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.
People are just not getting this are they? On one of the links I gave earlier ‘Science of Doom’ gives a possible explanation as to why.
MostCasualObserver says:
OK….but be aware that in this case the planet’s temperature will approach infinity because there is no heat loss.
Why? In fact it won’t; it will be at 253K and the surface of the planet will be much hotter (see later)
No it won’t; the outer surface of the shell will also be 253K (assuming it is a very thin conductive shell. For thick shells see http://scienceofdoom.com/2010/07/26/do-trenberth-and-kiehl-understand-the-first-law-of-thermodynamics/ )
Well, you are right about the 235 W/m2, but because you know this, the temperature of the sphere is not indeterminate. In fact you can work out precisely what it will be from the Stefan-Boltzmann Law. A blackbody which emits at 235 Watt/m2 must be at a temperature of 253K (we are told the shell is a blackbody). So the shell is at 253K.
We are told the planet is a blackbody so it doesn’t reflect anything. It absorbs everything.
Quote from article,
No, again from the Stefan-Boltzmann Law you can work out precisely how much. I leave that to you.
“Thanks, Will. Consider this question. Which one will result in a warmer earth, either looking up from the surface to see:
1. An sky which is radiating something like 300 W/m2 back to the surface from the presence of GHGs, or
2. Outer space, which is radiating … well … nothing back to the surface.”
Willis, consider this question. Why do you flip from one frame of reference to another half way through your question?
Earth as a whole v’s a localised phenomena. Surely you don’t think that the Earth as a whole is warmed by heat that is leaving the Earth via cloud formation do you Willis?
Surely even you can understand that this is merely localised warming resulting from global cooling?
Why do you have this tendency to double count?
Willis,
the flaw in your “shell game” is rather obvious. The shell never reaches the same temperature as the planet for the same reason that the surface of the sun is only 5,700º C while the core is 15,000,000º C.
The temperature sink of Space is the only true non-variable.
mkelly says:
February 8, 2013 at 11:04 am (Edit)
mkelly, I suggest you plot the temperatures starting with the surface at 235 W/m2 and the shell at 0 W/m2
At t1, the shell starts to warm. It gets to where it is radiating say 10 W/m2. That radiation goes both outwards and inward, so the surface temperature of the planet immediately jumps to the equivalent of 245 W/m2.
Ok, so by t2 the additional heat from the planet brings up the shell to radiate 20 W/m2 … and the surface goes to 255 W/m2.
I’m sure you can see the problem … the shell and the surface are never at the same temperature. They can’t be. Even if they were the same temperature when they were created, the surface would immediately start to warm because of the additional radiation coming back from the shell.
All the best,
w.
Alan B Eltor;
These men on this forum, – Willis, david hoffer, all the usual
“entropy calculations don’t tell the truth, charge is actually flowing backwards while all ‘odd’ charges are migrating to equalization, forward”
crowd –
>>>>>>>>>>>>>>
Like richardscourtney I really can’t figure out what the heck you are talking about. As for suggesting that entropy calculations don’t tell the truth, I challenge you to produce a quote by me that makes any such suggestion. Similarly, I said nothing about “charge”, nothing about which way charge flows let alone some being odd or not. You want to stick my name in a rant, by all means, but quote the specific words you are referring to, and explain what the terminology is that you are using.
jae says:
February 8, 2013 at 3:14 pm
Willis: You must have missed a couple of my comments. I’m not challenging the citations you provided.
>>>>>>>>>>>>>>>
You griped that neither Willis nor I had provided you the citations you demanded, Willis provided exactly what you asked for, and yet you still gripe about not getting the citations you demanded… but you aren’t challenging the ones Willis gave you. Are horses purple in your world?
richardscourtney says:
February 8, 2013 at 2:58 pm
Allen B. Eltor:
I am writing this as a sincere attempt to be helpful.
I have read your long post at February 8, 2013 at 2:28 pm three times, and I still fail to understand what you are trying to say. I suspect that if I don’t ‘get it’ then others also won’t.
Perhaps you are trying to say several things at once and they are getting jumbled up?
I commend you to make a shorter post which attempts to make a single point in as clear and concise a manner as you can. You can make subsequent posts when that issue has been addressed.
I hope this helps.
Richard
<<<<<Low energy distribution, aren’t telling the full story;
and that during the drift from unequal to equal energy density, there’s an entire undiscussed exchange of energy therefore energy charge, from the lower field, into the higher density one:
and that because mathematics can’t be found, describing this simultaneous and incalculable,
temporary reversal
of charge density distribution laws and mechanics means nothing to him. It can’t be proven it’s not true, so it’s not true, that energy MUST, ALWAYS, go from High->Low distribution when left to the devices of charge mechanics.”
That’s not science. That’s a mind adrift amid an ocean of radiative mechanical potentials, without the anchor of formal instruction on why energy can’t infuse light of a lower flux density, into interstice resonances that are already charged to a higher density.
It’s amateur hour on steroids and I can prove it.
Have him explain to you step by step about the magical sharing light bulbs,
that take each other’s light in, apply it to their own temperature, then give off a little of their own to the other bulb, each applying the others’ energy to raise their own temperature.
Try breaking the news to Willis and david hoffer and all these reverse-entropic, against-charge flow zombies that all the atomic interstices which could accept any energy, are already filled to higher energy density – charge – numbers.
===============================
You can’t charge a container with an average pressure of X pounds per square inch, then put on a hose, and insert air at LESS pressure.
================================
All entropic flow must obey classical pressure-differential calculations
in absolute algebraic terms. Period.
Without exception.
Or my electronic mathematics can’t predict and track it.
===============================
And mathematics can predict and track it,
as proven by the electronic radiation-age
Willis and david hoffer and all the other reverse-flow, against-charge, physics amateurs here
watch me, and the other radiative-energy management professionals around the world
make work.
Huffer:
Willis is responding as he should. With facts, reason, and integrity. YOU, on the other hand, have not added ANYTHING to this conversation, except insults. Fella, I’m beginning to think that that is because you CAN’T offer something of substance! Facts, not polemics, run science. Buck up!
REPLY: ya know, you aren’t blameless in this either with your own rhetoric. And if you want to get anywhere I suggest you stop insulting the man as “huffer” where his name is
HoffmanHoffer. I’m growing quite tired of it all.Take a 48 hour time out this time. There won’t be a third. – Anthony
I repeat my post from above, since Willis (and some others) seem to have missed it (or are ignoring it). I expect a response that is direct and not some type of strawman/sidestep/subject change. For once.
jae says:
February 8, 2013 at 8:40 am
“I suspect that no page numbers will be forthcoming to support the shell-game, because there is no support for the shell-game and its derivatives. I cannot weld steel with a propane torch. But according to the shell game, all I have to do is use 2 or 3 propane torches, adding their heat together, to do the trick. Sorry, no can do!!”:
I think it might help if people realized that it doesn’t matter how the shell is heated. That is, assume the sphere does not radiate energy outward. Assume it radiates at specific frequencies and is covered in a special reflective material that prevents outward radiation while allowing down-welling radiation to pass through. Now, also assume the sphere is connected to the shell by a conductor so all the energy has new route to get to the shell. This prevents the shell from overheating.
The bottom line is the equilibrium situation is exactly the same. The shell will still end up radiating 470 w/m2, half in each direction. Notice that the surface of the sphere now appears to the shell to be 0K as no radiation is being emitted. Hence, no seeming violation of the 2nd law while the surface warms just as it did in the case with outward radiation.
Now, what happens if we also coat the inside of the shell with our special reflective material? Since no radiation can be directed to the surface it must all go to space. The equilibrium will now be 235 w/m2 with no additional heating of the surface. Imagine that, it is the downward radiation that is heating the surface.
My cut and paste with WordPress always sucks as I’m putting posts together.
Hey, Anthony:
It is your blog, and what you do is OK by me. BUT, I want to say that you are not being fair at all here; you seem to be ignoring all the insults from Hoffer and concentrating on my replies only. For the record, you are being unfair. Twice, now!
Snip, if you wish, but you should still get the message, sir
AND, your “punishment” of 48 hours certainly does not achieve anything that I can see… I don’t give a “—-” about your punishment. Truth will prevail, censorship be damned. MORON.
Allan B Eltor;
You can’t charge a container with an average pressure of X pounds per square inch, then put on a hose, and insert air at LESS pressure.
>>>>>>>>>>>>>
I understand what you are trying to get at, but the analogy is not apt. Radiant energy is transmitted as photons, which have no mass, and are neither a wave nor a particle, but share the characteristics of both. If you’ve ever done a ripple tank experiment, one of the things you might have observed is that waves travelling in opposite directions can and do pass right through each other. At the point of intersection, you do indeed get some interesting effects such as areas of complete calm, and standing waves. But then the ripples appear again on the other side, seemingly unaffected by each other (of course there are some effects, but minor by comparison) seemingly undisturbed. You can simulate this quite easily by throwing two pebbles into a pond at the same time and watching what happens to the ripples as they intersect than then continue on past each other. My point is that radiated energy shares some of these wave type behaviours, which is part of why radiated energy from two surfaces don’t interact as they pass through each other either.
“REPLY: ya know, you aren’t blameless in this either with your own rhetoric. And if you want to get anywhere I suggest you stop insulting the man as “huffer” where his name is Hoffman. I’m growing quite tired of it all.
Take a 48 hour time out this time. There won’t be a third. – Anthony”
LOL. Heil Hitler!
[Reply: Labeling Anthony a “MORON” and equating him to Hitler is immature. Anthony has put up with mmore from you than I would have. I doubt there is anyone here who has any sympathy for your juvenile behavior. And you are wasting your time trying to comment for the next two days. I will simply delete anything you try to post. — mod.]
REPLY to “JAE”
Since you’ve just called me a moron and equated me to Hitler, let’s just dispense with the 48 hours and move straight to a permanent ban.
I have no use for people that hide behind acronyms and hurl insults. Be as upset as you wish, but all further comments from you will be deleted.
Anthony
Allen B. Eltor
Allen, that statement you made, right there on pressure, lights up a concept that dates way back to college taking astronomy but some clarity on that topic of radiative pressure and radiation density has faded. I’ve looked for it on the web but to no avail.
Are you saying in so many words that radiation density has a pressure and through quantum wave interaction there is even a reflection component of radiation on radiation depending on that radiation density at any point in space? That is the gist of that faint thought that I seem to have learned long ago and let it fade. I believe it was in astronomy.
Could you expound on that a bit. Some here are following your statements closely.
Could this touch a bit on diametrically opposed anti-polarity quantum waves of like frequency and cancellation of effect?
Anthony;
REPLY: ya know, you aren’t blameless in this either with your own rhetoric. And if you want to get anywhere I suggest you stop insulting the man as “huffer” where his name is Hoffman.
>>>>>>>>>>>>>>
Ouch ouch ouch. First Willis calls me out for failing to provide cites and now our host gets my name wrong. My feelings are hurt. Wait… I forgot… I’m a salesman. We don’t have feelings.
I’ll try and tone it down Anthony. My frustration with the repetitive claims over the course of several threads is showing. Apologies.
REPLY: I’m sorry Mr. Hoffer. I work with somebody named Hoffman, so it is an innocent slip. I’m getting really tired of all the sniping, as it is a waste of everybody’s time, especially me and the moderators, who have to decide whether to pass it or not. – Anthony
jae;
I cannot weld steel with a propane torch. But according to the shell game, all I have to do is use 2 or 3 propane torches, adding their heat together, to do the trick. Sorry, no can do!!
>>>>>>>>>>>.
Correct, you can’t do that. But there’s no shell game to be had in this example. Flames from propane torches are burning gasses. Stefan-Boltzmann Law doesn’t apply easily to gasses. Photons emitted from one flame may well pass right through the other flame because the molecules are already in an excited state and unable to absorb another photon. Physical surfaces don’t operate like that. This is part of the confusion regarding the ghe of the atmosphere. As seen from space, the “temperature” of the earth is about -18 degrees based on the radiated energy and SB Law. But when measuring that radiated energy, all we can do is measure the total emitted. Some may have come from the top of the atmosphere, and some from the bottom, and some from anywhere in between. We don’t actually know where any given photon came from. So what are we measuring the temperature of? The system as a whole, not any given point in the system. There’s just no surface to measure!
So, for solid matter, such as the model Willis has used, SB Law is a relatively easy calculation. For two surfaces radiating toward each other, it is still an easy calculation. For two gasses from propane torches radiating toward each other, it is a devil of a calculation and I will freely admit I don’t know how. My point however is that not having surfaces, the “shell game” as you call it doesn’t apply in the first place because there is no “shell”.
However, if we change the analogy a bit, try and melt iron with coal. Pretty tough to do. Put the coal in a blast furnace though, and no problem. Yes, part of the heat from the blast furnace is due to the air being driven through the burning coal, but if you take away the walls of the blast furnace, the blast alone isn’t enough to melt the iron. Put the “shell” on, and presto, melted iron.
Willis Eschenbach says:
February 6, 2013 at 2:33 pm
If I light a candle on the earth during the day, the sun ends up warmer than it would be if I didn’t light the candle. Of course the reverse is true as well, the candle ends up warmer than if there were no sun. Since NET heat flow is from the sun to the candle, no thermodynamic laws are broken … but that doesn’t mean that the light from the candle is not absorbed by the sun. It is definitely absorbed, and the sun ends up warmer because of that radiation.
Please tell me you are not serious. Wrong on so many levels. But first let’s perform a simple experiment. Take an ice cube out of the freezer with a pair of tongs to hold it. Place your hand above the ice cube. Feel any cold radiating from it? No. Does the ice cube “warm” up your hand? Why then would a candle warm up the sun?
Your candle/sun example violates the Second Law; the general Clausius statement of it being: “No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature. EM radiation can travel from a colder body to a warmer one, but it will not be converted to thermal energy, as it must obey the Second Law. This idea of “net” heat flow is not supported. There are two distinct radiation beams; the one from the candle, and the one from the sun, and the Second Law must apply to each.
Here is a u-tube video of the kind ripple tank experiment I was talking about previously. The physicist even explains how the ripples in the water simulate certain aspects of light. As one can easily see, the ripples pass through each other while travelling in opposite directions.
EM radiation can travel from a colder body to a warmer one, but it will not be converted to thermal energy, as it must obey the Second Law.
>>>>>>>>>>>>>>>
By violating the First Law?
http://en.wikipedia.org/wiki/First_law_of_thermodynamics
http://en.wikipedia.org/wiki/Conservation_of_energy
David,
I am not going to read those references and attempt to guess your point. Please give me an explanation in your own words.
I am assuming you think the ice cube will warm my hand, and the candle will warm the sun? Yes, no?
ThePhysicsGuy says, February 8, 2013 at 8:33 pm: “David,
I am assuming you think the ice cube will warm my hand, and the candle will warm the sun? Yes, no?”
===========================================================
I am really not an expert on “mutual warming theory”, but there is an interesting implication. If the temperature of the Sun was constant, let us say due to the internal solar processes, then the Willis candle has warmed the Sun. Well, the warmer Sun in turn warms additionally not just the candle, but also the surface of the Earth. Now, the warmer surface of the Earth warms additionally the Sun. The Sun is pleased with the Earth warming it and warms the Earth further. The Earth is getting warmer and warmer this way, so is the Sun and we have global warming on both the Earth and the Sun.
ThePhysicsGuy says:
February 8, 2013 at 8:33 pm
David,
I am not going to read those references and attempt to guess your point. Please give me an explanation in your own words.
>>>>>>>>>>>>>>>>
Your contention that EM radiation can travel from a colder body to a warmer one, but it will not be converted to thermal energy would violate the First Law of Thermodynamics and the Law of Conservation of Energy. I’ve provided wikipedia links to both for your reference.
ThePhysicsGuy says:
I am assuming you think the ice cube will warm my hand, and the candle will warm the sun? Yes, no?
>>>>>>>>>>>>>>>>>.
Yes to both. The ice cube however will warm your hand a lot less than, for example, a puppy, and a lot more, for example, than dry ice, which would be still more than outer space. Your hand will warm the ice cube a lot more than the ice cube warms your hand. The candle will warm the sun. For the candle to NOT warm the sun would be a violation of the First Law, Conservation of Energy, and Stefan-Boltzmann Law. All of these laws can be reconciled one way, and one way only: By the Second Law referring to the net energy flux which can be determined by subtracting the SB Law flux of the cooler body from the SB Law flux of the hotter body to arrive at the net flux.
On the other hand, I think Willis picked a bad example in terms of the candle and the sun because verifying by experimentation would be…. problematic. But as far as the laws of thermodynamics go, yes.
With a name like “ThePhysicsGuy” I was hoping you would get the physics right, but alas you messed up here.
“Please tell me you are not serious.”
In Willis’ example of the candle and the sun, the warming effect of candle around the sun would be so infinitesimally small that it could never be measured even in principle. But is does exist.
Imagine lighting a gazillion candles around the sun, so that the entire sphere surrounding the sun was covered with candle flames. This would throw a lot of light back to the sun, and would warm the surface of the sun (or “slow the cooling” if you prefer — the net result would be a surface temperature higher than 5780 K). If you remove the candles one at a time, the surface of the sun would cool until you removed the last candle and the sun returns to the original 5780 K.
“Take an ice cube out of the freezer with a pair of tongs to hold it. Place your hand above the ice cube. Feel any cold radiating from it? No. Does the ice cube “warm” up your hand?”
To make this analogy work, the rest of the room would have to be EVEN COLDER than the ice — dry ice perhaps. Then the “warm” ice would indeed help keep your hand warmer than it would have been when surrounded by dry ice. Dry Ice at -80 C radiates only about 80 W/m^2 toward your hand, but the regular ice (0 C) radiates about 315 W/m^2. Reducing the heat loss would definitely make your hand get frost bite more slowly.
” “No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature.
Process = IR radiation.
IR radiation does not transfer heat (net energy) from the colder object to the warmer.
There is no violation.
PS. See my note about conduction above before objecting to the word “net”.
http://wattsupwiththat.com/2013/02/06/the-r-w-wood-experiment/#comment-1219292
PPS. If you really want to go deeper and decide if the word “net” is implied, then try the more fundamental statistical mechanics definition of entropy, and see if entropy of the universe increases when SOME energy moves from cold to warm as long as there is MORE energy moving from warm to cold.
davidmhoffer says:
February 8, 2013 at 5:24 pm
“Like richardscourtney I really can’t figure out what the heck you are talking about.”
…. Similarly, I said nothing about “charge”, nothing about which way charge flows, let alone some being odd or not. ”
————————
You talk so much you don’t know WHAT you said.
” davidmhoffer says:
February 7, 2013 at 7:28 pm
“…In order for Stefan-Boltzmann Law and the Second Law to both be true, the second Law must refer to the net energy flux, and the energy flux must be two way (meaning from hot to cold and cold to hot at the same time).”
———-
So much for your word being worth the time it takes to read it.
———
You’re such a grasping amateur you don’t even know all energy computation is done in pure algebraic terms; that if it says one-way flow, then your false claim,
*of reverse undocumented, incalculable flow *
is you, the turbo-posting amateur, on the internet;
pretending you know things are happening that mathematics can not express.
To me.
Electronic energy in an entity is potential heat; heat in electronic engineering, is disruption, spurious emissions, and death to quality signal processing and radiative emissions;
so don’t try to claim you’re talking about “other radiative transfer”.
I calibrate analyze & maintain some of the most sophisticated RADIATIVE TRANSFER APPARATUS in the HISTORY of HUMANITY.
I earned that lofty position doing something called “Knowing the right answers when amateurs get confused and think energy spontaneously flows backward.”
I have my bona fides, that precious lambskin: after completion of four years’ BLISTERING work in 3-1/2, in the radiative transfer principles of everything mankind ever witnessed.
Not that he ever built. That he ever witnessed.
And no one ever witnessed “reverse flow undocumented by the calculations.”
There’s only you, and a very few other people –
who, incidentally are connected with the most mindlessly stupid, & egregious scientific postulates, maybe in the history of humanity –
making claim that “energy flows both ways”, when mathematics says CLEARLY: it’s flowing one.
———-
We have incredibly sensitive instruments; and incredibly innovative ways of deploying them, to trigger when previously unnoticed, or undesigned-for phenomena, show themselves.
Not one of the men I have ever worked with, has EVER said,
“Did you know that in spite of charge density demands of pressure-gradient energy transfer principles,
energy actually flows AGAINST the algebraic mandate?”
Not once.
Just you, the amateur on the internet,
and some crooked government employees whose science is notoriously bad.
I seeas I’ve returned you’re STILL turbo-posting, affecting pretense we’ve had some conversation about dropping rocks in swimming pools.
You just never know when to stop because you’re behind.
I’ve seen other working radiation-transfer professionals be horned out because turbo-posting amateurs can’t be confronted with the fact they laid claim to having special elevated knowledge
of incalculable and undocumentable –
and unmeasureable for that matter, “reverse” or “two-way flow” against CLASSICAL pressure-gradient energy migration demands.
You might think because you talk more than nearly anyone, the working professionals in radiative emissions are intimidated. I’m certainly not.
I could take you into the lunch room and let you tell that
“it’s incalculable and unmeasurable, but if you really understood, YOU’D SEE like ME: there’s magic BACKWARD flux-density assignments! REALLy! That’s the only way, all the EQUATIONS work out!”
You’re an embarrassment to me, and to my kind: the radiative transfer & emissions
expert professional kind.