Guest Post by Willis Eschenbach
Pushed by a commenter on another thread, I thought I’d discuss the R. W. Wood experiment, done in 1909. Many people hold that this experiment shows that CO2 absorption and/or back-radiation doesn’t exist, or at least that the poorly named “greenhouse effect” is trivially small. I say it doesn’t show anything at all. Let me show you the manifold problems with the experiment.
To start with, let me give a curious example of the greenhouse effect, that of the Steel Greenhouse. Imagine a planet in the vacuum of space. A residue of nuclear material reacting in the core warms it to where it is radiating at say 235 watts per square metre (W/m2). Figure 1 shows the situation.
Figure 1. Planet in outer space, heated from the interior. Drawing show equilibrium situation
This planet is at equilibrium. The natural reactor in the core of the planet is generating power that at the planet’s surface amounts to 235 W/m2. It is radiating the same amount, so it is neither warming nor cooling.
Now, imagine that without changing anything else, we put a steel shell around the planet. Figure 2 shows that situation, with one side of the shell temporarily removed so we can look inside.
Figure 2. As in Figure 1, but with a solid steel shell surrounding the planet. Near side of the shell temporarily removed to view interior. Vertical distance of the shell from the surface is greatly exaggerated for clarity—in reality the shell and the shell have nearly the same surface area. (A shell 6 miles (10 km) above the Earth has an exterior area only 0.3% larger than the Earth’s surface area.)
[UPDATE: Misunderstandings revealed in the comments demonstrated that I lacked clarity. To expand, let me note that because the difference in exterior surface area of the shell and the surface is only 0.3%, I am making the simplifying assumption that they are equal. This clarifies the situation greatly. Yes, it introduces a whopping error of 0.3% in the calculations, which people have jumped all over in the comments as if it meant something … really, folks, 0.3%? If you like, you can do the calculations in total watts, which comes to the same answer. I am also making the simplifying assumption that both the planet and shell are “blackbodies”, meaning they absorb all of the infrared that hits them.]
Now, note what happens when we add a shell around the planet. The shell warms up and it begins to radiate as well … but it radiates the same amount inwards and outwards. The inwards radiation warms the surface of the planet, until it is radiating at 470 W/m2. At that point the system is back in equilibrium. The planet is receiving 235 W/m2 from the interior, plus 235 W/m2 from the shell, and it is radiating the total amount, 470 W/m2. The shell is receiving 470 W/m2 from the planet, and it is radiating the same amount, half inwards back to the planet and half outwards to outer space. Note also that despite the fact that the planetary surface ends up much warmer (radiating 470 W/m2), energy is conserved. The same 235 W/m2 of power is emitted to space as in Figure 1.
And that is all that there is to the poorly named greenhouse effect. It does not require CO2 or an atmosphere, it can be built out of steel. It depends entirely on the fact that a shell has two sides and a solid body only has one side.
Now, this magical system works because there is a vacuum between the planet and the shell. As a result, the planet and the shell can take up very different temperatures. If they could not do so, if for example the shell were held up by huge thick pillars that efficiently conducted the heat from the surface to the shell, then the two would always be at the same temperature, and that temperature would be such that the system radiated at 235 W/m2. There would be no differential heating of the surface, and there would be no greenhouse effect.
Another way to lower the efficiency of the system is to introduce an atmosphere. Each watt of power lost by atmospheric convection of heat from the surface to the shell reduces the radiation temperature of the surface by the same amount. If the atmosphere can conduct the surface temperature effectively enough to the shell, the surface ends up only slightly warmer than the shell.
Let me summarize. In order for the greenhouse effect to function, the shell has to be thermally isolated from the surface so that the temperatures of the two can differ substantially. If the atmosphere or other means efficiently transfers surface heat to the shell there will be very little difference in temperature between the two.
Now, remember that I started out to discuss the R. W. Wood experiment. Here is the report of that experiment, from the author. I have highlighted the experimental setup.
Note on the Theory of the Greenhouse
By Professor R. W. Wood (Communicated by the Author)
THERE appears to be a widespread belief that the comparatively high temperature produced within a closed space covered with glass, and exposed to solar radiation, results from a transformation of wave-length, that is, that the heat waves from the sun, which are able to penetrate the glass, fall upon the walls of the enclosure and raise its temperature: the heat energy is re-emitted by the walls in the form of much longer waves, which are unable to penetrate the glass, the greenhouse acting as a radiation trap.
I have always felt some doubt as to whether this action played any very large part in the elevation of temperature. It appeared much more probable that the part played by the glass was the prevention of the escape of the warm air heated by the ground within the enclosure. If we open the doors of a greenhouse on a cold and windy day, the trapping of radiation appears to lose much of its efficacy. As a matter of fact I am of the opinion that a greenhouse made of a glass transparent to waves of every possible length would show a temperature nearly, if not quite, as high as that observed in a glass house. The transparent screen allows the solar radiation to warm the ground, and the ground in turn warms the air, but only the limited amount within the enclosure. In the “open,” the ground is continually brought into contact with cold air by convection currents.
To test the matter I constructed two enclosures of dead black cardboard, one covered with a glass plate, the other with a plate of rock-salt of equal thickness. The bulb of a thermometer was inserted in each enclosure and the whole packed in cotton, with the exception of the transparent plates which were exposed. When exposed to sunlight the temperature rose gradually to 65 oC., the enclosure covered with the salt plate keeping a little ahead of the other, owing to the fact that it transmitted the longer waves from the sun, which were stopped by the glass. In order to eliminate this action the sunlight was first passed through a glass plate.
There was now scarcely a difference of one degree between the temperatures of the two enclosures. The maximum temperature reached was about 55 oC. From what we know about the distribution of energy in the spectrum of the radiation emitted by a body at 55 o, it is clear that the rock-salt plate is capable of transmitting practically all of it, while the glass plate stops it entirely. This shows us that the loss of temperature of the ground by radiation is very small in comparison to the loss by convection, in other words that we gain very little from the circumstance that the radiation is trapped.
Is it therefore necessary to pay attention to trapped radiation in deducing the temperature of a planet as affected by its atmosphere? The solar rays penetrate the atmosphere, warm the ground which in turn warms the atmosphere by contact and by convection currents. The heat received is thus stored up in the atmosphere, remaining there on account of the very low radiating power of a gas. It seems to me very doubtful if the atmosphere is warmed to any great extent by absorbing the radiation from the ground, even under the most favourable conditions.
I do not pretend to have gone very deeply into the matter, and publish this note merely to draw attention to the fact that trapped radiation appears to play but a very small part in the actual cases with which we are familiar.
Here would be my interpretation of his experimental setup:
Figure 3. Cross section of the R. W. Wood experiment. The two cardboard boxes are painted black. One is covered with glass, which absorbs and re-emits infrared. The other is covered with rock salt, which is transparent to infrared. They are packed in cotton wool. Thermometers not shown.
Bearing in mind the discussion of the steel greenhouse above, I leave it as an exercise for the interested reader to work out why this is not a valid test of infrared back-radiation on a planetary scale … please consider the presence of the air in the boxes, the efficiency of the convective heat transfer through that air from the box to the cover plates, the vertical temperature profile of that air, the transfer of power from the “surface” to the “shell” through the walls of the box, and the relative temperatures of the air, the box, and the transparent cover.
Seems to me like with a few small changes it could indeed be a valid test, however.
Best regards,
w.
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I have never actually read the exact (German language) statement by Rudolph Clausius, of the Second Law of Thermodynamics; perhaps one of our German posters can dig it up and cite it for us.
I’m under the impression that it translates as:-
“No cyclic machine may have no other effect than to transport heat from a source at one Temperature, to a sink at a higher Temperature.”
Our German friends can verify (or not) that for us too.
Well I don’t know what cyclic machine translates to in German, or what it means in that language, but to me, it implies that heat may move in both directions; or else all bets are off.
And it implies that something other than nothing must happen in addition to heat going from cold to hot; like some work being done for instance.
In any case, Clausius, a Thermodynamicist of impeccable credentials, also used the second law to derive what we call “The Optical Sine Theorem”, which basically says that no optical system can create an image that has a higher radiance than the source object. It also says that no optical system can form an image that is brighter than the image formed by an “Aplanatic ” system; that being a system that is corrected for both spherical aberration and coma, the two simplest of the Seidel aberrations of optical images.
Clausius’ thermodynamic limit also applies to “non-imaging” optics, which is the main discipline behind illumination engineering.
In any case, a one way system, where energy or (heat at least) cannot flow both ways, would be outside the purview of Clausius’ statement.
ThePhysicsGuy:
At February 9, 2013 at 2:54 pm you say
That shows such an immense ignorance of heat, EM radiation and the Second Law that it is mindblowing.
There is no flow of heat: there is a flow of EM radiation in two opposing directions. The energy of the EM radiation is converted to heat when it is absorbed by the surfaces.
I provide a brief explanation of this in my post at February 9, 2013 at 2:38 pm. Please read it.
Richard
” donald penman says:
February 9, 2013 at 10:13 am
What insulation is in contact with the planetary nuclear core? How is this insulation affecting the temperature of the planetary nuclear core? If we put extra mass directly in contact with the nuclear core this would also increase the temperature of the core because heat is lost more slowly from the core.If instead of adding more mass to make a shell you took some of the mass close to the nuclear core and moved it out until it formed a shell as described would the nuclear core not cool down and the energy lost to space increase.”
On Earth:
” Away from tectonic plate boundaries, it is about 25°C per km of depth (1°F per 70 feet of depth) in most of the world.”
http://en.wikipedia.org/wiki/Geothermal_gradient
So if you added 1 km of dirt to earth it would have some effect upon our core temperature- not 25 C, but rather some immeasurable amount.
But bury a box 10′ underground, then put the 1 km dirt above it, and eventually the temperature in the box would rise by about 25 C. And 2 km of dirt- 50 C, etc.
You already have 50 km of rock insulating the mantle temperature and more than thousand km insulating the core. So maybe if added as much as 100 km of dirt it could make noticeable affect on the core temperature- but adding this same amount of mass to the core, should make more of a difference in increasing the core’s temperature.
Earth does not radiate 235 watts per square meter at the surface, instead it’s about 1/10th of a watt per square meter.
And we presently know of no planet [or Jupiter moon] which radiates this much energy.
But it conceivable that planet without plate tectonics and with earth’s level of geothermal energy at various times could emit this much energy.
Or some think so:
“Venus has no evidence of plate tectonics, so this theory states that the interior of the planet heats up (due to the decay of radioactive elements) until material in the mantle is hot enough to force its way to the surface. The subsequent resurfacing event covers most or all of the planet with lava, until the mantle is cool enough for the process to start over.”
http://en.wikipedia.org/wiki/Geology_of_Venus#Global_resurfacing_event
It’s also conceivable that Earth even with it’s plate tectonics did at some point in history emit this level of energy [for a geologically brief time period]- and certainly did during Earth’s planetary formation period- it emit more than this amount of energy- for hundreds of thousands of years.
So life being something like 3.8 billion year old- life may have experienced times when average geothermal energy of Earth for hundreds or thousands of years was this high.
But even this would not be so dire as Hansen’s idea of what could happen- Earth becoming like Venus.
And it should be obvious that such an event would heat Earth more than a greenhouse effect from CO2 [and methane] could reasonably be expected to do- though such dramatic increase geothermal heat would also in addition release vast amount of CO2 and methane.
But for a planet to constantly emit this much energy for billions of years would require it to be quite strange.
@dikranmarsupial
As an aside, my vacuum flask is actually made entirely of steel (stainless, but still a bit magnetic) and the inside is a satin finish rather than highly polished. It’s incredibly effective and informs me that heat loss from radiation is a lot less significant than heat loss from conduction and convection. Near boiling inside, barely more than ambient temp outside. Very handy when out with my telescope under the stars, wishing that greenhouse gas theory was just a bit more warming than imagined.
Bryan;
Its as simple as that.
The cold object does not increase the temperature of the warmer.
>>>>>>>>>>>>>>>>>>>>
If you remove the cold object and replace it with “nothing”, what will happen?
Hint: The temperature of “nothing” is -273 degrees C.
It is as simple as that.
“dikranmarsupial says:
February 9, 2013 at 7:17 am
Will, no, they are not equivalent. The system will reach thermal equilibrium whether or not the vacuum is maintained.
The nuclear core generates energy at a constant rate, the outer shell radiates heat at a rate proportional to the fourth power of its temperature. Thus as the temperature of the outer shell increases, there will come a point where it radiates as much energy as the core generates and thermal equilibrium is reached. This is true whether the energy is transmitted from the core to the shell by radiation, conduction or convection, so whether the internal vacuum is maintained is irrelevant.
Willis is correct, there is no paradox, there is no double counting of energy.
ThePhysicsGuy says:
February 9, 2013 at 7:33 am
Willis Eschenbach says:
February 6, 2013 at 5:01 pm
Here’s an example to consider. Mount a light bulb socket on a table. Insert a 60-watt bulb. Measure the temperature, we’ll say it’s 150°C. Try a 100-watt bulb, it’s hotter, say 200°C.
Now, mount another socket right next to the first one. Insert a 100-watt bulb in one and a 60-watt bulb in the other. What will the final temperatures of the bulbs be?
I say both of them will run hotter in that situation, because A warms B, and B warms A
You also stated that, “..of course it has been proven, many times over. It’s in every college thermo textbook”.. In 1850 Clausius stated, “Heat cannot of itself pass from a colder to a hotter body.” That statement (or similar version) of the Second Law of Thermodynamics has been in every college textbook since that time. Yet you indicate that the colder light bulb will warm the warmer light bulb. So, what will happen with your version of thermodynamics? The 150°C bulb will cause the 200°C. bulb to become warmer. But now the warmer 200°C+ bulb (B) will then warm bulb (A) even more, with the cycle going back and forth until both bulbs thankfully explode, before melting a hole to China.
Care to explain yourself?
REPLY: Seems like an easy experiment to perform to find out. I think I’ll actually do this – Anthony”
It might warm the lightbulbs. But will it warm the filaments?
The lightbulbs will mostly warm due lack of convection.
Having cooler air or simply using a fan will do wonders cooling
the lightbulbs.
But can 60 watts make the 100 watt brighter?
Can two 100 watts make both brighter?
The filaments can increase the bulb temperature.
So if had two filament- one 60 w and other 100 w
in same lightbulb the bulb temperature should increase.
Since the filaments in above example would further apart it
would have less effect. But main affect seems to the inhibition
of convection- which the main way they cool down.
Or light bulbs in a vacuum should be hotter- but not
burn brighter.
Bryan wrote:
“The hot object will radiate more intensivly than the cold.
Both hot and cold will absorb each others radiation.
But the hot object will lose internal energy and its temperature will drop.
The cold object will gain internal energy and its temperature will rise.
Its as simple as that.
The cold object does not increase the temperature of the warmer.”
You’re almost there, Bryan – but you’re just-not-quite-thinking-it-through.
In your thought experiment, in which you acknowledge the hot object will absorb the radiation from the cold object, add another scenario where an 80C object is in isolation. In that isolation situation the 80C object will cool faster than the 80C object next to the 20C object. Why is that? How can it be otherwise than that the 80C object is heated by the radiation from the 20C object?
Not convinced? Change your 20C object to one at 79.999999999C and think about it further.
Bryan:
I am copying all your post at February 9, 2013 at 3:17 pm so it is clear that I am answering all of it.
Your ‘simplification’ causes confusion of a simple problem. Forget it and consider the following when the energy source exists.
Firstly, in the considered cases of the hypothetical planet and the real Earth they have a source of energy input: i.e. the radioactive reaction heating of the hypothetical planet and the Solar energy heating of the Earth.
So they don’t cool from the radiation they emit. Any loss of “internal energy” is replenished by the energy source.
Secondly, you admit that “Both hot and cold will absorb each others radiation” so you admit that they each gain an additional energy input from each other.
The hotter object thus gets the energy input from the energy source AND the energy from the cooler object. So, the hotter object obtains an increase to its energy input (because its input from the energy source does not change but it also gets the energy from the cooler object). This additional energy increases its temperature until it gets hot enough to radiate away ALL the energy it is obtaining.
In other words, and using your phraseology, the cooler object heats the hotter one.
Richard
Anthony,
May I propose an additional twist to your light bulb experiment?
For those who insist that energy flux is a one way street, could you please get one of those 5,000 watt spot lights? If they are correct, you ought to be able to shut off the 60w and 100w light bulbs simply by pointing the 5,000 watt spot light at them.
Willis Eschenbach says:
February 9, 2013 at 2:37 pm
ThePhysicsGuy says:
February 9, 2013 at 7:33 am
Willis Eschenbach says:
February 6, 2013 at 5:01 pm
For Anthony, let me propose one change in the experiment. Take the measurements as the bulbs cool. This avoids the issue of the changing resistance of the filament of the bulb with temperature, and changing electrical power draw and the like. So … three stages.
Wiilis the best way to do the experiment might be to use dichroic halogen bulbs since they are designed to heat up the filaments with IR. It would depend on the exact orientation of the dichroic coating though.
Use two “Precise™ MR16 lamps” When they face each other they should each heat up and increase their respective light outputs. I’d suggest using a narrow angle lamp facing a wide angle one and the wide angle one should get brighter if I understand their tech drawings correctly. By varying the respective lamp powers you could get some good data.
Richardscourtney
Lighten up dude. No reason to be insulting. In using the term “flow”, I was restating what Willis said. “Flow” is just another term for “flux”, It is not uncommon to see the term “heat flux” used in radiation physics. Heat is transferred from the warmer body to the cold by thermal radiation, the radiation being absorbed by the the cooler body and converted to heat. I know the basics of heat transfer and EM radiation.
I am not the one who made the following statement:
“Greg house had claimed there was no “experiment” which shows a cooler object warming a hotter object. I pointed out that I own an apparatus – a microwave oven – which does exactly that by the action of a flow of energy providing electromagnetic radiation to the hotter object.”
The “experiment” is supposed to be in context of the 2nd Law. You can’t add an external source of “work” to it per the 2nd Law. So a microwave, oven, and refrigerator are incorrect examples.
Please go Wikipedia, look up Second Law of Thermodynamics, and read the portion under Clausius Statement for further clarification.
Phil. says, February 9, 2013 at 2:57 pm: “I refer you to the dichroic halogen lamps which improve the efficiency of the lamps by selectively reflecting IR light from the envelope to the light source, thereby increasing its temperature and increasing the proportion of visible emitted. It’s a perfect illustration of the effect discussed in this thread, the product can be purchased in stores,
http://www.gelighting.com/LightingWeb/emea/images/Halogen_MR16_IR_Lamps_Data_sheet_EN_tcm181-12732.pdf”
==============================================================
On this thread above we have the Wood experiment refuting the notion of reflected IR being capable of warming the source. The result in the Wood experiment was zero or negligible.
I can understand the manufacturer though, because an allegedly more efficient lamp can be better sold. I do not necessary think that the manufacturer intentionally lied, because this tale about “back radiation warming” has been around for many decades, so why not “use it”.
ThePhysicsGuy says:
February 9, 2013 at 4:26 pm
Richardscourtney
Lighten up dude. No reason to be insulting. In using the term “flow”, I was restating what Willis said. “Flow” is just another term for “flux”, It is not uncommon to see the term “heat flux” used in radiation physics. Heat is transferred from the warmer body to the cold by thermal radiation, the radiation being absorbed by the the cooler body and converted to heat. I know the basics of heat transfer and EM radiation.
In which case you will be aware that thermal radiation is also transferred from the cooler to the hotter, that radiation being absorbed by the the hotter body and converted to heat also. The only way around that being if the hotter ‘knows’ that a particular photon incident on it came from a colder source and ‘refused’ to absorb it!
Greg House says:
February 9, 2013 at 4:35 pm
Phil. says, February 9, 2013 at 2:57 pm: “I refer you to the dichroic halogen lamps which improve the efficiency of the lamps by selectively reflecting IR light from the envelope to the light source, thereby increasing its temperature and increasing the proportion of visible emitted. It’s a perfect illustration of the effect discussed in this thread, the product can be purchased in stores,
http://www.gelighting.com/LightingWeb/emea/images/Halogen_MR16_IR_Lamps_Data_sheet_EN_tcm181-12732.pdf”
==============================================================
On this thread above we have the Wood experiment refuting the notion of reflected IR being capable of warming the source. The result in the Wood experiment was zero or negligible.
Because of a poorly designed experiment which was incapable of determining that, as detailed above.
I can understand the manufacturer though, because an allegedly more efficient lamp can be better sold. I do not necessary think that the manufacturer intentionally lied, because this tale about “back radiation warming” has been around for many decades, so why not “use it”.
Mr House you’re an idiot and not worth wasting breath over! I guess the manufacturer managed to fool patent offices around the world too!
I note that you made no response concerning the definitive experiments with thermocouple shields, and the associated example calculations from heat transfer texts!
ThePhysicsGuy:
I am replying to your post at February 9, 2013 at 4:26 pm.
A factual statement is not an insult. And my post (at February 9, 2013 at 3:06 pm) did not object to your use of the word “flow”. As I stated, I was objecting to your use of the phrase “heat flow” (in your post at February 9, 2013 at 2:54 pm). As I replied
Clearly, you did not read it.
Instead, you chose to again display your ignorance of what you are trying – but failing – to talk about and wrote
I made that correct statement in response to Greg House claiming there is no demonstration of a cooler object heating a warmer one. A microwave oven does that.
And those of us who understand the subject ARE discussing situations which include “an external source of “work””. The examples are the hypothetical planet with radioactivity providing the heating, the Earth with the Sun providing solar energy, the proposed light bulb (or metal spheres) experiment with electricity providing the heating.
A microwave oven and refrigeration systems are good examples of how energy can be moved from a cooler object to a hotter one and then converted to heat with resulting heating of the hotter one. Indeed, the microwave oven is a very apt example because – like the GHE – it conducts the energy transfer by a flow of EM radiation.
Clearly, ThePhysicsGuy, your alias is a misnomer. You know no physics. You don’t understand the Thermodynamic Laws. And you think wicki. is all the information you need to hide your ignorance: it is not.
Several people on this thread have been explaining the realities to you. It seems your ignorance is so profound that you are incapable of understanding what they have written. Either that or you have chosen not to read the posts which explain your errors.
Richard
Greg House;
I can understand the manufacturer though, because an allegedly more efficient lamp can be better sold. I do not necessary think that the manufacturer intentionally lied, because this tale about “back radiation warming” has been around for many decades, so why not “use it”.
>>>>>>>>>>>
Well Greg, the manufacturer has to provide specifications for their products including the watts consumed and the lumins produced. There’s no room for lying because their competitors would sue them, various consumer groups would sue them, various regulatory bodies would slap them with fines and OEM customers would sue them.
Your marriage to your belief system in the face of compelling contrary evidence is beyond aggravating. I’m very sorry I spent a single moment of my life trying to honestly assist you in understanding the physics, and I consider it a mistake on Anthony’s part to continue allowing you to spout your drivel.
richardscourtney says, February 9, 2013 at 4:58 pm: “And those of us who understand the subject ARE discussing situations which include “an external source of “work””. The examples are the hypothetical planet with radioactivity providing the heating, the Earth with the Sun providing solar energy,”
=============================================================
Of course, you are entitled to discuss “situations which include “an external source of “work””, or the Sun providing solar energy and heating the mother Earth, absolutely. But I’d like to remind you that the trapped/back radiation allegedly warming the surface (according to the IPCC story) and refusing to do that in the real Wood experiment is a different thing. As the Wood experiment demonstrates, the trapped/back radiation won’t work. Other things work, wonderful, but this one does not.
dikranmarsupial says:
February 9, 2013 at 2:34 pm
He is wrong, but you are not right. Yes, a cool object can emit photons which are absorbed by the warmer object. But the warmer object sends back more photons, so overall, the warmer object always heats the cooler one.
But, that is not the dynamic playing out here. See comment above. It isn’t about the flows, it is about the delay in establishing equilibrium which holds back some of the flow to be stored behind the “barrier”. It is about Joules, not Watts.
Does anyone here plan to do any actual math, as I have, or do you just want to shout your opinions at each other? Never mind. It’s the latter. So much heat and so little light on this thread.
davidmhoffer says, February 9, 2013 at 5:31 pm: “Well Greg, the manufacturer has to provide specifications for their products including the watts consumed and the lumins produced. There’s no room for lying because their competitors would sue them, various consumer groups would sue them, various regulatory bodies would slap them with fines and OEM customers would sue them.”
===========================================================
By the way, I have not found these specifications in their data sheet: http://www.gelighting.com/LightingWeb/emea/images/Halogen_MR16_IR_Lamps_Data_sheet_EN_tcm181-12732.pdf. I assume, their reflector might redirect light and one can get a certain spot brighter, but not because of alleged higher temperature of the filament caused by trapped/back radiation. How trapped/back radiation works you can see above (the Wood experiment).
But I like your logic. So, because, let us say, the accused knew that he would be caught, prosecuted and put in jail, we can conclude that he could not have possibly committed the crime. Acquitted!
Of course, the same goes for climate scientists. Because they knew that they would be fired, disrespected and possibly sued they could not have possibly revived the old debunked notion about trapped/back radiation… (see above)! That trapped/back radiation warming must be true!
Bart;
Does anyone here plan to do any actual math,
>>>>>>
I’ve done the math for these bone heads in the past. It changes nothing. Thanks for yours though, I didn’t comment because I thought it stood on its own. to the cardboard box crowd though it is just a bunch of symbols. cardboard defeats science.
I assume, their reflector might redirect light and one can get a certain spot brighter, but not because of alleged higher temperature of the filament caused by trapped/back radiation.
>>>>>>>>>>>>>>>>>
Find me another type of light bulb that produces 6000 cd from 35 watts. with a reflector, without a reflector, with bunny rabbits stapled to it, I don’t give a shit. You’re only comeback is that the manufactuerer must have lied because your cardboard boxes say so. When you resort to reasoning that stupid, this ceases to be a science blog.
[Reply: Are you proposing censorship? — mod.]
Time for me to leave this conversation. As typically happens in these conversations, there is lots of wrong information (which is not getting any better), and lots of slightly wrong information (which is not worth nit-picking), and a little bit of correct information.
Thermodynamics is NOT easy, and statistical mechanics is tougher. “Soundbite Science” will not suffice for grappling with these issues. But at least the right information is in there, so that people might indeed be able to head in the right direction.
[Reply: Are you proposing censorship? — mod.]
I’m stating an opinion. What you guys do about it (or don’t do about it) is entirely up to you.
davidmhoffer says:
February 9, 2013 at 6:31 pm
Thanks. I hope you noted my point above though that, even though the GHE does raise the surface temperature above what it would be without the GHGs, there is no guarantee that an incremental increase in GHGs will lead to an incremental increase in surface temperature, i.e., the local sensitivity is not necessarily positive, much less significantly so.
davidmhoffer says, February 9, 2013 at 6:50 pm: “Find me another type of light bulb that produces 6000 cd from 35 watts. with a reflector, without a reflector, with bunny rabbits stapled to it, I don’t give a shit. You’re only comeback is that the manufactuerer must have lied because your cardboard boxes say so.
============================================================
I said specifically: “I do not necessary think that the manufacturer intentionally lied, because this tale about “back radiation warming” has been around for many decades, so why not “use it”.
A lot of people just believe things, especially when they are told that those things are scientific facts.
As for this allegedly “more efficient lamp”, as I said, their effect seems to be a redirection of light due to the reflector, that is all. You will get one certain spot brighter and the area around it darker than without the reflector.
And please, it was not just “my cardboard box” in the Wood experiment, it was an actual temperature measurement. Or can there be warming without an increase in temperature?