Guest Post by Willis Eschenbach
Pushed by a commenter on another thread, I thought I’d discuss the R. W. Wood experiment, done in 1909. Many people hold that this experiment shows that CO2 absorption and/or back-radiation doesn’t exist, or at least that the poorly named “greenhouse effect” is trivially small. I say it doesn’t show anything at all. Let me show you the manifold problems with the experiment.
To start with, let me give a curious example of the greenhouse effect, that of the Steel Greenhouse. Imagine a planet in the vacuum of space. A residue of nuclear material reacting in the core warms it to where it is radiating at say 235 watts per square metre (W/m2). Figure 1 shows the situation.
Figure 1. Planet in outer space, heated from the interior. Drawing show equilibrium situation
This planet is at equilibrium. The natural reactor in the core of the planet is generating power that at the planet’s surface amounts to 235 W/m2. It is radiating the same amount, so it is neither warming nor cooling.
Now, imagine that without changing anything else, we put a steel shell around the planet. Figure 2 shows that situation, with one side of the shell temporarily removed so we can look inside.
Figure 2. As in Figure 1, but with a solid steel shell surrounding the planet. Near side of the shell temporarily removed to view interior. Vertical distance of the shell from the surface is greatly exaggerated for clarity—in reality the shell and the shell have nearly the same surface area. (A shell 6 miles (10 km) above the Earth has an exterior area only 0.3% larger than the Earth’s surface area.)
[UPDATE: Misunderstandings revealed in the comments demonstrated that I lacked clarity. To expand, let me note that because the difference in exterior surface area of the shell and the surface is only 0.3%, I am making the simplifying assumption that they are equal. This clarifies the situation greatly. Yes, it introduces a whopping error of 0.3% in the calculations, which people have jumped all over in the comments as if it meant something … really, folks, 0.3%? If you like, you can do the calculations in total watts, which comes to the same answer. I am also making the simplifying assumption that both the planet and shell are “blackbodies”, meaning they absorb all of the infrared that hits them.]
Now, note what happens when we add a shell around the planet. The shell warms up and it begins to radiate as well … but it radiates the same amount inwards and outwards. The inwards radiation warms the surface of the planet, until it is radiating at 470 W/m2. At that point the system is back in equilibrium. The planet is receiving 235 W/m2 from the interior, plus 235 W/m2 from the shell, and it is radiating the total amount, 470 W/m2. The shell is receiving 470 W/m2 from the planet, and it is radiating the same amount, half inwards back to the planet and half outwards to outer space. Note also that despite the fact that the planetary surface ends up much warmer (radiating 470 W/m2), energy is conserved. The same 235 W/m2 of power is emitted to space as in Figure 1.
And that is all that there is to the poorly named greenhouse effect. It does not require CO2 or an atmosphere, it can be built out of steel. It depends entirely on the fact that a shell has two sides and a solid body only has one side.
Now, this magical system works because there is a vacuum between the planet and the shell. As a result, the planet and the shell can take up very different temperatures. If they could not do so, if for example the shell were held up by huge thick pillars that efficiently conducted the heat from the surface to the shell, then the two would always be at the same temperature, and that temperature would be such that the system radiated at 235 W/m2. There would be no differential heating of the surface, and there would be no greenhouse effect.
Another way to lower the efficiency of the system is to introduce an atmosphere. Each watt of power lost by atmospheric convection of heat from the surface to the shell reduces the radiation temperature of the surface by the same amount. If the atmosphere can conduct the surface temperature effectively enough to the shell, the surface ends up only slightly warmer than the shell.
Let me summarize. In order for the greenhouse effect to function, the shell has to be thermally isolated from the surface so that the temperatures of the two can differ substantially. If the atmosphere or other means efficiently transfers surface heat to the shell there will be very little difference in temperature between the two.
Now, remember that I started out to discuss the R. W. Wood experiment. Here is the report of that experiment, from the author. I have highlighted the experimental setup.
Note on the Theory of the Greenhouse
By Professor R. W. Wood (Communicated by the Author)
THERE appears to be a widespread belief that the comparatively high temperature produced within a closed space covered with glass, and exposed to solar radiation, results from a transformation of wave-length, that is, that the heat waves from the sun, which are able to penetrate the glass, fall upon the walls of the enclosure and raise its temperature: the heat energy is re-emitted by the walls in the form of much longer waves, which are unable to penetrate the glass, the greenhouse acting as a radiation trap.
I have always felt some doubt as to whether this action played any very large part in the elevation of temperature. It appeared much more probable that the part played by the glass was the prevention of the escape of the warm air heated by the ground within the enclosure. If we open the doors of a greenhouse on a cold and windy day, the trapping of radiation appears to lose much of its efficacy. As a matter of fact I am of the opinion that a greenhouse made of a glass transparent to waves of every possible length would show a temperature nearly, if not quite, as high as that observed in a glass house. The transparent screen allows the solar radiation to warm the ground, and the ground in turn warms the air, but only the limited amount within the enclosure. In the “open,” the ground is continually brought into contact with cold air by convection currents.
To test the matter I constructed two enclosures of dead black cardboard, one covered with a glass plate, the other with a plate of rock-salt of equal thickness. The bulb of a thermometer was inserted in each enclosure and the whole packed in cotton, with the exception of the transparent plates which were exposed. When exposed to sunlight the temperature rose gradually to 65 oC., the enclosure covered with the salt plate keeping a little ahead of the other, owing to the fact that it transmitted the longer waves from the sun, which were stopped by the glass. In order to eliminate this action the sunlight was first passed through a glass plate.
There was now scarcely a difference of one degree between the temperatures of the two enclosures. The maximum temperature reached was about 55 oC. From what we know about the distribution of energy in the spectrum of the radiation emitted by a body at 55 o, it is clear that the rock-salt plate is capable of transmitting practically all of it, while the glass plate stops it entirely. This shows us that the loss of temperature of the ground by radiation is very small in comparison to the loss by convection, in other words that we gain very little from the circumstance that the radiation is trapped.
Is it therefore necessary to pay attention to trapped radiation in deducing the temperature of a planet as affected by its atmosphere? The solar rays penetrate the atmosphere, warm the ground which in turn warms the atmosphere by contact and by convection currents. The heat received is thus stored up in the atmosphere, remaining there on account of the very low radiating power of a gas. It seems to me very doubtful if the atmosphere is warmed to any great extent by absorbing the radiation from the ground, even under the most favourable conditions.
I do not pretend to have gone very deeply into the matter, and publish this note merely to draw attention to the fact that trapped radiation appears to play but a very small part in the actual cases with which we are familiar.
Here would be my interpretation of his experimental setup:
Figure 3. Cross section of the R. W. Wood experiment. The two cardboard boxes are painted black. One is covered with glass, which absorbs and re-emits infrared. The other is covered with rock salt, which is transparent to infrared. They are packed in cotton wool. Thermometers not shown.
Bearing in mind the discussion of the steel greenhouse above, I leave it as an exercise for the interested reader to work out why this is not a valid test of infrared back-radiation on a planetary scale … please consider the presence of the air in the boxes, the efficiency of the convective heat transfer through that air from the box to the cover plates, the vertical temperature profile of that air, the transfer of power from the “surface” to the “shell” through the walls of the box, and the relative temperatures of the air, the box, and the transparent cover.
Seems to me like with a few small changes it could indeed be a valid test, however.
Best regards,
w.
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PJF, as far as I can see a vacuum flask essentially does only work by preventing heat loss from conduction and convection. They are often silvered to reflect heat back into the flask, but that is reflection, rather than back-radiation. If the flask were unsilvered and made of a material that did not absorb IR radiation, it would work on Earth, but not in space.
If an object cannot be “thermalised” (presumably this means “made warmer”) by IR radiation from cooler objects, the key question is how can the object know which IR photons come from cooler objects (and should not be absorbed) and which come from warmer objects (and should be absorbed). All objects emit photons with a distribution of wavelengths, the distribution depends on the effective temperature of the object, but that can only be determined by looking at the distribution of the wavelengths of photons it emits, not a single photon.
Well I’m rather glad that I exhausted my interest in this thread some time ago. Why David, and Richard are still here taking all this abuse, I cannot fathom.
I always learn some stuff here. I had no idea there were “Translational Temperatures” and “Rotational Temperatures”, but if Phil says there are, that is good enough for me. Evidently they relate to the internal resonance excitation modes of certain molecules, and hence belong in the realm of quantum mechanics of real materials.
I’ve been outside of academia for more than half a century, and left before developing quantum mechanics at my finger tips.
So I can’t actually cite any peer reviewed papers on experimental observations of isothermal bodies anywhere in the universe. So for all I know, there may not be ANY such thing as an isothermal object, anywhere in the universe. I also can’t cite any peer reviewed papers on experimental observations of 100 percent absorption of EM radiation by any physical body or bodies.
So I’d be happy to read such papers if those who know about them would cite references. I do still have inside connections at my alma mater, and can get papaers from behind paywalls, if I need them.
But based on what I have read, I would say that black body radiation theory, is a solution in search of a problem; since Planck’s black body radiation derivation explains no actual experimental observations anybody ever made, that I am aware of. It may involve matter consisting of “particles” which thermally interract (collide with each other) but it takes no acount of whether those particles comprise a solid, or a liquid or a gas; it doesn’t even involve any properties at all of any of those particles.
So we have a theoretical model of something that nobody has ever experimentally observed, since none of the ingredients exist anywhere.
The Stefan-Boltzmann Law is of course nothing but the definite integral of the Planck radiation law, over all frequencies or wavelengths. If I’m not mistaken, a form of S-B was derived from statistical mechanics, and ordinary classical thermodynamics; well just the T^4 part was.
The beauty of Planck’s formula, is that it derived a value for sigma, the S-B constant.
Actually the total radiated power density, is not sigma x T^4, as is most commonly stated.
The correct formula is Wtot = N^2.sigma T^4 Watts per metre squared, where of course N is the refractive index of the medium in which the total radiant emittance is measured or observed.
But this thread has lots of people bandying about the “S-B ;law” as if it is somehow connected to the so-called “greenhouse effect”; that being of course the atmospheric climatism involved greenhouse effect; not the fictitious greenhouse effect by which real greenhouses do not operate, and RW Woods experiment relating to that.
Some posters here clearly have English as a second language, or maybe a third language; Richard I’m sure is having a fit trying to translate some of the stuff here into English. Well David too, is as dumbstruck as I am reading some of the utterances.
As I have said before, I do not have QM at my fingertps where I can manipulate it for useful results, and I’m not a chemist, so I don’t have a lot of interest in the calculus of greenhouse absorptions.
It’s the non quantum mechanical thermal radiation that interests me most. So far as I know, there atre NO exceptions to its emission from thermally interracting particles at Temperatures above zero Kelvins.
Perhaps a single atom or molecule in free flight in space, does not emit thermal radiation; but that is not an exception, because such a non interractng isolated atom or molecule by definition is at zero Kelvins, so is not expected to radiate. And very sparse collections of molecules that collide less frequently, and at lower velocities, such as gases, would naturally be expected to emit (or absorb) less of the thermal radiation, than denser more frequently colliding collections of “particles”.
And those emissions are not related to atomic or molecular energy levels so they are not quantized; but they are emitted as “photons”, which may have any energy value. I’m not aware that Planck ever contemplated the exact physical cause of such radiant emissions. Heinrich Hertz and JC Maxwell simply derived it from accelerated electric charges; basic antenna theory, since a varying current is simply an accelerating electric charge.
I’ve already cited one paper on the “collision induced quadrupolar and hexadecapolar emissions in gases”. You can find it here at WUWT for yourselves. Dipole modes are not the onlt antenna radiation patterns.
Well material itself can only exist as atoms or molecules of a limited number of types or species; probably no more than 1,000 different types. Does that make chemistry quantum mechanics.
Photons of thermal radiation may have any energy from near zero eV to many GeV, with no missing values forbidden. That is NOT the case of the emissions that are involved in the “green house effect”, which may have only certain quantized values, and their broadened resonance lines.
And so long as we still have a 24 hour day on planet earth, this planet, will never be in thermal equilibrium.
davidmhoffer says
“But then out come the nay sayers with their version of science. Cold things can’t heat up warm things, because the Second Law of Thermodynamics says so.”
David
Why does every physics book and every physics department on the planet agree with Clausius.
There is no debate in physics about the direction of heat flow.
It ALWAYS flows spontaneously from a higher temperature to a lower temperature.
Why do advocates of the Greenhouse Theory find great difficulty with the thermodynamic meaning of HEAT?
dikranmarsupial wrote:
“PJF, as far as I can see a vacuum flask essentially does only work by preventing heat loss from conduction and convection. They are often silvered to reflect heat back into the flask, but that is reflection, rather than back-radiation.”
Curious. I chose the vacuum flask example for a couple of reasons. Willis’s steel shelled planet is essentially a vacuum flask in space. Willis also mentioned (later in the thread) multi-layered vacuum flasks that do benefit from increased back radiation to keep the contents warm for longer.
David
Why does every physics book and every physics department on the planet agree with Clausius.
There is no debate in physics about the direction of heat flow.
>>>>>>>>>>>>>>>>>
For the precise same reason that they agree with all the work that came after Clausius.
For the precise same reason that the 2nd Law has been reworded many times since Clausius.
For the precise same reason that they differentiate between heat flow and energy flux.
For the precise same reason that SB Law, Conservation of Energy and First Law can ONLY be reconciled with the 2nd Law by interpreting the 2nd Law as net flux. Either that, or we must choose which laws are right and which are wrong.
For the precise reason that 2nd Law interpreted as net flux means they are all right.
dikranmarsupial says, February 9, 2013 at 11:11 am: “Greg House – The “net transfer” is a consequence of viewing the transfer of heat as being mediated by the radiation and absorption of IR photons. Unless you can come up with a mechanism by which a cold body can know which directions it can safely emit a photon without it hitting a warmer body, or for a body to choose not to absorb photons from a cooler body (impossible as there is no way to determine the temperature of the body that emits an individual photon), then an exchange of photons (and hence energy) is inevitable.”
============================================================
No, the problem is that that your “consequence of viewing” is simply a conjecture. To become a scientific fact it has to be proven experimentally first.
Even Wikipedia says: “The modern photon concept was developed gradually by Albert Einstein to explain experimental observations that did not fit the classical wave model of light.” As you can see, experiments come first, then comes a term and knowledge about the properties of those “photons”. In other words, we know for sure about how “photons” behave in real world only from experiments. What is not known from experiments is just a conjecture or fiction.
And this “net” thing has apparently never been proven experimentally, hence it is not a scientific fact.
dikranmarsupial;
If an object cannot be “thermalised” (presumably this means “made warmer”) by IR radiation from cooler objects, the key question is how can the object know which IR photons come from cooler objects (and should not be absorbed) and which come from warmer objects (and should be absorbed).
>>>>>>>>>>>>>>>>>>>>>
A most excellent question!
May I offer a supplemental question?
If the photons from the cooler object are not “thermalised”, where does the energy associated with them go? Does it simply cease to exist, violating too many laws of physics to count by doing so? Does it rotate into another dimension, lurking there until a cold enough object to “thermalise” comes along at which point it leaps out and “thermalises” it? Perhaps it passes through a time warp and emerges at some point in the future as part of a Star Trek episode?
I’m certain Greg House must know the answer to both these questions, and that he derived same from an experiment with cardboard boxes.
MikeB says:
February 9, 2013 at 3:13 am
My math agrees. But, the ratio of the inner to the outer radius does matter.
This is the astounding thing about the formula: increasing the thickness of the shell while holding the mean radius constant reduces the surface temperature. This suggests that adding CO2 to the atmosphere does not necessarily result in an increase in surface temperature, and may even decrease it.
That leaves the GHE in place – the shell definitely heats the surface beyond what it would be without the shell. But, there is a limit, and increasing the depth of absorbing particles may actually have the opposite effect of incrementally decreasing the surface temperature.
There are at least two ways I see this claim potentially overreaching:
1) it could be too much of an idealization – as the shell becomes very thin, there is some point at which it does not intercept all the outgoing radiation from the planet
2) it does not account for any temperature gradient across the width of the shell
And, then there is my earlier objection that separating the shell by a vacuum does not really capture the dynamics on the Earth, where the atmosphere is in actual contact with the surface.
But, I think that at some level where the thickness is beyond critical yet the gradient is “small”, increasing thickness should decrease the surface temperature, at least for the idealized separation by a vacuum.
Things become even more complex as you add additional shells. One could represent an H2O shell, a CO2 shell, and a CH4 shell, for example. Then, one has to allow for transport of heat through conduction and convection. I do not believe anyone can reliably extrapolate the precise behavior of such a complicated system based on intuition and a very idealized back-of-the-envelope calculation for a grossly simplified analogous situation, which I strongly suspect is essentially what underlies the calculations of the climate establishment.
Words are imprecise, and intuition is fallible. Let’s have it mathematically.
Let’s not quibble and sidetrack the conversation from the meat on the table. You know what I mean.
tjfolkerts says:
February 9, 2013 at 6:41 am
Canceled out by neighboring emitters. The emission from one point on the shell to another is, on average, 180 degrees out of phase with that from another point at the same distance on a great circle passing through all three points. In the end, it all has to add up to the balance given. Not instantaneously, but time averaged, and in a very tight statistical band.
MiCro says:
February 9, 2013 at 8:11 am
No, it has a T1^4 – T2^4 term. You can factor that as (T1^2+T2^2)*(T1+T2)*(T1-T2), but T1 – T2 is not all there is.
davidmhoffer says:
February 9, 2013 at 10:30 am
That’s the problem with debating in words. Words are imprecise. Words can mask specious reasoning. Math is absolute.
Bryan says: “There is no debate in physics about the direction of heat flow.”
Just like there is no debate that “heat” is the net flow of energy due to a thermal gradient, and this this flow depends on the temperatures of BOTH objects involved.
Heat flow from a hot object to a cold object = “large”
Heat flow from a hot object to a not-quite-so-cold object = “not-quite-so-large”
So if you surround the earth by “not-quite-so cold” surroundings (ie the atmosphere), rather than “wickedly-cold” surroundings, then the heat flow from the earth’s surface will get reduced. “reduced heat flow out” of course means “more energy retained” which in turn means “the object warms up”.
The cool atmosphere does not “heat the surface” and no one (who is speaking in strict thermodynamic language) would say it does
The cool atmosphere DOES restrict the rate at which heat leaves the surface, which will lead to a warmer surface since the energy input from the sun remains fixed)
In other words, the atmosphere (in conjunction with the sun) “warms” the surface, but does not “heat” the surface.
Why do a̶d̶v̶o̶c̶a̶t̶e̶s̶ opponents of the Greenhouse Theory find great difficulty with the thermodynamic meaning of HEAT = net flow of energy?
PS. The “modern” interpretation of the 2nd law is something like “a system tends toward the most probable macrostate”. Entropy and the direction of net energy flow are a consequence of this. If this paragraph makes no sense to you, then you need to study statistical mechanics more before trying to teach others about the meaning of thermodynamics.
TJFOLKERS SAYS
“The cool atmosphere does not “heat the surface” and no one (who is speaking in strict thermodynamic language) would say it does”
Fine, you agree with me that David Hoffer made a mistake when he claimed that the second law allowed colder objects to ‘heat’ warmer objects.
But then you go and spoil it all by implying that statistical mechanics contradicts classical thermodynamics.
Of course it doesn’t!
Advocates of the Greenhouse Theory continue to muddy the water.
Despite 16 years of flat global temperatures and increasing atmospheric CO2.
Despite millions of years global temperatures leading and increasing atmospheric CO2 lagging.
No historical evidence backs this increasingly futile theory.
All this in spite of Woods historic hunch and subsequent experiment that the greenhouse theory was in error.
@Bryan
” There is no debate in physics about the direction of heat flow.”
Nor is there any debate that all objects above absolute zero radiate ir photons. And it’s photons that carry heat energy between objects in a vacuum.
Depending on the wave length, it is either absorbed and thermalized or re-emitted, reflected, or it doesn’t interact at all, based on the materials emissivity.
@MikeB, if I didn’t use the right terms, please let me know, and i’ll try to use them correctly the next time.
I’ve learned a lot about this stuff, but know I still have a lot to learn.
“based on the materials emissivity”
I think it should read emissivity and absorption spectrum.
“REPLY: Seems like an easy experiment to perform to find out. I think I’ll actually do this – Anthony”
Honestly, Anthony, you have better things to do. The true believers (TB) will ignore or “explain” away anything that goes against their religion, especially facts. Upthread, for example, Phil (February 8, 2013 at 11:26 am) cited an actual practical case, and was ignored. I pointed out another (February 8, 2013 at 7:09 am) and was–surprise!–ignored. A TB asks for textbooks, Willis cites chapter and verse, and the TB says, “See? You can’t deliver!” (I paraphrase.)
So why do I join these pointless discussions? Certainly not because I expect to change anyone’s mind (that leads only to frustration), but because:
1) It helps clarify my own understanding of the subject. If I can’t explain why a TB is wrong on a particular point, I need to learn more. I’ll note here that the Scienceofdoom blog has some nice articles related to this discusson.
2) I’m fascinated by our (human) capacity to build our own realities. This is my opportunity to explore one such “reality” in some depth (yes, extreme fear of CAGW is another).
3) Let’s face it, these guys are a hoot!. You just can’t make this stuff up. Their mangling of the 2nd Law of Thermodynamics, for example, is exceeded only in certain creationist arguments.
Rather than do yet another pointless experiment, Anthony, why not put the onus on the TBs, where it belongs? “Convenional” physics works just fine despite the TB insistence that it can’t; the TBs are the ones with something to prove, and everything to gain by doing so. In these threads I endlessly suggest that TBs actually perform a definitive experiment themselves, e.g. Dr. Roy Spencer’s “Yes Virginia” thought experiment. Strangely, nobody accepts the challenge, but the mere fact of the refusal is to me far more telling than any argument.
Ryan says:
February 7, 2013 at 1:53 am (Edit)
Ummm … ummm … where to start.
This thought experiment of the “steel greenhouse” is an extremely simplified conceptual model. It is the simplest conceptual model I know of that could actually work and that functions using the poorly named “greenhouse effect”.
I use it because you could actually build a scale model of it, and it would actually work using the actual (although poorly named) greenhouse effect. As a result, it explains why the greenhouse effect works, and it makes it clear that the essential mechanism of the greenhouse effect has nothing to do with and does not require the presence of:
1. CO2 or other GHGs
2. An atmosphere
3. Lapse rates
4. Effective radiation heights
To me this is a crucial step, to establish just what it is that we are calling the “greenhouse effect”, and how it works at its simplest. Not how it works in its particular incarnation here on Earth, but how the greenhouse effect itself actually works.
Should we be concerned if “IT IS COMPLETELY WRONG FOR PLANET EARTH!”?
My friend, it’s not a model of the planet Earth, nor was it meant to be. It is a simplified system that likely doesn’t exist anywhere in the universe, a radioactive planet surrounded by a steel shell.
However, this kind of simplified physics model can teach us lots of things that are useful for understanding how a greenhouse effect really works, whether on Earth or anywhere. For example, it shows us that the shell needs to be thermally isolated from the surface, or the greenhouse effect doesn’t work at all … which makes it immediately relevant to the Woods experiment, for obvious reasons … which is why I brought it up.
w.
PJF Reflecting the IR is generally more eficient than absorbing and re-radiating it, as more than 50% can be reflected back. If it were constructed from steel, rather than glass, so that the walls did absorb IR radiation, then it would work as Willis mentions. They key point I was trying to make is that vacuum flasks do work principally by preventing conduction and convection, so whether it works in space very much depends on the materials it is made out of, rather than that it is vaccum flask per se. Sorry to have confused the issue.
And, will people stop arguing about “a colder object cannot heat a warmer one”? That is not what is happening. The flow is always from higher temperature to lower. The argument is not about flow of energy (Watts). It is about accumulation of energy (Joules).
It is exactly like the water analogy I described above. The colder object does not heat the warmer one, it slows down the cooling of the warmer one and, in the time it takes to equilibrate, the warmer object accumulates more energy.
It is so annoying, this confusion between the time rate of change of energy in Watts versus the storage of energy in Joules. The water barrier does not act as a source of water, either. But, it does cause the water to pool up behind it. That’s all that is going on. It is very simple.
johnmarshall says:
February 9, 2013 at 3:25 am
I have said many times that I will not have any truck with a scumball who calls me a liar and accuses me of trying to deceive people. I am not that man.
You are just the latest in a string of officious douchebags who wants to announce your entry into the discussion by starting off with a sleazy, uncited, untrue, and deeply offensive insult.
You can apologize, or you can get ignored from here on out.
Your choice.
w.
Greg House:
At February 9, 2013 at 9:58 am you say to me:
Aha! I am starting to understand your confusion! But you are starting to ‘get it’.
The finger doesn’t do it: the finger only triggers it.
The air conditioner moves heat from the cool air inside your house and pumps that heat to the warmer air outside of your house thus heating the warmer air outside the house.
Other than that, your example is good. Indeed, it is a form of refrigeration system and I used refrigeration systems as an example myself in my post at February 9, 2013 at 3:30 am
Richard
David Socrates says:
February 9, 2013 at 4:05 am
Perhaps … or it might relate to the fact that Tallbloke banned me from his site for saying that a perpetual motion machine wouldn’t work, and I haven’t been back since. He’s still welcome here and visits my threads, but I won’t go to a site where people are banned for their scientific views. He put himself in the same boat as RealClimate and Open Mind, I won’t increase either their or his page count by one, they censor scientific discussion, and that to me puts them beyond the pale.
YMMV, of course,
w.
Greg House – it has not been demonstrated experimentally that photons of IR emitted by a cooler object are never absorbed by a warmer object either, so that isn’t a scientific fact according to your criterion either. This cannot be determined by observing the temperature of macroscopic objects, as you would get the same result whether there was a directional transefer of heat proportional to (T1 – T2) or a bidirectional transfer proportional to T1 in one direction and T2 in the other. If you know of an experiment where the behaviour of the photons themselves is observable, then please provide the details.
Do you agree that heat energy is transmitted by photons, yes or no?
ThePhysicsGuy says:
February 9, 2013 at 7:33 am
First, PhysicsGuy, you fail to distinguish between the NET energy flow (called “heat”), and the two individual energy flows from A to B and B to A, which are not called heat.
For Anthony, let me propose one change in the experiment. Take the measurements as the bulbs cool. This avoids the issue of the changing resistance of the filament of the bulb with temperature, and changing electrical power draw and the like. So … three stages.
1. Put a 100w bulb in a socket, and let it reach equilibrium. Turn it off, and take the temperature every 30 seconds or so as it cools.
2. Repeat the experiment with a 60w bulb.
3. Repeat the experiment with two sockets and two bulbs. First, move the sockets far apart, and let the bulbs come to equilibrium. Turn both off, and immediately move the sockets near to each other. Then measure the cooling rates of the bulbs.
This setup avoids a lot of variables in the first one. The only remaining conflating variable is the fact that both are air-cooled … still, that shouldn’t make any appreciable difference.
I’m glad you’re talking about doing the experiment, Anthony. I was gonna do it, but I don’t have an IR thermometer …
w.
Bryan:
At February 9, 2013 at 12:33 pm you ask David Hoffer
Allow me to provide an answer.
They don’t: indeed, they do understand “the thermodynamic meaning of HEAT”.
The problem is that those who want to disbelieve in the real, physical and observed GHE have no understanding of electromagnetic (EM) energy and how that energy can be converted to heat.
EM radiation is quantised; i.e. it exists as discrete particles called photons. Each photon has an energy and carries a wavelength. If a photon is absorbed by something (e.g. the Earth’s surface) then its energy is absorbed by the something. That energy is added to the something and (unless it is a specific molecule which gains rotational or vibrational energy) that addition of energy raises the temperature of the something. And that happens whatever the temperature of the absorbing something.
Richard
richardscourtney
One of the problems in dealing with a two object heat exchange is that the hotter object is often in addition heated continuously by an external source.
This can cause confusion.
Lets simplify the situation and miss out further heating of the hotter object.
Picture an object at say 80C and another neighbouring object at 20C
If the only heat transfer is by radiation then this is what happens.
The hot object will radiate more intensivly than the cold.
Both hot and cold will absorb each others radiation.
But the hot object will lose internal energy and its temperature will drop.
The cold object will gain internal energy and its temperature will rise.
Its as simple as that.
The cold object does not increase the temperature of the warmer.
MiCro says:
February 8, 2013 at 12:17 pm
Phil. says:
February 8, 2013 at 10:56 am
“Its translational temperature will stay the same its rotational and vibrational temperatures will increase depending on which particular energy levels have just been populated. The excited state will now attempt to lose that excess energy by a combination of collisional deactivation and radiation. Which one is favored will depend on conditions, near the earth’s surface collisions are the major route, up near the tropopause it’s radiation.”
Thanks, if the photon is thermalized, will the molecule’s kinetic temperature increase proportional to the energy of the photon?
The process of thermalization is the collisional exchange of vibrational energy in the excited molecule with its neighbors eventually all the vibrational energy ends up as kinetic energy (translational) shared among all the molecules involved (the atmosphere). Near the Earth’s surface mean time between collisions is less than a nanosec, whereas the characteristic time for emission by CO2* molecules is several orders of magnitude greater.
Willis/Anthony I don’t think the experiment will work as it stands, because the 100W bulb won’t absorb much of the heat emitted by the 60W bulb (and vice versa), but will instead simply pass through the bulb or be reflected off it, so there won’t be much of a change in temperature. Metal spheres heated by e.g. the heating elements from soldering irons would be a better bet (painted black to minimise reflection/maximise absorption). Also, unless the experiment is performed in a vacuum, an objection may be that the transfer of heat is by conduction of heat through the air. I don’t mean to be awkward, its just I want the experiment to be successful and convincing.
Willis,
Yes there are two EM radiation flows from A to B and from B to A. But there is not a “net” heat flow. Heat only goes in one direction, from hot to cold per the 2nd law.
MiCro says:
February 9, 2013 at 8:29 am
Anthony, if the bulbs are say 6″ apart, measure between them, then 3″ on the outside of both, and with a say 60w and 100w bulbs you will get 3 different readings. Also note the filaments will both be similar temps, since they both output similar color spectrums (color temps).
I refer you to the dichroic halogen lamps which improve the efficiency of the lamps by selectively reflecting IR light from the envelope to the light source, thereby increasing its temperature and increasing the proportion of visible emitted. It’s a perfect illustration of the effect discussed in this thread, the product can be purchased in stores, yet every time I mention it it’s ignored and we still get the comments of the idiotic Greg House etc.
http://www.gelighting.com/LightingWeb/emea/images/Halogen_MR16_IR_Lamps_Data_sheet_EN_tcm181-12732.pdf
“Halogen IR Technology
Standard incandescent and halogen lamps lose approximately 76% of the input energy by radiating heat, and convert only 8% into useful light. The PreciseTM IR halogen capsule has multiple layers of very durable, thin, interference film which redirects heat, which would otherwise be wasted, back onto the lamp filament. This increases the filament temperature and allows it to give off more visible light for the same input power.
The increased burning efficiency provides the same light performance with a significantly reduced power input, alternatively allows a longer lamp operating life or a combination of both.”