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pschopf at 08:58 AM on 17 July 2014Atlas of Mortality and Economic Losses from Weather, Climate and Water Extremes
Small point: The WMO is the World Meteorological Organization - not Association.
Moderator Response:[JH] My bad. Fixed. Thank you.
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DSL at 08:35 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
And again, Devon, the question Postma refused to answer: is the energy being radiated from Object A (at 10C) absorbed by Object B (at 50C)?
As soon as we start talking about "heat," the discussion gets silly. The term "heat" needs to be described in terms of energy exchange. The term "net" needs to be incorporated. The rhetorical ploy is to try to push the "argument" into terms that the general public understands: "cold things don't warm hot things." The reality is that all things absorb/emit radiation. That is, "all things exchange energy." Thus, cold things do indeed supply energy to warm things, but the warm things get rid of more energy than they absorb.
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DSL at 08:29 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
Devon, do all things above absolute zero radiate?
If so, then all such things are both emitting and absorbing radiated energy. "Cold" things emit less energy than "hot" things, but all things are relative, and all things absorb energy.
A block of ice is radiating energy (else you wouldn't be able to see it). A block of red hot iron is radiating. Put the two in the same room. Now put an identical block of red hot iron in an identical room but without the ice. Which block of iron cools down more rapidly?
It's not that the room is heating the iron above its initial temperature. Rather, it is decreasing the cooling efficiency of the iron relative to the situation with the block of ice.
In the same way, the atmosphere is absorbing energy radiated by the surface. Some of that radiation is then radiated downward, back to the surface. The surface is not a static energy source (well, the geothermal component is, but it's relatively small). It is continually (except at night) warmed by the sun. Energy once emitted by the surface returns to the surface and brings it to a temperature greater than solar energy alone would produce. Keep in mind that downwelling radiation has been measured from the surface directly for about twenty years. It's real. The atmosphere is radiating toward the surface. -
devon at 07:31 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
Could anyone give a real life example of cold heating hot, or energy being created from nothing as the K&T energy budgets suggest.
So, if cold could in fact heat the already hot, then this would have been established as a theory and a universal fact of physics. I would invite someone here to give a real life example! -
KR at 05:20 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
Energy can't flow from a cold source to a warmer object? But most of the world around me is cooler than my body temperature, and that means Aaaahhh, I'm blind!
Utter nonsense, Postma, utter nonsense.
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CBDunkerson at 05:04 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
Of course energy can't flow from a cold source to a warmer source... I mean, that's why when you open a freezer the light from the bulb in the back can't travel from the cold area at the back of the freezer to the warmer area of your eyes. :]
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Dikran Marsupial at 03:54 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
"It can be misinterpreted by inferring that a cold source can heat a warmer source. "
This essentially explains why JPostma wouldn't give a direct answer to my question, but it isn't the one he thinks it is. If some proposition X can be used to derive some conclusion Y, then if one accepts X then one must also accept Y, unless they can find an error in the chain of reasoning that links X to Y. However, the possibility that there may be a specious chain of reasoning that links X to Y is not a good reason to refuse to explicitly state that you accept X if you actually do. Refusal to do so is basically cowardice, not being confident that you will be able to demonstrate an error in the later chain of reasoning.
Unfortunately I suspect JPostma knows at some level that if he maes a genuine attempt to engage in the thought experiment, he might not be able to spot a flaw in the chain of reasoning, but is unwilling to accept that his understanding of Y might be incorrect.
This is why I often try to ask skeptics a series of carefully posed questions that are designed to define their position as clearly and unambiguously as possible, so that if they are right, we can all see if, and if they are wrong, then that will also be obvious. However in my experience, this is generally met by evasion. JPostma has provided a particularly clear example of this.
Moderator Response:[TD] I believe Robert Murphy has asked the next question that Dikran would have asked. JPostma, please give a simple yes or no answer to that question, and only then give any explanation that you'd like to give.
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John Hartz at 03:50 AM on 17 July 2014Rupert Murdoch doesn't understand climate change basics, and that's a problem
Also see:
Scientists Take Issue With Rupert Murdoch’s Remarks on Climate Change by Eric Roston, The Grid/Bloomberg News, July 16, 2014
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Robert Murphy at 03:47 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
Mr. Postma, when the photon from the cooler source hits the warmer source (which you acknowledge does happen), does the warmer source gain energy? Does the photon have any effect on the warmer body?
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Rob Honeycutt at 03:45 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
"It is unfortunate that so much time has to be wasted on BS like this."
Unfortunate, but also illuminating to watch cognitive dissonance on steroids.
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Dikran Marsupial at 03:42 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
"It can be misinterpreted by inferring that a cold source can heat a warmer source. "
O.K., give me an example of how that can be done.
BTW, the net flow of heat energy is always from the warmer object to the cooler, so it is pretty irrelevant anyway.
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JPostma at 03:39 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
"give me an example of how it could be misinterpreted"
It can be misinterpreted by inferring that a cold source can heat a warmer source. Of course, this is why the heat flow equation must be referenced with the answer, so that such misinferrence doesn't arise.
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KR at 03:39 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
Postma provides a great demonstration of denial here, in particular of the first law of thermodynamics, conservation of energy.
At equilibrium, which may include averaged short term variations around that equilbrium, the energy leaving a mass (surface, atmosphere, red-faced denier) equals the energy entering that mass (photons, convection, junk food). Input = output, in other words. If there is an imbalance, as with the current climate and radiative imbalance, internal energy will change and with it temperature.
The lowest radiative input to a mass, the lowest contribution to that masses temperature, would be the mass surrounded by absolute zero, with the 3K universal background radiation of deep space a close second.
Any object warmer than absolute zero, regardless of whether it is warmer or cooler than that mass, radiates photons according to its emissivity, and any of those photons absorbed by the mass in question increase it's energy. That added energy makes the mass warmer, until (via radiation, convection, conduction) the increased temperature causes the mass to lose energy at the same rate it gains it.
Radiation from anything above absolute zero will warm an object compared to its absence - and the warmer the radiating object the warmer the receiving object will become, no matter what their relative temperatures are. The starting point is zero, and even a relatively cool radiator provides a non-zero addition to nearby masses energies. Cool objects do, indeed, cause nearby warm objects to rise in temperature.
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Basic physics. Which Postma denies. His arguments are meaningless, Q.E.D. Not surprisingly, even many of the pseudo-skeptics disavow Postma and the 'Slayers', finding their imaginary physics an embarrassment...
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PhilippeChantreau at 03:37 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
It is unfortunate that so much time has to be wasted on BS like this.
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JPostma at 03:36 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
"give me an example of how it could be misinterpreted"
It will be misterpreted precisely by avoiding the heat flow equation, and thus thinking that perhaps the cooler object heats the warmer object because the cooler emission "hit" it. But of course, this is why it is important to reference the heat flow equation.
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JPostma at 03:34 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
More than avoid ambiguity, which is what would occur with a simple "yes or no" answer, I have provided the heat flow equation and some discussion of it to help clarify the fact that opposing emission is what determines equilibrium, and that the warmer object will not be heated, i.e., raise in temperature, due to the cooler object. I am sorry that I can not provide ambiguous answers, and that my explanation of the answer seems like ambiguity to you. But it is simple: cold does not heat hot, opposing emission determines equilibrium, etc.
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Dikran Marsupial at 03:31 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
JPostma writes "The problem is that answering only with a "yes or no" leaves the door open to misterpretation and sophist obfuscation of the physics"
Sorry, but that is hilarious. An answer of "yes" to the question "Does A emit any photons that strike B, "yes" or "no"?" leaves the door open to misunterpretation? Give me a break! Better still, give me an example of how it could be misinterpreted!
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JPostma at 03:29 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
"So, Mr Postma, for the last time, in the thought experiment described above, does object A emit any photon that strikes object B? Anything else than yes or no will be considered obfuscation by me."
The problem is that answering only with a "yes or no" leaves the door open to misterpretation and sophist obfuscation of the physics, because without the heat flow equation the meaning of the answer can be interpreted arbitrarily. Thus, I include the heat flow equation to help make it clear what the result of the physics is: heat not flowing from cold to hot, which means the warmer source does not increase in temperature.
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PhilippeChantreau at 03:26 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
I see the answer cam while I was writing my post. That yes has numerous consequences, which essentially negate your argument. "Following it up with Physics" is a fancy name for obfuscation. The condescending tone is rather amusing. I'm done here, that yes was all I needed to measure the depth of Mr Postma's need to receive Physics education, not give it
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Dikran Marsupial at 03:25 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
JPostma, I had repeatedly asked you just to reply "yes" or "no" to avoid all ambiguity, you chose not to, even when doing so would be a tacit admission that you were being deliberately obstructive. I am always happy to discuss science, and be proven wrong if that is what I am, however I don't have endless enthusiasm for obfuscation, and I don't think it unreasonable for people to give a direct "yes" or "no" answer when explicitly asked for it.
You have missed your chance to clearly explain your position. My thought experiment was designed to very clearly distinguish between two possible explanations of your position, but you chose not to co-operate.
Being willing to give direct answers to direct questions should be taken for granted in a scientific discussion, where the aim is to determine the truth. The fact that you needed input from the moderator should tell you that your behaviour was unproductive. Note the moderator did not delete your post, just gave advice.
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JPostma at 03:25 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
Q = sigma*(Th^4 - Tc^4) shows quite precisely that emission from a cold source does not heat up a warmer source. The emission from the cold source is what balances with the emission from the warm source in order to determine equilibrium - the emissions cancel out, leaving equilibrium, whether or not we wish to discuss whether the photons from the cold source "strike" the hot source...there is a greater physics occuring than that, as found in the heat equation, which shows how emission actually balances or cancels out at equilbrium.
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PhilippeChantreau at 03:22 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
This is clear indeed, and rather boring. This is just a repeat of G&T nonsense and playing on words to spread confusion. We've talked at length about this years ago. Net heat transfer is different from energy transfer. Energy, under the form of photons, can be transferred from any object to any other object, regardless of temperature, as long as it is above absolute zero. Net heat, which in Thermodynamics is usually referred to as just "heat" can only go from hot to cold.
Postma's condescending tone does not match his unability to answer the very simple question posed by Dikran. Thers is no need to give additional information or "drown the fish" in a sea of words. The thought experiment described by Dikran is simple. The question is does A emit photons that strike B, not does A heat B or does A emit photons. The answer from Postma should be yes or no.
So, Mr Postma, for the last time, in the thought experiment described above, does object A emit any photon that strikes object B? Anything else than yes or no will be considered obfuscation by me.
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JPostma at 03:20 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
Emission from 'A' in all directions equates precisely to 'yes' how is that not clear? But it is of course important to follow that up with some physics, such as the Stefan-Boltzmann Law and heat flow equation, which indicates that emission from the cold object will not warm the warmer object.
I find it surprising that there is an assumption that Dikran is the director of this conversation and that I must state answers only in the way he or she desires them - there is of course no point in conversation if my role here is only to say what Dikran requests of me.
I am dissapointed both at Dikran and the moderator for thus prejudicial behaviour. I had thought to be free to discuss physics.
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Dikran Marsupial at 03:15 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
JPostma, it is a shame you take that attitude. Had you just replied "yes", you would have doen yourself a big favour as it would have revealed where an apparent misunderstanding about your position has arisen. Your loss entirely.
"yes" is only three letters, you should ask yourself why you were unwilling to write them, rather than carry on with this obstructive behaviour.
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JPostma at 03:10 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
I have precisely stated the answer for you Dikran. This is a silly game you play...as if you wish to move my mouth for me, or some dominance thing. It is silly and not necessary.
'A' emits photons in all directions. However, emission from 'A' does not heat a warmer object, as the heat flow equation shows that heating only occurs from hot to cold.
I am sorry if you wish to attempt to find a different answer than this, but this is the answer. You can not deny the Stefan-Boltzmann Law heat equation. Are things clear for you yet, on the physics? Do you wish to rather simply state your desired end-point, rather than directing me how to move my mouth for you?
Moderator Response:[TD] JPostma, you are refusing to answer simple questions with simple answers. You are merely repeating your statements regardless of the question, which constitutes sloganeering, which violates the rules of engagement for this site.
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Dikran Marsupial at 03:04 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
JPostma, I'm sorry, you are being deliberately obstructive. I could not have taken more effort in posing the question for you in a way that made a simple, direct, unequivocal, unambiguous answer possible, but yet again you refuse to do so.
Here is your last chance to show that you are genuinely engaged in this discussion and are not just obfuscating:
Does A emit any photons that strike B, "yes" or "no"?
If your answer is anything other than "yes" or "no", it will be a tacit admission on your part that you are being deliberately obstructive.
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JPostma at 03:03 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
See the later paper here @PC. A higher temperature than the maximum solar insolation temperature is not observed...which means that back-radiation/trapping does not contribute to surface heating.
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JPostma at 03:00 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
I am offering you as much information as I possibly can. The answer is stated precisely by the heat flow equation which I have given you.
Let me repeat:
"As the example you presented was of a sphere in space, then the radiant emission power is given by P = A*sigma*T^4, and this goes out into space in all directions, as a uniform wavefront basically."
and
"We simply apply the heat flow equation using the Stefan-Boltzmann Law. The result is that heat flow is from hotter to cooler, of course. Thus, the cooler object heats up. The cooler one does not heat the warmer one, of course, and the equation for heat flow obviously does not imply this. Equilibrium is given when the respective emissions cancel eachother out, as in Q = sigma*(Tf^4 - Ti^4). Photons are of course emitted and the equation does not say that no photons are emitted from the cooler object, as you have attempted to conjecture. It is the balance of emission which determines equilibrium, not lack of emission."
'A' emits photons in all directions. However, emission from 'A' does not heat a warmer object, as the heat flow equation shows that heating only occurs from hot to cold.
I do sincerely hope this is clear for you.
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PhilippeChantreau at 02:56 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
The link to J.Postma's paper in the OP appears broken. Considering how confident Mr Postma is of the need for everyone here to receive his teaching on Thermodynamics, I was curious to take a look at it and check the science publication where it appeared.
I am very surprised by this statement: " the expected results of the artificial radiative greenhouse effect, using any argument or model for it, do not manifest, thus confirming their artificiality."
The radiative transfer models do predict with remarkable accuracy not only the downwelling IR at the surface (both quantitatively and in wave length distribution) of which real time measurements are easily accesible for some locations, but at various other altitudes too. What data is there to support the statement that the expected results do not manifest?
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Dikran Marsupial at 02:54 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
JPostma, yet again you avoid giving a straightforward "yes" or "no" answer, so I shall rephrase it.
Does A emit any photons that strike B, "yes" or "no"?There can only be two answers to this question, either it does or it doesn't, so any unwillingness to give a straight "yes" or "no" will be an indication of deliberate obstruction on your part, as far as I am concerned.
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JPostma at 02:52 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
Again I do apologize for answering with more information, but it is necessary to simply and kindly provide clarification of the physics, such that some erroneous interpretation that a cold source can heat a warm source is prevented from being assumed or inferred, which would of course be ridiculous in any case.
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JPostma at 02:49 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
We simply apply the heat flow equation using the Stefan-Boltzmann Law. The result is that heat flow is from hotter to cooler, of course. Thus, the cooler object heats up. The cooler one does not heat the warmer one, of course, and the equation for heat flow obviously does not imply this. Equilibrium is given when the respective emissions cancel eachother out, as in Q = sigma*(Tf^4 - Ti^4). Photons are of course emitted and the equation does not say that no photons are emitted from the cooler object, as you have attempted to conjecture. It is the balance of emission which determines equilibrium, not lack of emission.
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Dikran Marsupial at 02:43 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
JPostma, please give a simple "yes" or "no" answer to questions where this is possible, in order to avoid ambiguity. I will take that answer as a "yes".
Now consider a second black body object "B" which is introduced into the scenario. B is a plane of infinite extent and infinite conductivity (so it is the same temperature everywhere). B is maintained at a temperature fractionally higher than that of A. B is placed to the right of A, and in close proximity, but not actually touching, so the only form of heat transfer possible will be radiative.
Now, as B is warmer than A, is it your contention that A will not radiate any photons that will strike B (i.e. A will only emit photons to its left and none to its right)?
This is a question where a "yes" or "no" answer is possible. If "yes", then no further comment is necessary as we understand eachother; if "no", please clearly state that your answer is "no" and then provide an explanation.
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JPostma at 02:33 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
As the example you presented was of a sphere in space, then the radiant emission power is given by P = A*sigma*T^4, and this goes out into space in all directions, as a uniform wavefront basically.
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JPostma at 02:31 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
Of course, thermal radiant emission goes out in all directions not otherwise blocked to it. I am glad that we can state such basic physics concepts.
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Dikran Marsupial at 02:26 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
JPostma wrote "even though I have statedd "yes", by supplying the actual Stefan-Boltzman equation, which of course you had referenced initially"
supplying the Stefan Boltzmann equation does not equate to an unambiguous answer of "yes" as my question also involved the direction in which the photons were radiated, which is not covered by the Stefan-Botzmann law (only the power). So despite the amount of words that have been written, you still have not answered my question, which was:
Lets try an even more simple thought experiment. Consider a spherical blackbody object (A) at a temperature slightly above absolute zero, in a total vacuum with no other sources of radiation. Do you agree that the object will radiate photons in random directions according to the Stefan-Boltzmann law? A "yes" or "no" answer ought to be possible here, if "no" please state this explicitly and explain why.
Note I have highlighted the part of the question you have not yet addressed to make sure you don't miss it this time.
This is not a game (at least on my part). The reason for wanting a direct answer was to make sure there could be no misundertanding of your position. "yes" and "no" are utterly unequivocal, which is why I designed my question to be answerable with a simple "yes" or "no".
So please, for the record, is your answer to my question (including the bit about the direction in which photons are radiated) "yes" or "no"? -
JPostma at 02:20 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
Dikran, I am afraid that the equivocation is on your part, as you refuse to acknowledge the direct answer, and even the simple "yes" as you have requested it, which I have supplied to you. This seems to be some game you are playing now, and have been for some time in fact. I am sorry that I do not wish to play such a simple thing with you. I have in fact more fully answered the question than stating "yes", even though I have statedd "yes", by supplying the actual Stefan-Boltzman equation, which of course you had referenced initially. I am sorry if these satisfactions still do not fulfill you.
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Dikran Marsupial at 02:16 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
JPostma, why not simply reply "yes", rather than equivocate? You are giving the impression of trying to avoid discussing the thought experiment by being unwilling to give simple direct answers to simple direct questions. There is no need for equations for such a simple question where "yes" fully answers the question.
So, is it a "yes" or a "no"?
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JPostma at 02:14 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
I have used the Stefan-Botlzmann law to give you the answer, as you referenced it, so I am sorry if it was missed within the body of text.
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JPostma at 02:13 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
I did answer that precisely, in fact, to quote:
"An object radiates power P = A*sigma*T^4 where A is the surface area and T is the temperature. If you have two objects in a simplified geometry then you get the heat flow as the difference between their emissions, with heat flowing only from the warmer to the cooler."
Perhaps I am giving too much information, and it distracts you from being able to infer the answer. The answer, is of course, and I am sorry if you do not see the answer in the physics and in the math, that yes, and object radiates power as function of its surface temperature. I do hope this helps, and I hope that you can begin to recognize the greater answer in the math and physics.
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Dikran Marsupial at 02:09 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
"All of the questions have been answered. "
no, you have not answered mine, posed at 71, here it is again for your convenience
Lets try an even more simple thought experiment. Consider a spherical blackbody object (A) at a temperature slightly above absolute zero, in a total vacuum with no other sources of radiation. Do you agree that the object will radiate photons in random directions according to the Stefan-Boltzmann law? A "yes" or "no" answer ought to be possible here, if "no" please state this explicitly and explain why.
This ought to be uncontentious, so you should just be able to reply "yes", in which case we can move on to the next step.
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JPostma at 02:05 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
All of the questions have been answered. I am honestly trying to help you guys out, because it seems to be clear that a few basic thermodynamics and physics concepts might be unknown to you. I truly do not mean to sound patronizing or anything like that in making that statement, that is not my intention at all. So I do apologize for that. I have simply been assuming a certain level of ability to understand physics and thermodynamic concepts, such as the basic equation for heat flow under radiant emission, equilibrium conditions, etc.
The lump of iron is heated by the furnace. The iron does not heat the furnace. The heat flow equation shows exactly how and when the iron is radiating, as would the most basic conceptual understanding of the physics stated in words.
I hope this helped, and again, apologies for assuming certain things to be understood.
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Dikran Marsupial at 01:59 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
JPostma, I will not rise to your ineffectual attempts to be patronising ("Please let me know if you still require my help with this"), as I said the question is very easily answered under the modern statistical intepretation of thermodynamics, the fact that you can't (or won't) give a direct answer to MA Roger's question says it all.
Now, perhaps you would like to answer my question (it is the start of a thought experiment, but I thought it would be best to establish something we both ought to be able to agree on as a solid foundation)?
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JPostma at 01:55 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
I have just stated the answer, directly. I am sorry if you missed it again.
Please familiarize yourself with the heat flow equation Q = sigma*(Tf^4 - Ti^4). Is the answer still not clear? It also answers your latest question. Please let me know if you still require my help with this, and I will try to make it clear. An object radiates power P = A*sigma*T^4 where A is the surface area and T is the temperature. If you have two objects in a simplified geometry then you get the heat flow as the difference between their emissions, with heat flowing only from the warmer to the cooler. I hope this helps.
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Dikran Marsupial at 01:47 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
JPostma, repeating "The iron will radiate what is required of it" doesn't make it any less evasive.
Lets try an even more simple thought experiment. Consider a spherical blackbody object (A) at a temperature slightly above absolute zero, in a total vacuum with no other sources of radiation. Do you agree that the object will radiate photons in random directions according to the Stefan-Boltzmann law? A "yes" or "no" answer ought to be possible here, if "no" please state this explicitly and explain why.
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JPostma at 01:34 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
Here is the answer, in words: "The iron will radiate what is required of it". I am sorry, I thought you might appreciate the meaning of that, but I suppose I shouldn't do that.
So, if you look at the equation, Q = sigma*(Tf^4 - Ti^4), it is precisely clear when the iron radiates. It couldn't be any more clear. I am sorry if it is not clear to you. The answer is "as soon as the iron has a temperature". Thus, the iron can come to equilibrium with the source furnace, which is acheived when Q = 0, and of course, the heat flor equation shows that heat only flows from the furnace and into the iron while the iron is being heated; the iron will thus not heat the furnace, of course.
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Dikran Marsupial at 01:28 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
JPostma wrote "If you truly require me to answer "when" the iron starts to radiate, it is troublesome.",
Well yes, the fact that it is troublesome under your understanding of thermodynamics is exactly the point. It isn't at all troublesome for the modern statistical view of thermodynamics. The fact that you can't state the answer to such a straightforward question, and have to resort to such transparent equivocation, really says it all, and should give you pause for thought (to say the least).
Sorry again MA Roger, it was a very well thought out question. I suspect though that JPostma will be unable to see that the fact it is troublesome is a sign that his understanding of thermodynamics is decades out of date.
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JPostma at 01:24 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
"I'll end up using more energy to bring the frozen turkey to 60C than I will with a thawed turkey."
The cold turkey, of course, does not heat the oven. A colder turkey takes more time to heat, indeed, but neither a cold turkey nor a colder turkey heats up the oven.
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JPostma at 01:22 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
As words seem to creating only more confusion, perhaps the maths of heat flow will help? Heat flow in an ideal radiative situation is Q = sigma*(Tf^4 - Ti^4) where Tf is the temperature of the furnace and Ti is the temperature of the iron. Equilibrium is acheived when Q = 0, i.e. when heat flow equals zero, when there is no more heat flow. This means that the iron is not gaining or losing any energy but has constant energy. The end point is that Ti = Tf, the iron has come to the temperature supplied by the furnace. Of course, heat flow is only from the furnace into the iron. If the furnace was shut off and cooled faster than the iron, then the iron could heat the furnace somewhat and slow what would have otherwise been its cooling rate.
The question as to "when" the iron starts to radiate was what was precisely set up as the strawman. If you truly require me to answer "when" the iron starts to radiate, it is troublesome. But the answer is obvious, it has been stated, and the answer is specified directly in the heat flow equation here, and also by the simple understanding of heat flow and the ability to attain equilibrium.
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DSL at 01:16 AM on 17 July 2014Joseph E. Postma and the Greenhouse Effect
Joe: "The cold iron does not heat the furnace."
The cold iron does not radiate? If it radiates, then the net exchange is simply altered. The exchange never stops. Does the temperature of the iron in the initial state matter for the temperature of the furnace? Yes, Joe. Yes it does. A frozen turkey in my oven will be more of a net gainer of energy than a thawed turkey. I'll end up using more energy to bring the frozen turkey to 60C than I will with a thawed turkey.
In other words, it's all relative. The thawed turkey has heated the oven relative to a condition with a frozen turkey.
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