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.
Greg House,
“No, this is not true, anyway the “greenhouse effect” as presented by the IPCC is about particular “greenhouse gases” making the Earth warmer specifically by by returning back some of the Earth’s outgoing IR radiation, it is not about the whole atmosphere. They do not tell people, however, that this 150 years old hypothesis was debunked 100 years ago by the Wood experiment.”
1 GHE theory in general predates the IPCC significantly.
2 Willis shows quite convincingly why the Wood experiment demonstrates/debunkes NOTHING.
3) IPCC claims the GHE is about particular gasses returning back some of the Earths’ outgoing IR radiation. Others here claim that the above applies to the whole atmosphere and others still claim it is a gravitational effect not radiative. Some here even claim simple convetive distribution of heat fully explains the difference.
I did note that the mechanism by which the atmosphere keeps the surface warmer than it would be without one was in dispute. However, a simple comparison with the Moon shows that whatever the mechanism is the atmosphere clearly affects the Earths average surface temp.
‘Thought experiments’ may fit comfortably into either Philosophy or Mathematics, but they are not a proper part of Physics. [Sorry, Einstein!]
When others try to repeat Willis’ ‘thought experiment’ and get different ‘results’, it is quite clear that there is confusion of mind somehere. I’m not the one to point my finger at anyone in particular, or waste my time trying to disentangle where the mistaken thoughts might be, because all (including me) are equally susceptible to mistakes in this regard.
One of the commonest mistakes in ‘thought experiments’ is introducing unspecified assumptions without realizing you are doing it. Merely stating that one “knows” what a certain hypothetical object will do in specified circumstances reeks of hubris.
So, Willis, please:
1. Build your radioactive, model ‘planet’;
2. Measure its equilibrium temperature;
3. Enclose it in a steel sphere of specific size (without contacting the ‘planet’);
4. Re-measure the temperatures of the two spheres (good luck with that!); and
5. Report the results.
All else is imaginary, with real numbers, i.e. Mathematical games. The gulf between ‘what we think will happen’ and ‘what actually happens in a real experiment’ is often surprising and should be educational.
“The emissivity of various materials is given here. Reflectivity is equal to [ 1 – emissivity ]. The reflectivity of steel is about 10%-15%.”
Ummm, I think you need to go back and re-read that table again!
Anyway, I’m glad you admit the shell cannot be made of steel at least. Rusty steel perhaps (iron oxide will get you closer to being a black body) but not steel. So now you know we are talking about black bodies – definitely not steel, not anything like steel.
Lets consider what ACTUALLY happens with a CO2 molecule when a IR photon hits it. Remember that IR is not HEAT. It is an electromagnetic radiation with energy that can be converted to heat in certain circumstances. So a photon with energy in the IR spectrum reaches a CO2 molecule. What happens to it? Well if it is within the scope of the covalent bonds of the CO2 molecule it could be absorbed by the electrons in the bonds causing the electrons to be lifted to a higher energy level. But here is the crux of the matter – if the electron is already at that higher energy level then an incoming IR photon WILL NOT BE ABSORBED (at least not at the same frequency) – the photon will just travel straight through. But also bear in mind that lifting an electron to a higher energy level does not in itself cause heat to occur – the electron must drop back to its lower energy level and the assumption is that the electron then loses energy and that energy (probably) ends up at some point exhibiting itself as heat (just following the laws of entropy) but maybe not right away because the CO2 may just emit IR of a different wavelength (i.e. not heat as such) which may then be absorbed (or then again may not) by some other object which then gets a bit warmer.
Does any of this actual description of what happens to CO2 with an incoming photon strike you as being in any way related to ideal black bodies suspended in space over planet earth with no atmosphere when the Earth is being heated from the inside from an essentially infinite energy source of no known specific reaction temperature? Because it seems the model you are creating is in no way related to reality and therefore not useful, before we even begin to get into the details of whether your description of the maths is plausible.
davidmhoffer says:
After reading all of these, I think you’re one of the ones who have it right.
Willis, Working from a flux rate (watt’s/sq) alone makes the analogy harder to follow. S-B is based on a difference in Temps, with that an an emissivity, you can calculate a flux rate. In the case of this example the 100/40 temps (fhhaynie’s calculation) for the inner/outer shells(if the steel was a blackbody) sounds about right. The net flux from the surface should then match your 235w flux (net) from the inner to the outer, the back radiation is why the inner sphere is warmer than the outer.
Ryan says:
February 7, 2013 at 3:10 am
“I do wish Willis would wake up and realise that IR is not “heat” nor does it have a “temperature”.”
It has an equivalent temp. CO2’s main absorption bands are ~4u @ur momisugly ~850F, and then 10u-15u 60F- -112F But the 10u-15u band is very weak, and much of it is overlapped by water vapor.
Greg House, Please, please go to the hardware store, and buy yourself a handheld non contact IR thermometer. Once you can explain/understand how that works, you’ll at least get that S-B is well tested, and how everything radiates IR, even if it’s cold. Every things temp is the net of incoming radiation and out going radiation. Just as described by S-B.
My Ryobi IR thermometer on a clear sky 35F day, reads the skies temp at it’s minimum scale -40F, or it’s maximum scale of 608F. The less than -40F is the back radiation, instead of being the absolute zero of space, this lowers the temp difference between the earth and space slowing the cooling of the surface. It’s this -40F that all of the hubbub is about. But after digging through 120 million surface temp records, it doesn’t reduce reduce the amount of night time cooling any. The 608F is unexpected, but I think it’s just a reflection of solar IR off the atm. If there is a temp difference from increasing CO2, IMO it’s from this. Since a CO2 molecule will only absorb, and emit a single photon at a time, more Co2 will reflect more IR.
Lastly Dr Spencer did attempt to do this experiment.
Konrad. says:
February 7, 2013 at 1:29 am
“Does the surface emit IR? Yes. Does the atmosphere absorb some IR? Yes. Does the atmosphere radiate some IR back to the surface slowing it’s cooling rate? Yes. Is there a radiative GHE? Yes. Is the NET effect of radiative gases in our atmosphere warming? NO!”
This is a nice summary, but IMO the last answer is a possible slight warming, I put a max of maybe 25% of the measured trend, a slight beneficial warming to go along with all of the plant food. Plus if the Sun is getting ready to go on vacation, we’ll want all the extra warming we can get.
As a side note, we’ve had snow on the ground all but 4 or 5 days since a few days before Christmas in NE Ohio, it’s been quite a while since that’s happened!
Please explain how the low energy photons from the colder body gets absorbed by the atoms of the hot body already in higher excited states.
micro;
What the snow did was increase the insulation factor for the house, it’s the same way igloo’s can be a lot warmer inside than out, while made of snow. Also note, snow has a lot of trapped air, which is a very good insulator.
>>>>>>>>>>>>>>>>>>
Well gee, the air in the snow would be the same temperature as the snow, would it not? And how, exactly, do you think insulation works? It works by absorbing radiated energy and re-emitting some portion of it back.
So the summary of this thread reads pretty much like this:
Willis:
here’s how the Woods experiment was set up.
Whole Bunch of People Who Have Studied and Worked with Thermodynamics:
here’s about 20 different reasons why that experiment could not possibly demonstrate any meaningful data in regard to the radiative ghe
Whole Bunch of People Who Have NOT Studied or Worked with Thermodynamics:
you guys don’t know what you are talking about
I’ve scored the whole discussion and the winner is….
Lewandowsky
I’d like to make a quick comment about willis’ light bulb example and the assertion that a nice cool 60 watt bulb can increase the temperature of a hot 120 watt bulb.
Instead of light bulbs, imagine 2 batteries, one with 12 volts of electrostatic potential, one with 6. If you connect the two batteries in parallel (maybe through a resistance), the 6 volt battery doesn’t increase the potential of the 12 volt battery. That’s just obvious. Current only flows in one direction, from high to low. The potential of the 12 volt battery drops, and the potential of the 6 volt battery increases.
The exact same thing happens with a 120 and 60 watt bulb. The 60 watt bulb takes heat from the 120 watt bulb, cooling it.
Nevertheless, I completely agree with Willis, the 120 watt bulb is hotter because the 60 watt bulb is there. That’s because a 60 watt light bulb is a much poorer heat sink than the alternative, some nice cool air, which will absorb heat than waft away and be replaced by more cool air. The hot bulb replaces cool air and so the 120 watt bulb has a harder time getting rid of its heat and it runs hotter still.
So the 60 watt bulb is like a 6 volt battery and the air is like, say, a 1.5 volt battery. Less current flows from a 12 volt battery into a 6 volt battery than it would into a 1.5 volt battery, and as a result the voltage you’d measure at the junction of a 12 and 6 would be higher than at a 12 and 1.5. So in a sense, the 6 volt battery “increases” the potential of the 12 volt battery, but only in comparison to the alternative.
Now, before anyone jumps on me, let me state my awareness that electrical flow may not be a perfect analogy to heat flow. Electricity really really (pretty much really) only flows in one direction. What about heat? Does heat flow in one direction, or does it flow back and forth? Who Cares.
Quantum mechanics says that radiation scatters in all directions without preference. The presence of a hot light bulb doesn’t resist the flow of photons the way a positive electrical field would resist the flow of electrons, so the 60 watt bulb scatters in the direction of the 120 watt bulb exactly the same amount as it would in the direction of empty air or an ice cube or a vacuum. But thermodynamics says that heat only flows one direction, from hotter to cooler. That’s because in thermodynamics, which was come up with before corpuscles and turns out to be statistical law, you’re really talking about a net flow. In thermodynamics it’s ok to think of the 60 watt bulb as a heat sink, cooling the 120 watt bulb, even though the reason it’s a poorer heat sink, taking less heat away from the 120 than air would, is actually because it gives more back. Whether you think of heat as a flux flowing in only one direction, or as discrete bits of radiation bouncing between them back and forth, it really shouldn’t matter; as long as you just stay consistent about it you’ll get the same answer.
In other words, I think most, or at least a lot of, the arguing here is irrelevant and just people defending their initial semantic choices. But as for me and my own personal morality, I think it’s probably best when talking about “heat” to stick with thermodynamics convention and think of it as a flux and say it only flows from hot to cold, and if you’re talking about “radiation” it’s ok to say it goes everywhere.
Micro.
You can’t measure back radiation with an IR thermometer. They are specially designed to work within the IR atmospheric window, which means that they won’t see radiation from the atmosphere.
CO2’s absorption band relevant to the greenhouse effect is around 15 microns. It is not weak!
In fact the small amount of CO2 in the atmosphere will absorb 95% of radiation at 15 microns within a distance of 1 metre
lou says:
February 7, 2013 at 7:05 am
“Please explain how the low energy photons from the colder body gets absorbed by the atoms of the hot body already in higher excited states.”
Ok, I’ll have a go. Vibrating atoms of the “hot” body are not all moving at the same speed. Because they are colliding off each other, their movements are random. Some atoms will be moving at a speed higher than the temperature of the body would suggest, others moving slower. The slower moving atoms are able to absorb the photons from cooler body.
Though, having written that, I am not sure whether you are conflating two different things – the temperature of the body which consists of the average KE of the atoms, and whether the atoms exist in highly excited states. The latter means their electrons have moved to higher energy shells, and is not a function of their KE (temperature). Higher temperatures do not necessarily mean that their atoms exhibit highly excited (electronic) states – it merely refers to the average speed of motion of the atoms. This has nothing to do with whether or not the atom can absorb a photon, imo.
@ur momisugly Willis,
Figure 2 is WRONG!
There is no 470W/m^2.
.
Even IF the steel shell was a perfect reflector and insulator – not allowing any losses. The internal volume will still be 235W/m^2.
.
What this theory does not cover is the rate of loss against the rate of heating. The effect of shutting the oven (or fridge) door.
.
To comprehend more clearly with fewer issues please see across temperatures as a “potential difference” the same mindset to Voltage differences. Then consider the rate of heating and cooling as Amps. The outcome is, of course, in Watts.
davidmhoffer says:
February 7, 2013 at 7:11 am
Fair enough.
@MattS
I quoted John Christy for a reason, as I think he knows a little bit about atmospheric physics hence it is “a real head scratcher” the atmosphere isn’t warming per the “theory”. Which theory was he speaking of? Atmospheric CO2 has been increasing. The upper atmosphere should be warming at a higher rate than the surface. The stratosphere should be cooling because we’re told the troposphere “traps” the heat. The opposite is true, so the idea of an “enhanced” GHE is wrong. If the basic physics according to Christy is the
then something is wrong with the “theory”.
We are constantly lectured that increasing CO2 levels WILL warm the atmosphere; it is all about basic high school physics we’re told. Well, I don’t think it is about basic physics. I think it is about how the climate system works apart from “basic physics”. The physics is not well understood in my view.
We’re also told OHC increases in the past 30 years is due to AGW. There is zero empirical evidence that even a doubling of atmospheric CO2 can warm the oceans to any measurable degree. Instead, we’re told there is mixing by ocean waves that magically transports the top few microns of “heated” water affected by “back radiation” from a few extra ppm’s of atmospheric CO2. Sure, everyone stirs their hot coffee to make it warmer. How can anyone try to pass off the idea that anything other than SW solar radiation can warm the oceans and expect thinking people to believe that is frankly, insulting.
We’re also told water vapor is a positive feedback. Ok, then explain why deserts with very little water vapor become much warmer than tropical regions with much higher water vapor content. There seems to be a limit to how much the earth can warm, and water in its various forms appears to be the mediator.
I understand radiation is the only method by which heat is transferred to space. However, it is convection that dominates during the day when the sun shines which cools the surface by transporting heat to the upper regions of the troposphere. The surface of the earth would be uninhabitable if not for convection. I question whether the “basic physics” properly account for these convective processes.
I am a bit disappointed Willis did not address Tim Ball’s reference to http://principia-scientific.org/supportnews/latest-news/34-the-famous-wood-s-experiment-fully-explained.html . I claim no scientific expertise, only that I have been following the “greenhouse effect” meme for the last 25+ years and it is not working as advertised. Obviously the physics are not all accounted for or that well understood.
If the Woods experiment did not demonstrate/disprove anything, then neither did Arrhenius.
Finally, we’re told the earth’s atmosphere is not like a real glass greenhouse, yet I have seen dozens of references by various government science agencies and universities describing the earth’s atmosphere being like a real glass greenhouse in which the troposphere “traps” the heat thereby cooling the stratosphere.
MikeB says:
February 7, 2013 at 7:24 am
They have to measure IR in the same range the meter can measure temps in, 608F to -40F is the full scale range of the meter. Now the meter’s emissivity setting is set to .95, and the sky might not be .95, so the temp displayed could be wrong, but it has to measure IR in the 4.5u to 12.5u range. I would love to have an IR spectrometer, but it is out of my price range. And it obviously reads a IR signal from the sky. Besides NASA thinks it can be done.
CO2’s absorption band relevant to the greenhouse effect is around 15 microns. It is not weak!
In fact the small amount of CO2 in the atmosphere will absorb 95% of radiation at 15 microns within a distance of 1 metre
Here’s the spectrum of Co2 from NIST. 15u IR has an equivalent temp of -112F, that might keep Co2 from freezing, but it isn’t going to keep all the water on the planet from freezing.
Sorry meant to quote this
Willis Eschenbach says, February 7, 2013 at 2:57 am: “Since the Wood experiment is totally unlike and has nothing to do with the greenhouse effect, it couldn’t reveal a dang thing about the greenhouse effect …”
==========================================================
I provided a quote from the IPCC explanation of the “greenhouse effect” and the link to it on this thread, you must have overlooked it.
The Wood experiment deals exactly with the underlying mechanism of the IPCC “greenhouse effect” (effect of trapped/back radiation) and demonstrates that it is bupkis.
cd says:
February 7, 2013 at 1:12 am
CD you are correct. I was sloppy in my expression at that point. I addressed that topic in later posts.
Bart says:
February 6, 2013 at 11:54 pm
P_shell_outgoing = sigma*T_shell^4*(4*pi*(R_inner^2+R_outer^2)
This is incorrect. There is no net energy flow from the shell to the core. The shell is not a generator. With no net energy there is no power flow from the inner surface. Net energy transfer back is zero. At best what you add is thermal resistance (what the rest of the world call insulation). But you need to be careful here because there is something called a critical thickness. This is the thickness of insulation that actually increases heat transfer due to greater surface area.
There’s waaaay too much here to respond to it all, but let me pick a few random bits to reply to.
“But here is the crux of the matter – if the electron is already at that higher energy level then an incoming IR photon WILL NOT BE ABSORBED (at least not at the same frequency) – the photon will just travel straight through.”
NO! That is more or less what happens for visible light, but not for IR. IR absorption is due to quantized VIBRATIONS of the molecules, not EXCITATIONS of electrons. The molecule can have multiple units of vibrational energy, so even if a molecule is already vibrating, it can pick up additional vibrational energy by absorbing another photon.
“… because all (including me) are equally susceptible to mistakes in this regard. ”
NO! People with more education and experience and intelligence are less susceptible to making mistakes. Not everyone is created equal when it comes to analyzing science.
“Energy does in fact flow from the cooler to the warmer body, but at the SAME TIME, even MORE energy flows from the warm to the cooler body.
YES! As an additional example, consider heat conduction between two pieces of metal at different temperatures — which is really just the sum total of the energy transferred at a molecular level by collisions at the interface between the two blocks of metal. Some of the collisions will involve faster-than-average atoms in the cool block hitting slower-than-average atoms in the warm block. ENERGY can and will occasionally be transferred from the cool block to the warm block but such collisions, but more often the transfer will be from warm to cool. The NET transfer of energy (ie the “heat”) will always be from warm to cool.
“Still clinging on to the “cold objects can make warmer objects warmer still” garbage?”
I am (unfortunately) not amazed that people still cling to this objection. It is very seductive to people to don’t really understand thermodynamics or the statistical nature of the 2nd Law.
> “heat” does not move from cool to warm.
> “energy” can and does move from cool to warm (just not as much as moves from warm to cool).
Others have given other examples, but try this one. Take a shed and put a 1,000 W heater inside. On a very cold day (say-20 C) the interior might reach 10 C. Now put the shed in contact with other cold air (say -10 C). Now the interior might reach 15 C.
You can slice and dice and discuss this anyway you want, but the simple fact is that the presence of “cold material around the heated object” (at -10C) was an integral part of “warming” the interior of the shed. We used merely “sort of cold surroundings” instead of “really cold surroundings” and the result was a warmer interior.
Just like using “sort of cold surroundings” (the atmosphere) can warm the surface compared to having “really cold surroundings” (2.7 K outerspace).
“if I repeat this process n times, can we say that the t ° of the core can become infinite?
weird is it not?”
Yes, it is weird, but it is not wrong! There is is a significant limitation to the thought experiment that would prohibit infinite temperatures (besides the mere fact that the whole thing would eventually melt). The heater itself must continue to provide the required power even when the surroundings are hot — ie the heater itself must be HOTTER than the surroundings so heat can still flow from the hotter heater to the cooler surroundings. Limitations on the actual heater used would eventually impose an upper limit. For example, if the sun (5780 K) is used as the heater, then the surface could never get above 5780 K no matter what sort of insulating layers (or mirrors or lenses) you tried to use.
davidmhoffer says:
February 7, 2013 at 7:11 am
And how, exactly, do you think insulation works? It works by absorbing radiated energy and re-emitting some portion of it back.
As a general statement I disagree. Most insulation tries to prevent convection and conduction.
Regardless of the Woods experiment; or irregardless, as the case may be, The Steel Shell experiment Willis presents simply doesn’t work.
Willis asserts his nuclear powered planet (sans shell) emitting 235 W/m^2, is in equilibrium.
Not true, it is cooling down, and eventually be quite cold when the finite amount of radioactive materials has all decaysed. OK so it may be a long time.
So now add the steel shell, and wait for it to reach equilibrium, which Willis asserts it does.
Well now we know for sure that it is not in equilibrium, since it is not all isothermal. The shell is presumably cooler than the planet. Furthermore, if it WAS in equilibrium, then Kirchoff’s Law would apply, and the shell would be absorbing and receiving the exact same spectrum of whatever radiation was in the cavity between the planet and the steel shell.
So a fundamental assumption of the experiment cannot be met.
Whatever the radiation flying around this shell game, it certainly is not black body radiation, so SB law and Planck’s radiation law do not apply here.
The Planck derivation was for the radiation escaping from a small hole in a THERMALLY ISOLATED ISOTHERMAL cavity, from which NO energy can escape through the insulating walls.
Steel of course is not a black body radiator, nor is any other known material.
The electromagnetic radiation spectrum, extends from down to but not including zero frequency (DC) and up to but not including infinite frequency, and a black body must perfectly absorb ALL of that radiation completely.
NO physical material has zero reflectance for EM radiation; that would reuire it to have unity refractive index like the vaccuum, and no material does.
Black body radiation itself is fiction, in fact it is fictionally fictional. It involves the physical properties of no physical material whatsoever. No electronic energy states, or any other trappings of quantum mechanics of real materials is involved in the derivation of the black body radiation formulae.
It is a quite hypothetical radiation spectrum that does not and cannot exist, and the knowledge of no material properties of real physical materials, was used in its derivation.
Yes it is an extremely useful and improtant result, but Willis’s shell planet, is not emitting BB radiation.
@davidmhoffer: February 7, 2013 at 5:34 am
Well davidmhoffer, you seem so ripe for being sold one of these new inventions today, run out and talk to one of the I.P.C.C. salesmen, get yourself one! I know, not quite as nifty as when you were heating your home with the backradiation from snow or you were staying so toasty warm in an igloo with all of that parabolic backradiation, but close.
http://www.ilovemycarbondioxide.com/IPCC_oven.HTML
☺
Seriously davidmhoffer, you and Willis have got some big problems here.
The shell as shown is outputting more energy, 470‹J/s/m2›, per time per area than is even available from the nuclear source in the first place, there is only one energy source in Fig. 2 and it’s maximum output through the sphere’s surface is 235‹J/s/m2›. Ever heard of the 1st law of thermodynamics? Don’t you see your problem? This is getting embarrassing for W.U.W.T.
Ever heard that an energy source can never heat itself?
I think now you too have mixed possible power with flux.
But this does shine a bright light on exactly where the Stefan-Boltzmann equation is being used and misapplied in climate science. Reflected energy from a source, in any manner, can never then be added back to the source’s original hypothesized flux by SB and then fed back recursively into the SB equation to hypothesize a new higher temperature of the source itself. This is where the I.P.C.C.’s Free Energy Oven fails.
This is also where Trenberth slips in that extra 23‹W/m²› in his “global” energy budget to boost the “backradiation”.
It is like taking a sheet of polished aluminum foil and holding it up to a wall radiating by SB per T at 400‹W/m²› and 400‹W/m²› is reflected back to the wall by the foil at the temperature of the foil so now you have 800‹W/m²› at the wall so of course the wall’s T must rise… if you can really see that this happening and being something physically real then, rather than trying to teach radiative energy science to all of the scientists and engineers coming here and to comment, maybe you should just stick to selling computer systems and Willis to construction.
However, David, if a cloud or the atmosphere gases absorb solar energy that came from an energy source external to the Earth’s climate system, the sun, then that energy can be deemed an original energy source in the system and that energy can be handled isotropically with down welling LW radiation that can raise the surface’s temperature if it ever increased.
I myself have been wrong in this respect too in the near past, to TFK09’s energy budget graphic, the only down welling source radiation is one-half of the 78, or 39, absorbed by the atmosphere and clouds plus the 161+1, or 162 absorbed by the surface. That gives the surface 200‹W/m²› of absorbed energy. From that the other one-half of the 78, or 39, goes to space plus the 98 sensible and evapotranspiration plus 102‹W/m²› net up radiated from the surface that gives 239‹W/m²› as expected, TFK’s one off. The net LW flux is the 102 from the surface, up, minus the one-half of the 78, or 39 down, giving the 63‹W/m²› net that you can see day in and day out of the net surface IR graphs at the ESRL.NOAA site for various locations. Finally even that now makes sense and balances and matches the data on the government climate and weather sites. I learn something new every day so I guess Willis’s so wrong post ends up with a good ending, for me anyway.
@tjfolkerts “Others have given other examples, but try this one. Take a shed and put a 1,000 W heater inside. On a very cold day (say-20 C) the interior might reach 10 C. Now put the shed in contact with other cold air (say -10 C). Now the interior might reach 15 C.”
This has absolutely nothing to do with the cold air making the warm air warmer!!! Once again this is due to conduction. Heat inside the shed will be conducted to the outside. The rate of conduction is dependent on the thermal gradient. The thermal gradient is dependent on the outside air temperature. Make the outside air warmer and the rate of conduction will decrease so the interior will be warmer. You can also slow the rate of conduction by covering the shed in rockwool. You cannot, however, significantly warm the shed by covering it with tin foil (although many a claim has been made that you can….)
I think it would be useful to back up one step and have -just- the argument about a hot plate and a not-so-hot plate. There are plenty of examples in textbooks to quote. Having the detailed discussion about how individual atoms in the “cold plate” can be hot and how individual atoms in the “hot plate” can be cold would be useful. Perhaps even a discussion over spectroscopy and how exceedingly hot things can -still- absorb IR photons – particularly if they’re molecules and happen to have vibrational modes as well.
And how many of the general thermodynamics rules are over NET energy transfers – while -statistical- thermo really lays out what temperature actually means atomically.
Until this particular discussion is firmly laid out, the entire “Steel Greenhouse” discussion inevitably latches onto misunderstandings.