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.
Discover more from Watts Up With That?
Subscribe to get the latest posts sent to your email.
Bryan says:
February 11, 2013 at 11:23 am
So, let me ask this, If you have a pot of water, next to a roaring fire, that’s heated by IR, and the pot boils, is that not IR doing work? This may just be a terminology issue, where I’m not using the terms correctly.
I thought this picture of the James Webb Telescope was appropriate for this discussion.
http://upload.wikimedia.org/wikipedia/commons/1/1c/James_Webb_Telescope_DC.jpg
Bryan;
I would add the net energy flow is called heat and it is capable of doing thermodynamic work.
>>>>>>>>>>>>>>>>
The thing is Bryan, you are very close to getting it right. If you’d stop telling other people how stupid they are, you might actually learn something.
Heat cannot do thermodynamic work. Heat is the product of thermodynamic work. Or more correctly, one possible product. You are also using the word “flow” where you should be using the word “flux”. MiCro is on the right track.
David says
“Heat cannot do thermodynamic work. Heat is the product of thermodynamic work. Or more correctly, one possible product”.
More ‘creative’ thermodynamics here, which is unfortunately wrong.
MiCro
“If you have a pot of water, next to a roaring fire, that’s heated by IR, and the pot boils, is that not IR doing work? ”
The net flux of IR would be from fire to pot
If the pot was very shiny or if the pot was black would determine how much net IR was absorbed.
That’s one of the reasons why its important to stress that IR radiation is not in itself heat.
The usual situation is mostly absorption by the pot and conduction to the water.
This increases the temperature of the water and the water gains internal energy from the heat supplied .
If a heat engine were placed between fire and pot it would be capable of driving a piston or doing work.
However in this case the thermal energy from a high temperature is increasing the thermal energy of the water so strictly speaking no mechanical work is done
All in accordance with the laws of thermodynamics
Bryan says:
February 11, 2013 at 1:06 pm
It’s energy. But why does an IR lamp make you warm, where visible light which has more energy does not warm you?
Bryan
Replace the pot with a photo voltaic cell. Assume the PV cell is 100% efficient. Use the electricity to charge a battery.
Question:
Where is the “heat”?
Answer:
There is none. There never was. Since the work done by the energy flux went entirely into charging the battery, no heat was generated. If no heat was generated, there was none to flow in the first place. And don’t start whining that the battery would heat up from being charged or anything else stupid like that, you know very well that represents only a fraction of the energy transfer we’re talking about. Nor should you whine about a 100% efficient pv cell not existing, you know that’s not the point too.
And how about responding under your full name. What are you afraid of?
“””””…..MikeB says:
February 10, 2013 at 7:50 am
george e. smith
You are correct to say that in practice true blackbodies may not exist.
A blackbody is a hypothetical object which has a value for absorptivity and emissivity of one, for all wavelengths. Such an object would therefore absorb all electromagnetic radiation which fell on it and would also emit radiation over all wavelengths in a spectrum conforming precisely to Planck’s distribution formula.
.
Consequently, do you understand that in the steel shell model it does not matter if it the shell is a real blackbody or not? If it is not then the maths gets a bit more complicated but the principle is exactly the same.
Then you say
Planck’s black body radiation derivation explains no actual experimental observations anybody ever made, that I am aware of.
But Planck’s formula was produced to explain experimental observations? Let’s all agree that it was a big improvement on the ultra-violet catastrophe. If anything it was the theoretical derivation of Planck’s Law which was suspect. In order to derive this formula Planck had to make an unwarranted assumption. He assumed that energy was not infinitely divisible but came in ‘packets’ rather like atoms in matter. He offered no justification for this assumption, except that it gave the desired answer. Planck and everyone else at the time and for years afterwards regarded this as nothing more than a mathematical device, a trick needed to get an answer that matched observation. We now know that in fact Planck had laid the foundations for quantum mechanics……”””””
Well MikeB, most often in science, observers (experimentalists) make real world observations and/or measurements on some real world phenomena. Otherwise it does not belong in the purview of science.
That means there is some physical system, that is exhibiting the phenomenon. People do not obseve phenomena that are not exhibited by objects or systems that do not exist; and hence can exhibit nothing to obseve.
Black bodies do NOT exist, and I cited simply physics reasons why they cannot.
Now the real world is often too complicated for us to possibly know or see what is going on let alone explain it.
So we create “models” to which we assign certain properties and behaviors which we CAN study/compute/whatever, using mathematics; all of which is as fictional as our models. We compare our calculations of the expected performance of our model to the observed performance of the real system we are trying to explain. Any time we find a discrepancy between what we reliably observe, and what our models claim should occur, we then alter our description of our model, until it predicts better results that are closer to the real system.
We NEVER assign properties to our model, which we know a priori, the real system cannot possibly have; because that is a fatal difference between the reality we observe, and the modelled calculations prediction.
Nobody ever observed ANY object or system which can absorb any amount of EM radiation at any possible frequency or wavelength, which we believe covers all vallues except zero and infinity.
Nobody ever made observations of any real object or system emitting EM radiation at any possible frequency or wavelength.
Nobody ever made any observations of any object or system where everything in the system was at a single Temperature, with no heat able to move (net) in any direction in such an isothermal system. The sun is NOT isothermal and as it happens, it does not emit a Planck formula black body spectrum. A whole lot is known about what Temperatures various bits of the sun are and what that implies.
Purportedly, Planck’s actual derivation of the radiation law, requires an equally fictional object which also cannot exist; namely an isothermal cavity with perfectly reflecting conductive walls, that also do not conduct “heat”. Nothing can escape from this fictional object except radiation from a “small hole” in the otherwise perfectly reflecting walls.
Now in the real actual observable world where real observations and measurments can be made, most good reflectors of EM radiation, and especially over wide frequency ranges, turn out to be metals, and as it happens metals are also among the best conductors of “heat”
In particular, Silver, which has the highest known visible radiation reflectance among all metals, also has the highest electrical conductivity. Optically transparent media (in the bulk) typically are not electrically conductive.
A few exotic things, like Indium Tin Oxide, are both optically transparent, and electrically conductive (weakly) in thin films.
My point is the hypothetical black body, of Planck’s theory, is assigned properties which we knwo that nothing real can possess, and in his historic development of the Planck radiation formula he studies a cavity model, which also has properties no real cavity has been observed to have.
Hey let’s face it, the Planck et al theory of BB radiation is both remarkable and of great practical utility; and yes, I concede that it led to the development of quantum theory; but one thing it did NOT do is limit Planckian thermal radiation to only certain allowed frequenciesor wavelengths, as happens with atomic and molecula line/band spectra, that truly are the consequence of specific quantum numbers of real atoms and molecules.
Planck only said that the radiation must consist of packets (photons) whose packet energy was related to the frequency via E = hf. We can take this to mean that each cycle of the frequency of the photon wave packet, has a quantity of action (energy x time) of value h .
But since no restrictions are placed on the value of f, photons of any possible energy value are allowed. In that sense BB radiation is not quantized in the sense that atomic spectra radiations are, where only certain frequencies are permitted.
So the history suggests, that when researchers were unable to explain the results of real observations on real materials, and systems; they chose to construct a completely fictional ideal, and non existing object; a “Black Body” to which they ascribed properties no object has exhibited.
Yes it is useful in approximating what real objects seem to do.
I’m not aware that Planck’s derivation gives any insight into what happens when you don’t have an isothermal cavity , with perfectly reflecting perfectly conducting walls.
So I see no basis for asserting that Willis’s steels shell planet, which is not isothermal, and not in thermal equilibrium, emits anything like black body radiation.
I can’t see merit, in constructing mind exercise objects, which we know do not exist, such as steel shell planets; it teaches us nothing about real planets with real atmospheres that do permit thermal interractions.
Plancks surprise assertion of packets of energy E = hf came in 1900. Einstein’s similar assertion to explain the photoelectric effect came in 1905, and won him his Nobel Physics prize.
To this day, I am not aware of any classical explanation for the observations of the photo-electric effect.
Plancks 1900 declaration of energy packets (of all possible energy values), is no where near as arbitrary and unwarranted, as Nils Bohr’s quite irrational suspension of the Hertz/Maxwell observations that accelerated electric charges always radiate during acceleration.
It seems they still do despite Bohr, and Arnold Sommerfeld, who nevertheless gave us so much insight into observed atomic line spectra.
So quantum mechanics bailed them out by effectively dispensing with the orbital acceleration, rather than the requirement that radiation not happen sometimes.
The early chemists could be said to have quantized chemistry, by saying you could only get matter in certain sized chunks with only a limited number of sizes; maybe 92, later expanded to maybe 1000 or so different chunks. Planck placed no restrictions on the size of the chunks, only that it (energy) was related to the infinitely variable frequency f via h. so he didn’t really quantize thermal radiation in the sense that Bohr/Sommerfeld and other had for atomic line spectra, or the molecular spectr that haunt the GHG phenomenon.
MiCro;
But why does an IR lamp make you warm, where visible light which has more energy does not warm you?
>>>>>>>>>>>>>>
What do you mean by “more energy”? Per what? 10 w/m2 of IR and 10 w/m2 of visible light are each…rough math…approximation….round off….. hmmm…. 10 w/m2 😉
The visible light on the other hand tends to reflect off of you instead of being absorbed. Well, that’s what people like Bryan will try and tell you anyway. Everyone knows that Plato proved in 5th century BC that light only works by interacting with the rays from your eyes.
@ur momisugly davidmhoffer
Per photon.
davidmhoffer
Here’s an equation.
Its from a thermodynamics book so you wont have seen it before.
Q = dU – W
Q= heat which flows into the system
dU = the change in internal energy of the system
W = work done by the system
So if heat is added to the system by water say absorbing IR radiation from a higher temperature emitter, then the water temperature will rise (dU) minus any work done by the system (W)
The above equation is a restatement of conservation of energy in a form suitable for discussing thermal processes.
Your post of February 11, 2013 at 2:20 pm contains so many mistakes I must assume its you having a deliberate ‘wind up’
On the other hand your head must be so full of new correct information as a result of this thread that it will take you some time to assimilate it.
Friends:
I am to leave on another of my frequent but irregular absences which take me out of communication. Hence, I cannot continue to participate in this discussion. I hope that leaves none of my points ‘hanging’ and if it does then I have genuinely overlooked them.
Richard
Here’s an equation
>>>>>>>>>>>>>>>>>>>>>
OOOOOOOOOOOOOOOOH!
Bryan has an equation!
No answer to my question, but he’s got an equation.
And still anonymous.
“”””””…..davidmhoffer says:
February 11, 2013 at 3:02 pm
MiCro;
But why does an IR lamp make you warm, where visible light which has more energy does not warm you?
>>>>>>>>>>>>>>
What do you mean by “more energy”? Per what? 10 w/m2 of IR and 10 w/m2 of visible light are each…rough math…approximation….round off….. hmmm…. 10 w/m2 😉
The visible light on the other hand tends to reflect off of you instead of being absorbed. Well, that’s what people like Bryan will try and tell you anyway. Everyone knows that Plato proved in 5th century BC that light only works by interacting with the rays from your eyes……”””””
The following is an exact quote from:- “The Science of Color.” published by The Committee on Colorimetry, of the Optical Society of America, which Society is one of the founding bodies of The American Institute of Physics.
Specifically it is from Chapter 7 titled “Psychophysics of Color.”.
Item #1 after a brief introduction is:- “Definition of Light.”
“””””…..The relations between radiant energy, light, and visual sensation can be defined in the following generalized and qualitative manner: Light is the aspect of radiant energy of which a human observer is aware through the visual sensations which arise from the stimulation of the retina of the eye……””””
It then goes on to say:- “Light as thus defined is a psychphysical concept. Light is not identified with either radiant energy or visual sensation. Light is one of many conceivable aspects of radiant energy.”
Ergo, Light by definition is visible; so there is NO ultra-violet light or NO infra-red light Both of those are invisible.
This distinction is fortified by the fact that we use completely different units for “light” from what are used for other EM radiation phenomena., and we use Photometry rather than Radiometry to talk about light quantities.
So “Luminous Energy” in Talbots replaces Joules in Radiometry.
Luminous flux in lumens replaces Watts
Luminous intensity in Candela replaces Watts per steradian
Illuminance in lux (Lumens per m^2) replaces irradiance ; and so on.
Which is why I have said, “We get no light from the sun; we conjure it all up in our eyes and brains from certain Radiant energy frequencies.”
Likewise, we get no “heat” from the sun; heat is just one possible form we can convert incoming solar radiant energy into.
Photons know nothing about Temperature. Any single photon arriving from the sun, which can be uniquely detected by certain photo-multiplier tubes, tells us absolutely nothing about the Temperature of the sun. Temperature is a macroscopic property of large assemblages of material particles which physically interract with each other (collide).
If we collect a very large number of single photons from the sun and determine their statistics, it then becomes possible to approximate the emitting surface Temperature of the sun, by assuming that Planck’s radiation formula works well enough; but since that surface is not isotropic, the actual spectrum differs from Planckian black body radiation.
Lots of people have gotten Radiation and heat so much identified as being the same thing, that we still teach students that radiation is a method of transporting heat. Well so is a grocery shopping cart; just fill it with some bags of coal or charcoal.
But then on a recent visit to my Alma Mater Physics Department, one of the top Physics Professors told me, that they still teach that E = I x R is Ohm’s law.
It isn’t of course; Ohm’s law says “R is a constant” E = IR is simply a definition of R in a linear system.
Most electrical circuits; and virtually all the ones in your computer, do not obey Ohm’s law
So we should reserve use of “light” to that psycho-physical aspect of visible EM radiation, and not confuse it with the radiation itself. Otherwise people will never know what we are talking about.
MiCro says:
February 11, 2013 at 5:36 pm
@ur momisugly davidmhoffer
Per photon.
>>>>>>>>>>>>>>>
yikes!
No idea. I’d have to go research that.
george e smith? comment?
I suspect ratio of photons simply reflected vs absorbed would be a big part of the equation. Albedo varies with frequency.
george e smith says;
one of the top Physics Professors told me, that they still teach that E = I x R is Ohm’s law.
>>>>>>>>>>>>>>
I think most universities do. But even 30+ years ago they also made sure we understood that it was an approximation. Under a lot of circumstances it isn’t even a good approximation. An incandescent light bulb for example has an R of about 1 ohm at room temp, and 100 ohms at operating temp. Only a couple orders of magnitude….
Then we discovered there’s a small capacitance and inductance associated even with a bit of wire….
Then some moron came up with these silly semi-conductors where there’s holes in the darn things and the holes flow rather than the electrons… well sorta kinda… gotta start the explanation somewhere, that’s as good a place as any…. but E=IR is pretty much out the window by then.
george e. smith says (February 11, 2013 at 2:31 pm): “I can’t see merit, in constructing mind exercise objects, which we know do not exist, such as steel shell planets; it teaches us nothing about real planets with real atmospheres that do permit thermal interractions.”
How about something that should be constructible, like Dr. Spencer’s “Yes, Virginia” thought experiment?
http://www.drroyspencer.com/2010/07/yes-virginia-cooler-objects-can-make-warmer-objects-even-warmer-still/
If the experiment is ever performed, what do you predict the result will be? In the presence of the cooler bar, the electrically heated bar is
a) warmer
b) cooler
c) the same temp
as it is in the absence of the cooler bar?
davidmhoffer says:
February 11, 2013 at 2:20 pm
Ummm … no. Energy cannot be created or destroyed. However, it can be converted from one form to another. Light, for example is energy in the form of radiation. It can be converted into a variety of different forms. It doesn’t contain those forms of energy, but it can be converted into those forms.
For example, in a solar cell, part of the energy that strikes it (in the form of radiation) it is converted into another form, thermal energy. Another part of the radiation is converted into a second form, electricity.
Note, please, that “heat” is the NET flow of energy to and from an object. All objects are constantly losing energy and gaining energy. It is the only the NET energy that is called “heat”.
To summarize:
1. Light (or infrared) is energy in the form of radiation. It does not “contain heat”, that statement makes no sense.
2. Energy in the form of radiation can be converted into other forms of energy, including mechanical energy (e.g in a Crooke’s radioscopy), chemical energy (in a leaf), atomic bond energy (raising the energy levels the absorbing substance), or electrical energy (solar cell), or thermal energy.
3. Whenever radiation is absorbed by anything, it changes form, it is no longer energy in the form of radiation.
4. None of these energy conversions are perfectly efficient, so at least some of the energy is lost as thermal energy going into the objects or the environment.
w.
willis;
Ummm … no. Energy cannot be created or destroyed. However, it can be converted from one form to another.
>>>>>>>>>>>>
that was my point.
consider a line from say 1 million km from the sun (A) to 1 million km from earth (B) . Is there an energy flux from A to B? yes. Over 1300 w/m2 at B. Is there heat? No. There is no heat until the energy flux is absorbed by matter. At which point it might be turned into heat. Or electricity. or something else. But if it is turned into heat, then the heat is the product of the work being done. No work, no heat. From the path of A to B, no work is done, no heat is created, no heat flows.
” MiCro says:
February 11, 2013 at 1:32 pm
Bryan says:
February 11, 2013 at 1:06 pm
That’s one of the reasons why its important to stress that IR radiation is not in itself heat.
It’s energy. But why does an IR lamp make you warm, where visible light which has more energy does not warm you?”
The simple answer is IR lamp is designed to give heat. The difference between a once standard
100 watt incandescent bulb and IR lamp is filament would larger and not glow as bright, and use the same or less watts [power]. So if increased the volts to IR lamp it would glow brighter [probably burn out cause not designed to handle more power- and not be bright as light bulb].
IR lamps may have reflective backing [spotlight and other light bulbs can also have reflective backing to give directed light] if these are IR lamps you mean, having reflective backing is another reason it can heat thing better than light bulb.
But your typically spotlight also deliver a fair amount of heat to area they are pointed at.
Up the thread, there is the halogen reflector lamps:
http://www.gelighting.com/LightingWeb/emea/images/Halogen_MR16_IR_Lamps_Data_sheet_EN_tcm181-12732.pdf
A problem with “normal” halogen reflector lamps is they give a lot heat-
So above ad it says they reflect this heat back the bulb to give energy [so the vaporized
tungsten gets redeposited on filament “better”]. Which apparently allows one *design*
higher filament temperature than “normal” halogen reflector lamps [and reduces the
amount heat- which is problem they trying to minimize. The coating probably also reduces
UV- which could be problem for some uses- other uses may want the UV].
As for the shuttle using old specs and not changing them, this common with all
spacecraft- if not broken don’t fix- is strong tendency due high cost development
and where one mistake causes hundreds millions dollars in damage- and/or loss
of crew.
Some more on Apollo 13:
“There were 2,181 ampere hours in the LM batteries. Ground controllers carefully worked out a procedure where the CM batteries were charged with LM power. All noncritical systems were turned off and energy consumption was reduced to 1/5, which resulted in having 20 percent of LM electrical power left when Aquarius was jettisoned. ”
So car battery has 220 ampere hours- so power of about 10 car batteries. That is all power which was available.
“Water was the main consumable concern. It was estimated that the crew would run out of water about five hours before Earth re-entry, which was calculated at around 151 hours. ”
http://www.nasa.gov/mission_pages/apollo/missions/apollo13.html
So 2,181 ampere hours to last 151 hours plus had to charge the Command Module with this charge.
The say, above: The temperature dropped to 38 degrees Fahrenheit and condensation formed on all the walls.
Other sources give different number:
“Further examination of Biomedical Results of Apollo reveals that actual cabin temperatures dropped to a low of 43°F ”
http://history.nasa.gov/SP-4029/Apollo_18-39_CModule_Cabin_Temperature_History.htm
And:
“The temperature dropped to around 40 degrees by the end of the trip. ”
http://www.lpi.usra.edu/lunar/missions/apollo/apollo_13/return/
Wow, this thread is still going ?
Ok, here’s some more empirical evidence regarding two light bulbs in close proximity and the alleged “cross-heating” that some folks predict will happen.
You can purchase commercially an integrating sphere that will accept multiple light bulbs through ports in the exterior. By turning on or off different combinations of these light bulbs it is possible to create a range of radiance values leaving an exit port (radiance is the amount of light per unit of solid angle (how much light is in a cone leaving the exit port)). If you add a shutter between one of the lamps and the entrance to the sphere you can get a variable light source with a pretty constant spectra distribution. This is pretty standard instrumentation used to test digital camera sensor chips and radiometers.
Very high quality power supplies with highly regulated current levels are used to excite the lamps. So we are controlling exactly the amount of power entering the sphere (assuming the lamps are on all the time rather than in a pulsed mode).
So we have two hypothesis;
1) The lamps have no discernible effect on each other and the total light output is a simple linear sum of the individual output of each lamp. I.E. we can measure lamp #1’s output (with lamp #2 OFF) and likewise measure lamp #2’s output (with lamp #1 OFF) and add those spectral measurements together and get what the total will be when both lamps are ON.
OR
2) Lamp # 1 (when ON) heats lamp #2 (which is also ON) and the light output of lamp #2 goes up (it’s an incandescent lamp, it’s output is proportional to the temperature of the filament) and the total when both lamps are ON is somehow higher than the sum of the individual lamps when they are ON at separate times.
I can save you the time doing this experiment, within the accuracy of the measurements THERE IS NO DISCERNIBLE “cross heating” effect between the lamps, hypothesis #1 is correct. It is also in conformance with the laws of thermodynamics.
Now please note that I specifically mentioned running the lamps in a constant current mode with a fixed power supply. If you where to run them in constant voltage mode (i.e. just plug them into a wall outlet), then yes indeed one lamp will heat the other by a slight amount. If you hold your hand within a few inches of an incandescent light bulb your hand will warm by maybe 10-20 degrees F. This heating will reduce the resistance of the filament slightly (10 degrees F compared to about 3100 K color temp is not much). And the light bulb will output more light, BUT IT WILL ALSO REQUIRE MORE CURRENT (AND ENERGY) from your power supply. So the folks that believe hypothesis #2 are mistaking a system that consumes more energy with a system (hypothesis #1) that outputs no more light with a fixed energy supply.
If you do this experiment with an energy supply (a battery) instead of a power supply you might measure just a bit more light output but for a shorter time duration (the battery will run down faster) and have NO NET ENERGY GAIN.
You can probably find some more application notes on the web from the folks that sell integrating spheres with light bulbs inside. They have been around for decades.
Cheers, Kevin.
Willis,
The Second Law of Thermodynamics says nothing about a “net” flow regarding heat. The flow of heat is in one direction only, from warmer to colder. A cool body will simply not warm a warmer body under any circumstances without external work applied.
SkepticGoneWild says (February 11, 2013 at 8:27 pm): “The Second Law of Thermodynamics says nothing about a “net” flow regarding heat. The flow of heat is in one direction only, from warmer to colder. A cool body will simply not warm a warmer body under any circumstances without external work applied.”
Just so there’s no semantic confusion about your statement (we’ve seen plenty of that upthread), could you look at Dr. Spencer’s “Yes, Virginia” thought experiment and predict the result if the experiment is ever run for real?
http://www.drroyspencer.com/2010/07/yes-virginia-cooler-objects-can-make-warmer-objects-even-warmer-still/
There’s a number of post talking about Dr Spencer’s thought experiment.
Well, it isn’t just a thought experiment.
I should have included this quote from the above link
My IR thermometer testing (and my work with the NCDC data set) shows most of the IR is from water vapor. My testing @35F, the sky’s temp was less than -40F. In the NCDC data I found the average annual night time drop in temp is ~18F, almost an exact match to the ~18F increase during the day. I also found a small sample set where with low humidity, low wind speeds, the swing was a matching (daily rise/fall) 40F. I’ve also found that when looked at on a daily average basis, the difference changes with the length of day, as the day starts to get longer, rising temp is more than falling. This difference peaks about April where it starts to fall, crossing zero in August, with the maximum cooling (fall larger than rise) about November.
KevinK says:
February 11, 2013 at 8:20 pm
I’m not sure this is actually what we want to test. It’s not that it would change the temp of the filament (though this might happen in a carefully constructed test), it’s that the air temp next to bulb 1, will be warmer when you turn bulb 2 on.
davidmhoffer says:
February 11, 2013 at 3:02 pm
So, @10W/m2 at 10u wavelength, takes 20x the flux as it would at .5u
Here’s the spectrum of the Sun and Earth in watt’s/M2/uM, note the Sun’s output is far higher than that of the Earths.
And the sky looking up is a lot colder than the earth.
MiCro says, February 12, 2013 at 7:04 am: “There’s a number of post talking about Dr Spencer’s thought experiment. Well, it isn’t just a thought experiment.”
==========================================================
This one is not a thought experiment, but has a problem with the thought.
If there are a few factors involved, you can not just choose one of them and declare it to be the cause. Instead, you need to create two states where the only (significant) difference between them is that very factor you intend to prove to be the cause. Like professor Wood did, see above.
Greg,
Can you provide any explanation for why the chamber is colder than the outside temperature other than the sky being colder than the surrounding air temp?