation is rising…..
In a couple of previous posts, I mentioned the persistent effect of *convection* in creating an an adiablatic lapse or adiabatic gradient in the atmosphere. As right as I still think that is, I notice that I may have been misleading myself, just a bit, by also emphasizing the low thermal conductivity (or the high molecule-to-molecule resistivity) of gases, in quite the way that I did? The thing to understand is that the molecule to molecule resistivity of all atmospheric gases, including oxygen and nitrogren is so *good*, that *other things*, like convection and radiative transfer, tend to take over, when it comes to actually moving heat energy from the ground upward. Also, it would seem that the basic dry adiabatic lapse rate is essentially what you might call an “onset of convection” condition; the lapse rate is what happens when the most basic convective effect has *stopped*. The idea is that there is a negative feedback, with convection stopping and starting, and with a temperature gradient being maintained thereby.
I’ve now read the “Steel Greenhouse” thought experiment that Willis Eschenbach had mentioned before in his current article “Can a Cold Object Warm . . . “. My initial reaction is that the thought experiment is simply trying to establish something that I myself find intuitive, that “some” gases, at least, should be expected to either absorb and/or emit infrared radiation, just like other forms of matter, e.g., steel, tend to do. Shouldn’t I expect that even the regular gases like nitrogen, those gases being a form of matter, would also engage in “warm” infrared emissivity? My intuition, at least, says that emissivity is in general a kind of universal material process. If someone shoves a warm bag of nitrogen gas out into deep space somewhere, why wouldn’t we expect the nitrogen as such to emit IR out to the cold microwave background of the universe?
Now, if I search the internet using terms like “infrared emissivity of nitrogen”, I can easily find seemingly nonsensical statements like “oxygen, nitrogen and the rare gases (such as argon) do not emit radiation in the IR range”. At the same time, if I ‘Duckduckgo’ search a more pointed question, say, “Does clear nitrogen gas emit thermal infrared radiation or doesn’t it “, I get more interesting results! Try for instance, the third hit down on the page of hits that results from *that* question, that is the web page https://chiefio.wordpress.com/2016/02/17/nitrogen-active-in-the-ir-a-ghg/ .
The blogger at the above link is quoting a “harvard.edu”/ American Astronomical Society journal article from *1944*. This is about detecting *IR radiation of approximately one micrometer*, coming from *molecular nitrogen in the night sky*. Now, if such results as were reported in that journal article are true, why, you can just *watch* my bag of nitrogen cool down. Take *that* dudes!
My challenge, now, is to all “warmistas”, “lukewarmers” and anyone else who ever thought that CO2 or other minor gases were especially important in this regard. Try asking your favorite web searcher a pointed question, such as “Does clear nitrogen gas emit thermal infrared radiation or doesn’t it “. If it turns out that GHG theory boosters can then hardly deal with any informed results they may get, shouldn’t they be rethinking their position?
David Blenkinsop November 29, 2017 at 1:39 am
I’ve now read the “Steel Greenhouse” thought experiment that Willis Eschenbach had mentioned before in his current article “Can a Cold Object Warm . . . “. My initial reaction is that the thought experiment is simply trying to establish something that I myself find intuitive, that “some” gases, at least, should be expected to either absorb and/or emit infrared radiation, just like other forms of matter, e.g., steel, tend to do. Shouldn’t I expect that even the regular gases like nitrogen, those gases being a form of matter, would also engage in “warm” infrared emissivity? My intuition, at least, says that emissivity is in general a kind of universal material process. If someone shoves a warm bag of nitrogen gas out into deep space somewhere, why wouldn’t we expect the nitrogen as such to emit IR out to the cold microwave background of the universe?
Unfortunately this is where your intuition lets you down, a college freshman class in physical chemistry would tell you that the infrared spectra of gases is the result of quantized energy transfers between discrete rotational/vibrational energy levels, and that such transfers require a dipole which homonuclear diatomic molecules such as O2 and N2 don’t possess. In your thought experiment we wouldn’t expect the N2 to emit IR into space, the bag however would emit IR into space and thus cool the N2 down via conduction.
Now, if I search the internet using terms like “infrared emissivity of nitrogen”, I can easily find seemingly nonsensical statements like “oxygen, nitrogen and the rare gases (such as argon) do not emit radiation in the IR range”. At the same time, if I ‘Duckduckgo’ search a more pointed question, say, “Does clear nitrogen gas emit thermal infrared radiation or doesn’t it “, I get more interesting results! Try for instance, the third hit down on the page of hits that results from *that* question, that is the web page https://chiefio.wordpress.com/2016/02/17/nitrogen-active-in-the-ir-a-ghg/ .
The blogger at the above link is quoting a “harvard.edu”/ American Astronomical Society journal article from *1944*. This is about detecting *IR radiation of approximately one micrometer*, coming from *molecular nitrogen in the night sky*. Now, if such results as were reported in that journal article are true, why, you can just *watch* my bag of nitrogen cool down. Take *that* dudes!
Unfortunately as indicated above you don’t understand elementary physical chemistry, if you did you would understand that N2 when exposed to very energetic UV photons can either be electronically excited to higher electronic states (does not require a dipole) or excited so much that it splits into two N atoms. Those N atoms can recombine to form an electronically excited N2 (after losing some excess energy). In the decay of those electronically excited states some IR is emitted, this is not the same as what happens in the troposphere where such highly energetic photons have been filtered out (the processes the paper refers to occur at ~200km above the earth, the thermosphere where it can be as hot as 1500K). In fact the mechanism proposed in that paper is that the N atoms recombine to form N2 molecules in the ground vibrational state of the electronically excited triplet state (B3Пg), this then decays with the emission of the observed IR to the electronically excited singlet state (A3∑+u). My guess would be that there is first an intersystem crossing to the singlet state then a vibrational transition to the v=0 level. None of these processes are accessible in the troposphere.
My challenge, now, is to all “warmistas”, “lukewarmers” and anyone else who ever thought that CO2 or other minor gases were especially important in this regard. Try asking your favorite web searcher a pointed question, such as “Does clear nitrogen gas emit thermal infrared radiation or doesn’t it “. If it turns out that GHG theory boosters can then hardly deal with any informed results they may get, shouldn’t they be rethinking their position?
To which the answer is: not in our atmosphere.
If instead you asked ‘does nitrogen in a near vacuum at 1500K irradiated with far UV emit in the IR?’, the answer would be yes.
Because gravity creates a pressure and density gradient so the higher molecules are spaced further apart which allows creation of non sensible potential energy from sensible kinetic energy.
My comment at 4.02am was directed at Berenyi Peter but was misplaced in the thread.
talldave2 said:
“This is another instance of being basically correct but with framing problems — objects radiate exactly the same way no matter where they are. The atmosphere isn’t “in the way” so much as it is radiating back at the warmer objects (even though it’s colder than they are!), especially when there are big lumps of water vapor in it.”
Please explain how conduction works if objects radiate exactly the same way no matter where they are.
Heat is just kinetic energy which is molecular movement/vibration. If a molecule at the surface is moving/vibrating at a rate commensurate with an S-B temperature of 255k and then encounters gas molecules to which it passes 33k ‘worth’ of that energy instead of emitting a photon then how can it possibly radiate at 255k with that conducted energy then missing?
You can suggest that the removed energy is immediately replaced by fresh insolation so that the temperature remains at 255k and should emit commensurately.
The trouble is that within a convecting atmosphere the conduction process is constant so that although the surface temperature remains at 255k any fresh insolation is itself whisked away into conduction and convection so that the surface will only radiate to space at 232k despite being at 255k.
That deals only with convective cycle one when the atmosphere first lifted off the ground.
As soon as convective cycle one completes with the subsequent descent to the surface then it is the case that the previously missing conducted 33k is being reintroduced to the surface in addition to continuing insolation ‘worth’ 255k which then raises surface temperature from 255k to 288k
But you still have that constant upward flow of conducted energy elsewhere so that the radiation to space
from the 288k surface temperature is still limited to 255k as per observations.
So, surfaces beneath a convecting atmosphere do not radiate exactly the same way as if there were no convecting atmosphere.
For those who keep claiming that reflected radiation cannot heat the lamp that emitted the radiation, Anthony did the experiment here. So you are not just arguing against well-established thermodynamic theory … you’re arguing against experimental results as well.
w.
PS—Do NOT trust anything published by Principia Scientific International without first running the numbers and reviewing the logic … and you’ll still have to wash your hands afterwards.
“PS—Do NOT trust anything published by Principia Scientific International without first running the numbers and reviewing the logic … and you’ll still have to wash your hands afterwards.“
How vicious and vindictive of you.
Tony November 29, 2017 at 4:26 pm
Tony, clearly you’re not familiar with the website …
In any case, you may (or may not) recall, I said specifically that I did NOT think that you are either vicious or vindictive, viz:
So now, you are using my NOT calling you either vicious or vindictive as an excuse to call me both … dude, you are a piece of work.
w.
Tony November 29, 2017 at 4:26 pm
One more thing. Tony, here’s a quote from a typical post on Principia Scientific International … it starts off as follows:
And THAT is why I advised washing your hands after you go there. Talk about vicious and vindictive … like I said above, it appears you have no idea what the site is all about.
w.
Willis, Mr Skolnick was taken down by the Canada bar association:
“The Law Society of British Columbia (LSBC) has now ruled that green activist Andrew Skolnick’s official complaint concerning Dr. Tim Ball’s libel attorney, Michael Scherr and science writer, John O’Sullivan, was baseless.
Andrew Skolnick had filed the complaint against Dr. Ball’s legal team as part of a coordinated attack stage managed by lawyer Roger McConchie, representing disgraced climatologist, Michael Mann.
Specifically, the LSBC has affirmed there is no evidence to support Skolnick’s malicious allegation that anyone “knowingly asserted something for which there is no reasonable basis in evidence.”
http://johnosullivan.livejournal.com/41331.html
Sunsettommy November 29, 2017 at 10:21 pm
Well, kinda. ONE of Mr. Skolnick’s claims was found to be baseless … but if you read the link I gave, that claim isn’t listed. Instead, you’ll find lots of other claims with good citation and documentation.
And in any case, PSI published a post calling me psychotic and drug-addled … what does the Canada Bar Association have to say about that kind of scummy sliming? Where is the noble Mr. O’Sullivan in that? And why are you defending a man who does that?
w.
Willis, I was just pointing out another of your double standards. Your article begins with a request for civility, but your own civility stops when it comes to any mention of “extreme skeptics” or their arguments. And, “well he started it” is not a defence (nor would it most likely be true)
Tony called Willis “vicious and vindictive,” because Willis said PSI is untrustworthy.
That’s absurd. Nothing Willis said was vicious and vindictive.
Even what I sad (that the PSI people are “stark, raving, clinically insane”), though harsher than what Willis said, was not vicious and vindictive.
If you want an example of real viciousness & vindictiveness, look no further than your despicable PSI pals. Accusing President Bush of staging the 9-11-2001 terrorist attack, and murdering nearly 3000 of his fellow Americans with “micro-nukes,” as PSI’s Joe Olson did, that is truly vicious and vindictive, as well as stark, raving, clinically insane.
If Tony actually cared about civility, he would apologize to Willis, and condemn his vicious, vindictive, dishonest PSI pals.
“Tony called Willis “vicious and vindictive,” because Willis said PSI is untrustworthy.“
Or it could have been because of the ongoing campaign to smear PSI that WUWT has mounted over the last x number of years…lol
Wllis,not trying to defend John, just to point out that Andrew made a serious allegation that was found to be false.
Skolnick had other legal problems too that he lost,which included getting dismissed:
“CMS lawsuit
The AMA dismissed Skolnick when Correctional Medical Services, one of the for-profit health care companies criticized in the “Death, Neglect and the Bottom Line” article, threatened JAMA and the Post-Dispatch with litigation.[21][22] [23]
Skolnick also sued CMS, claiming their responses to the articles were defamatory, but a summary judgement ruled in favor of CMS, the defendants.”
https://en.wikipedia.org/wiki/Andrew_A._Skolnick
We remember, Willis, how it was made evident that the increase was caused by a different factor. And we wonder…..
Made evident how? Evident to whom?
The responses here from knowledgeable people — including some who have taught thermodynamics — all agree with Willis’ analysis. Everything he wrote in the top post agrees with standard textbooks on the subject. It is right that you start to wonder why you can’t convince people with your ‘evidence’.
Brett Keane November 29, 2017 at 8:12 pm
Brett, I have no clue what you are talking about. What increase? What factor?
What I wonder is why you have trouble following a polite request, viz:
Why is that so hard? As it stands, you are talking to the wall, because your shorthand is impenetrable.
w.
https://wattsupwiththat.com/2017/11/29/study-no-acceleration-in-global-warming-climate-sensitivity-to-co2-too-high/
You wil not believe the empirical evidence held in NASA solar system data, so you will also disrespect the findings of Dr Christie above. Disrespecting me peronally, your problem not mine.
But the tone is like that of most warmista, when confronted by truth. That makes me wonder about the standards of this debate, and what is going on behind it. Tony above nailed it. Never mind, truth will out…..
Brett, this is unintelligible. First off, there are dozens of commenters here. So who is the “you” whom you are addressing? Because if it is me, your claims are ridiculous. I respect Dr. Christie greatly, he’s one of my scientific heroes.
Next, you don’t know what I or anyone else will or won’t believe, or will or won’t disrespect. You’d have to be a mind reader to do that, and you’re not one. You’re just pulling claims out of your fundamental orifice.
Next, you say:
This is an absurd blanket generalization. And in any case, who are the “warmista”, and what “tone” are you referring to?
Seriously, you need to QUOTE THE WORDS THAT YOU ARE DISCUSSING, because as it stands, you could be out in space talking to aliens for all the good you are doing us.
w.