Light Bulb Back Radiation Experiment
Guest essay by Curt Wilson
In the climate blogosphere, there have been several posts recently on the basic principles of radiative physics and how they relate to heat transfer. (see yesterday’s experiment by Anthony here) These have spawned incredibly lengthy streams of arguments in the comments between those who subscribe to the mainstream, or textbook view of radiative heat transfer, and those, notably the “Skydragon Slayers” who reject this view.
A typical statement from a Slayer is that if “you have initially a body kept at a certain temperature by its internal source of energy”, that if another body at a lower temperature is placed near to it, that the radiation from this colder body could not increase the temperature of the warmer body, this being a violation of the 2nd Law of Thermodynamics. They continue that if this were possible, both objects would continually increase the other’s temperature indefinitely, which would be an obvious violation of the 1st Law of Thermodynamics (energy conservation).
This is part of a more general claim by Slayers that radiation from a colder body cannot transfer any energy to a warm body and lead to a higher temperature of the warm body than would be the case without the presence of the colder body.
It occurred to me that these claims were amenable to simple laboratory experiments that I had the resources to perform. A light bulb is a classic example of a body with an internal source of energy. Several Slayers specifically used the example of reflection back to a light bulb as such an example.
In our laboratory, we often have to do thermal testing of our electronic products so we can ensure their reliability. Particularly when it comes to power electronics, we must consider the conductive, convective, and radiative heat transfer mechanisms by which heat can be removed from these bodies with an “internal source of energy”. We have invested in good thermocouple measurement devices, regularly calibrated by a professional service, to make the temperature measurements we need.
We often use banks of light bulbs as resistive loads in the testing of our power electronics, because it is a simple and inexpensive means to load the system and dissipate the power, and it is immediately obvious in at least a qualitative sense from looking at the bulbs whether they are dissipating power. So our lab bench already had these ready.
If you want to isolate the radiative effects, the ideal setup would be to perform experiments in a vacuum to eliminate the conductive/convective losses. However, the next best thing is to reduce and control these to keep them as much alike as possible in the different phases of the experiment.
So, on to the experiment. This first picture shows a standard 40-watt incandescent light bulb without power applied. The lead of the thermocouple measuring device is taped to the glass surface of the bulb with heat-resistant tape made for this purpose. The meter registers 23.2C. In addition, a professional-grade infrared thermometer is aimed at the bulb, showing a temperature of 72F. (I could not get it to change the units of the display to Celsius.) Note that throughout the experiment, the thermocouple measurements are the key ones.
Next, the standard North American voltage of 120 volts AC (measured as 120.2V) was applied to the bulb, which was standing in free air on a table top. The system was allowed to come to a new thermal equilibrium. At this new equilibrium, the thermocouple registered 93.5C. (The IR thermometer showed a somewhat lower 177F, but remember that its reported temperature makes assumptions about the emissivity of the object.)
Next, a clear cubic glass container about 150mm (6”) on a side, initially at the room temperature of 23 C, was placed over the bulb, and once again the system was allowed to reach a new thermal equilibrium. In this state, the thermocouple on the temperature of the bulb registers 105.5C, and the outer surface of the glass container registers 37.0C (equivalent to body temperature).
The glass container permits the large majority of the radiative energy to escape, both in the visible portion of the spectrum (obviously) and in the near infrared, as standard glass is highly transparent to wavelengths as long as 2500 nanometers (2.5 microns). However, it does inhibit the direct free convection losses, as air heated by the bulb can only rise as far as the top of the glass container. From there, it must conductively transfer to the glass, where it is conducted through the thickness of the glass, and the outside surface of the glass can transfer heat to the outside ambient atmosphere, where it can be convected away.
The next step in the experiment was to wrap an aluminum foil shell around the glass container. This shell would not permit any of the radiative energy from the bulb to pass through, and would reflect the large majority of that energy back to the inside. Once again the system was allowed to reach thermal equilibrium. In this new state, the thermocouple on the surface of the bulb registered 137.7C, and the thermocouple on the outer surface of the glass registered 69.6C. The infrared thermometer is not of much use here due to the very low emissivity (aka high reflectivity) of the foil. Interestingly, it did show higher temperatures when focused on the tape on the outside of the foil than on the foil itself.
Since adding the foil shell outside the glass container could be reducing the conductive/convective losses as well as the radiative losses, the shell was removed and the system with the glass container only was allowed to re-equilibrate at the conditions of the previous step. Then the glass container was quickly removed and the foil shell put in its place. After waiting for thermal equilibrium, the thermocouple on the surface of the bulb registered 148.2C and the thermocouple on the outside of the foil registered 46.5C. The transient response (not shown) was very interesting: the temperature increase of the bulb was much faster in this case than in the case of adding the foil shell to the outside of the glass container. Note also how low the infrared thermometer reads (84F = 29C) on the low-emissivity foil.
Further variations were then tried. A foil shell was placed inside the same glass container and the system allowed to reach equilibrium. The thermocouple on the surface of the bulb registered 177.3C, the thermocouple on the outer surface of the foil registered 67.6C, and the infrared thermometer reading the outside of the glass (which has high emissivity to the wavelengths of ambient thermal radiation) reads 105F (40.6C).
Then the glass container was removed from over the foil shell and the system permitted to reach equilibrium again. The thermocouple on the surface of the bulb registered 176.3C and the thermocouple on the outside of the foil registered 50.3C.
All of the above examples used the reflected shortwave radiation from the aluminum foil. What about absorbed and re-emitted longwave radiation? To test this, a shell of black-anodized aluminum plate, 1.5mm thick, was made, of the same size as the smaller foil shell. A black-anodized surface has almost unity absorption and emissivity, both in the shortwave (visible and near infrared) and longwave (far infrared). Placing this over the bulb (without the glass container), at equilibrium, the thermocouple on the bulb registered 129.1C and the thermocouple on the outside of the black shell registered 47.0C. The infrared thermometer read 122F (50C) on the tape on the outside of the shell.
The power source for this experiment was the electrical input. The wall voltage from the electrical grid was steady at 120.2 volts. The electrical current was measured under several conditions with a professional-grade clip-on current sensor. With the bulb in open air and a surface temperature of 96.0C, the bulb used 289.4 milli-amperes of current.
With the bulb covered by a foil shell alone and a surface temperature of 158.6C, the bulb drew slightly less, 288.7 milliamperes.
Summary of Results
The following table shows the temperatures at equilibrium for each of the test conditions:
| Condition | Bulb Surface Temperature | Shell Temperature |
| Bulb open to room ambient | 95C | — |
| Bulb covered by glass container alone | 105C | 37C |
| Bulb covered by glass container and outer reflective foil shell | 138C | 70C (glass) |
| Bulb covered by outer reflective foil shell alone | 148C | 46C (foil) |
| Bulb covered by inner reflective foil shell inside glass container | 177C | 68C (foil) |
| Bulb covered by inner reflective foil shell alone | 176C | 50C |
| Bulb covered by black-anodized aluminum shell alone | 129C | 47C |
Analysis
Having multiple configurations permits us to make interesting and informative comparisons. In all cases, there is about a 35-watt (120V x 0.289A) electrical input to the system, and thermal equilibrium is reached when the system is dissipating 35 watts to the room as well.
I used a low-wattage (40W nominal) bulb because I had high confidence that it could take significant temperature increases without failure, as it has the same package design as much higher-wattage bulbs. Also, I would not be working with contraband high-wattage devices 😉
The case with the glass container alone is the important reference case. The glass lets virtually all of the radiant energy through, while inhibiting direct convection to the room ambient temperature of 23C. Conductive/convective losses must pass from the surface of the bulb, through the air under the container, to and through the glass, and then to the room atmosphere, where it is conducted/convected away. Under these conditions, the bulb surface temperature is 105C, which is 10C greater than when the bulb can conductively dissipate heat directly to the room atmosphere.
Compare this case to the case of the larger foil shell alone. The foil shell also inhibits direct conductive/convective losses to the room atmosphere, but it will not inhibit them to any greater extent. In fact, there are three reasons why it will inhibit these losses less than the glass container will. First, the material thermal conductivity of aluminum metal is far higher than that of glass, over 200 times greater (>200 W/(m*K) versus <1.0 W/(m*K)). Second, the foil, which is a small fraction of a millimeter thick, is far thinner than the glass container, which is about 4 mm thick on average. And third, the surface area of the foil is somewhat larger than the glass container, so it has more ability to conductively transfer heat to the outside air.
And yet, the surface of the bulb equilibrated at 146C under these conditions, over 40C hotter than with the glass container. With conductive/convective losses no less than with the glass container, and very probably greater, the only explanation for the higher temperature can be a difference in the radiative transfer. The glass container lets the large majority of the radiation from the bulb through, and the foil lets virtually none of it through, reflecting it back toward the bulb. The presence of the foil, which started at the room ambient of 23C and equilibrated at 46C, increased the temperature of the bulb, which started at 105C on the outside (and obviously warmer inside). The reflected radiation increased the temperature of the bulb, but did not produce “endless warming”, instead simply until the other losses that increase with temperature matched the input power of 35 watts.
Interestingly, the foil shell without the glass container inside led to a higher bulb temperature (148C) than the foil shell with the glass container inside (138C). Two layers of material around the bulb must reduce conductive/convective losses more than only one of them would, so the higher temperature must result from significantly more reflected radiation back to the bulb. With the glass inside, the reflected radiation must pass through two surfaces of the glass on the way back to the bulb, neither of which passes 100% through.
Another interesting comparison is the large foil shell that could fit outside of the glass container, about 160mm on a side, with the small foil shell that could fit inside the glass container, about 140mm on a side. With the large shell alone, the bulb temperature steadied at 148C; with the smaller shell, it steadied at 176C. With all direct radiative losses suppressed in both cases, the difference must come from the reduced surface area of the smaller shell, which lessens its conductive/convective transfer to the outside air at a given temperature difference. This is why halogen incandescent light bulbs, which are designed to run hotter than standard incandescent bulbs, are so much smaller for the same power level – they need to reduce conductive/convective losses to get the higher temperatures.
All of the above-discussed setups used directly reflected radiation from the aluminum foil. What happens when there is a barrier that absorbs this “shortwave” radiation and re-emits it as “longwave” radiation in the far infrared? Can this lead to higher temperatures of the warmer body? I could test this using black-anodized aluminum plate. Black anodizing a metal surface makes it very close to the perfect “blackbody” in the visible, near-infrared, and far-infrared ranges, with absorptivity/emissivity (which are the same at any given wavelength) around 97-98% in all of these ranges.
With a black plate shell of the same size as the smaller foil shell, the bulb surface temperature equilibrated at 129C, 24C hotter than with the glass container alone. Once again, the thin metal shell would inhibit conductive/convective losses no better, and likely worse than the glass container (because of higher material conductivity and lower thickness), so the difference must be from the radiative exchange. The presence of the shell, which started at the room ambient of 23C and increased to 47C, caused the bulb surface temperature to increase from 105C to 129C.
Another interesting comparison is that of the smaller foil shell, which led to a bulb surface temperature of 176C and a shell temperature of 50C, to the black plate shell of the same size, which led to a bulb surface temperature of 129C and a shell temperature of 46C. While both of these create significantly higher bulb temperatures than the glass container, the reflective foil leads to a bulb surface temperature almost 50C higher than the black plate does. Why is this?
Consider the outside surface of the shell. The foil, which is an almost perfect reflector, has virtually zero radiative absorptivity, and therefore virtually zero radiative emissivity. So it can only transfer heat to the external room by conduction to the air, and subsequent convection away. The black plate, on the other hand, is virtually the perfect absorber and therefore radiator, so it can dissipate a lot of power to the room radiatively as well as conductively/convectively. Remember that, since it is radiating as a function of its own temperature, it will be radiating essentially equally from both sides, there being almost no temperature difference across the thickness of the plate. (Many faulty analyses miss this.) The foil simply reflects the bulb’s radiation back to the inside and radiates almost nothing to the outside. This is why the infrared thermometer does not read the temperature of the foil well.
The electrical voltage and current measurements were made to confirm that the increased temperature did not come from a higher electrical power input. The current measurements shown above demonstrate that the current draw of the bulb was no higher when the bulb temperature was higher, and was in fact slightly lower. This is to be expected, since the resistivity of the tungsten in the filament, as with any metal, increases with temperature. If you measure the resistance of an incandescent bulb at room temperature, this resistance is less than 10% of the resistance at its operating temperature. In this case, the “cold” resistance of the bulb is about 30 ohms, and the operating resistance is about 415 ohms.
Let’s look at the dynamic case, starting with the thermal equilibrium under the glass container alone. 35 watts are coming into the bulb from the electrical system, and 35 watts are leaving the bulb through conductive losses to the air and radiative losses to the room through the glass. Now we replace the glass with one of the metal shells. Conductive losses are not decreased (and may well be increased). But now the bulb is receiving radiant power from the metal shell, whether reflected in one case, or absorbed and re-radiated back at longer wavelengths in the other. Now the power into the bulb exceeds the power out, so the temperature starts to increase. (If you want to think in terms of net radiative exchange between the bulb and the shell, this net radiative output from the bulb decreases, and you get the same power imbalance.)
As the temperature of the bulb increases, both the conductive losses to the air at the surface of the bulb increase (approximately proportional to the temperature increase) and the radiative losses increase as well (approximately proportional to the 4th power of the temperature increase). Eventually, these losses increase to where the losses once again match the input power, and a new, higher-temperature thermal equilibrium is reached.
I originally did these tests employing a cylindrical glass container 150mm in diameter and 150mm high with and without foil shells, and got comparable results. In the second round shown here, I changed to a cubic container, so I could also create a black-plate shell of the same shape.
It is certainly possible that improvements to these experiments could result in differences of 1 or 2C in the results, but I don’t see any way that they could wipe out the gross effect of the warming from the “back radiation”, which are several tens of degrees C.
All of these results are completely in line with the principles taught in undergraduate engineering thermodynamics and heat transfer courses. The idea that you could inhibit net thermal losses from an object with an internal power source, whether by conductive, convective, or radiative means, without increasing the temperature of that object, would be considered ludicrous in any of these courses. As the engineers and physicists in my group came by the lab bench to see what I was up to, not a single one thought for a moment that this back radiation would not increase the temperature of the bulb.
Generations of engineers have been taught in these principles of thermal analysis, and have gone on to design crucial devices and infrastructure using these principles. If you think all of this is fundamentally wrong, you should not be spending your time arguing on blogs; you should be out doing whatever it takes to shut down all of the erroneously designed, and therefore dangerous, industrial systems that use high temperatures.
Conclusions
This experiment permitted the examination of various radiative transfer setups while controlling for conductive/convective losses from the bulb. While conductive/convective losses were not eliminated, they were at least as great, and probably greater, in the cases where a metal shell replaced the glass shell over the bulb.
Yet the bulb surface temperature was significantly higher with each of the metal shells than with the glass shell. The only explanation can therefore be the radiative transfer from the shells back to the bulb. In both cases, the shells were significantly cooler than the bulb throughout the entire experiment, both in the transient and equilibrium conditions.
We therefore have solid experimental evidence that radiation from a cooler object (the shell) can increase the temperature of a warmer object (the bulb) with other possible effects well controlled for. This is true both for reflected radiation of the same wavelengths the warmer body emitted, and for absorbed and re-radiated emissions of longer wavelengths. The temperature effects are so large that they cannot be explained by minor setup effects.
Electrical measurements were made to confirm that there was not increased electrical power into the bulb when it was at higher temperatures. In fact, the electrical power input was slightly reduced at higher temperatures.
This experiment is therefore compatible with the standard radiative physics paradigm that warmer and cooler bodies can exchange radiative power (but the warmer body will always transfer more power to the cooler body). It is not compatible with the idea that cooler bodies cannot transfer any power by radiative means to warmer bodies and cause an increase in temperature of the warmer body.
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UPDATE: The Principia/Slayers group has post a hilarious rebuttal here:
http://principia-scientific.org/supportnews/latest-news/210-why-did-anthony-watts-pull-a-bait-and-switch.html
Per my suggestion, they have also enabled comments. You can go discuss it all there. – Anthony
bring a 6 inch diameter pot of water to 100 degrees in a room with ambient temp of 60 … turn off the heat and suspend a very thin (12″x12″) sheet of glass 12 inches above the pot and measure how fast the water returns to 60 degrees …
now repeat heating the same pot to 100 and suspend a 12″x12″x12″ block of aluminum 12 inches over the pot …
so now we should have the same convective cooling going on with the addition of the block radiating energy back into the pot of water …
based on the experiment described in the article I would say it would now cool slower since the block must be radiating energy back into and thus “re-heating” the water …
the experiment in the article has too many uncontrolled variables for my taste … convection being the primary one …
[snip]
Go away Myrrh and come back when you can explain why Maxwell’s equations do not apply to visible light just as they do for infra-red and radio waves. (hint: you won’t be able to)
https://en.wikipedia.org/wiki/Maxwell%27s_equations
I think what he was saying is that there is a difference between vibrational, rotational, and other modes of absorption and emission.
Light of different wavelengths produces different effects when it interacts with matter.
This is why visible light is transmitted through air with little to no distortion while other wavelengths are partially or completely blocked.
Similarly this is why a microwave oven is able to warm your food, while the same microwaves in a different use are able to transfer information through your walls from your wifi router to your computer.
Photons at visible wavelengths do not produce the same sort of rotational or vibrational heating as microwave and infrared do, respectively, visible light involves electrons hopping between energy levels.
Physics, yo, it works.
Hello Baa. The difference in this case is that the foil on the inside is a smaller box,giving less scattering to clear space and increasing the radiative power to the bulb surface. It’s like looking at the mirror from the bulb’s perspective and moving the mirror closer, the bulb in the mirror gets bigger = more power directly reflected to the bulb surface.
The difference would be through conductionof the heat in the foil to the glass, therefore the glass acting as a large heat sink to the outside world.
But once again, there’s the difference of the foil on the inside being the smaller box and greater reflector, whereas on the outside it’s larger and the IR can be scattered more easily within that space. Note also that the glass on the inside will absorb some of the IR as well, transferring that heat to the foil on the outside through conduction.
The different permutations of this experiment affect all 3 things differently – reflectivity, convection and conduction – which is why they were all taken into account.
Max:
This is a straw man argument since it a addresses and refutes a point that was never made.
While the predominant effect is blocking convective heat loss, the point was the other effect exists too.
Perhaps, but it doesn’t matter, since the claim above was visible light doesn’t heat objections. You still get absorption of photons, regardless of the mechanism, and a heating up of the matter that did the absorption.
As you said, physics works.
OK you folks who think that visible light doesn’t heat things. All light that strikes an object, and doesn’t reflect, is absorbed and results in heat. Energy is conserved.
JeffC, you say, “the experiment in the article has too many uncontrolled variables for my taste … convection being the primary one …”
One of the key aspects of the experimental design was to control for convection. That is why the cases with the metal shells were compared to that of the glass shell, and not to the bulb in open air. In all cases, direct convection to the room was suppressed.
I stated that I could not come up with any reason the conductive/convective losses would be less with the metal shells than with the glass shells, and several why they should be greater. (So the higher temperatures with the metal shells must be from radiative effects.) No one in the over 200 comments so far has even suggested a reason why the metal shells might have lower conductive/convective losses.
JeffC says (May 28, 2013 at 5:38 pm): “the experiment in the article has too many uncontrolled variables for my taste … convection being the primary one …”
What experiment did you read? The one I read explicitly controlled for convection. Any “leftover” convection would only have diminished the quite dramatic “back radiation” effect.
Uh, no, I’m pretty sure the point I was responding to was about IR blocking being important in greenhouses when it is at best a secondary or tertiary effect.
You can build a greenhouse just fine with glass that doesn’t block IR, you can not build a greenhouse without stopping convection, the point I responded to was implying that the IR blocking of the glass was worth considering, it would be with the variation I suggested, where there would be a sheet of glass between the bulb and the mirror, without otherwise stopping convection.
graphicconception says (May 28, 2013 at 4:37 pm): “My concerns include:
o) the use of glass in two places while still expecting IR to pass;”
Let your heart not be troubled. If IR passes through, fine. If it’s reflected, fine. If it’s absorbed, the glass temp increases and it loses more heat via convection and, yes, IR.
“o) the lack of consideration for any of the surfaces to absorb radiation and become warmer as a result resulting in conduction/convection;”
All of the surfaces become warmer when added to the experiment, and all of them lose heat via convection and radiation. The “inside” foil/glass/plate surfaces radiate/reflect radiation toward the bulb, toward the bench, and toward each other. Can you explain your concern in more detail?
“o) in the real world, the gases warmed by conduction at the surface will emit increased amounts of IR as a result. So GH gases, 1% of total, will absorb and re-radiate, but all gases, 100%, will radiate because of their temperature. Should this be ignored?”
Since the infrared emissivities of, for example, gaseous nitrogen and oxygen are orders of magnitude less than, for example, water vapor and carbon dioxide gas, yes, infrared radiation from these gases can be ignored, despite their greater abundance.
usurbrain, you say, “Then I wasted all the money I spent on that high dollar 3M Sun Control Film that only blocks 15% of the visible light and 97% of the IR?” What did I say that would even imply that? Do you think that the absorption in the visible spectrum does not affect the heating of the object?
Actually, you have a good test in your sun room since you have blocked out the IR part of the sun’s spectrum. Under similar solar conditions, cover everything in white fabric, then in black fabric. See if you can tell the difference.
Stated as it is that is blatantly incorrect in science! I’m ashamed of your lose terms with word Anthony and this is what keeps this misunderstanding to continue ad infinitum. If you would have included the words that this only occurs if and only if the warmer mass is an energy source, creating new energy into the system you would have been correct. Surely you are not doing this to generate traffic and comments for your site? Taken point blank you are being as bad as the ‘slayers’!
[Reply: You might want to note that Anthony is not the author of this test or article – mod]
Max, you say, “It started out dissipating 35 W to the room, it wound up dissipating 35 W to the room. So, congratulations, you’ve invented the lamp shade while busily confirming your own preconceptions, bravo! :D”
But in the different cases, temperatures internal to the system (especially the bulb surface temperature) are radically different: 105C for the glass shell, 129C for the black metal shell, and 176C for the reflective metal shell. In all cases with a 35W input from the electrical system and 35W losses to the room (with these numbers actually being slightly smaller when the temperatures are higher). That is the key point of the results.
And this is what the Slayers say is impossible. Postma says the idea that putting a highly absorptive/emissive shell around a body with an internal power source could increase the temperature of that body “is so intellectually offensive, so mentally incompetent, that it beggars the imagination.” (That’s a direct quote.) And yet that is what the black-anodized aluminum shell in the experiment does.
Curt says May 28, 2013 at 11:49 am “wikeroy, you say: “I did the basic course in thermodynamics 30 years ago. I went to the [attic] and looked through the books. Couldnt find the word [backradiation]. Probably some new definition.”
In engineering heat transfer texts, it’s usually referred to with a term like “radiative exchange” between bodies. Right near the beginning of the chapter on radiative heat transfer in any of these texts, you will see an equation for the radiative heat transfer between two bodies with some constants multiplying the term (Th^4 – Tc^4), where Th is the absolute temperature of the hotter body and Tc^4 is the temperature of the colder body. The Tc^4 term is the “back radiation”, whatever you choose to call it.”
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This is not true. The Tc^4 term refers to the temperature Tc of the colder body, it is not the “back radiation” and has nothing to do with back radiation. “Th” is the temperature of the hotter body. The term (Th^4 – Tc^4) refers to the heat transfer from the hot to the cold body and does not imply any “back radiation”. There is no room for any “exchange” in this term, only transfer.
Wikeroy has hit the nail on the head.
Nice bit of work. Clearly shows cooler object adding energy to a warmer one. (and, I’m pretty sure the warmer object does not know or care whether that energy is reflected or radiated).
Michael Moon says: May 28, 2013 at 8:52 am
The Temperature of the Filament is the only temperature that matters here. Did the Filament warm from back-radiation? Did heat actually “flow uphill?”
Geez Michael, I for one will bet you a dollar the filament will be at least a bit warmer if the bulb is so much hotter.
Gary Hladik-
“I remember chris y gave this example back in 2009, in the “Steel Greenhouse” thread.”
How did you remember that? I didn’t even remember that! I do however remember Willis’ excellent post on the steel greenhouse.
“It seems the Slayers/trolls at WUWT have been “in the dark” about this light for over three years. :-)”
🙂
Can I get class credits for digesting the education offered by rgbatduke?
Notes for critics of this experiment.
1. Nobody disputes that convection is significant in moving heat away from the bulb.
2. With the glass cube around the bulb, temperature increases until energy loss through convection, conduction and radiation is equal to the electrical energy input. This temperature will remain steady until something else changes.
3. Foil is wrapped around the glass cube and the temperature increases until energy loss through convection, conduction and radiation is equal to the electrical energy input. This temperature will remain steady until something else changes.
4. The glass cube is removed and the foil cube put back around the bulb. Temperature increases until energy loss through convection, conduction and radiation is equal to the electrical energy input. This temperature will remain steady until something else changes.
Everyone understands that the glass or foil cube slows convection so the bulb heats. Can any critic of the experiment explain the difference between glass only /glass and foil /foil only? Do you think the foil was a better insulator (of conduction or convection) than the glass? The difference is radiation effects.
This actually proves very little , the filament of a light bulb can reach temperatures in accesses 2500c which is over 4500 in Fahrenheit .
Radiation is one of the three methods of heat transmission. Radioactivity is different phenomenon. Your expt. does not prove what you claim. Heat is always transmitted from body with higher temperature to the body of lower temperature, like water from higher to lower height to level up the height.
rgbatduke says: “The conductive insulation capability of aluminum foil is zilch — try using a piece as a potholder if you doubt that.”
Now that is funny, I don’t care who you are!
Well-done (pardon pun) there rgbatduke.
From the article:
It looks as though there is more than one layer of Al foil especially at the top where the side panels are folded on one another creating many layers. With all of those alternating layers of foil and air, it creates a better blanket than the single layer of glass or blackboard does.
Please can you reveal to us Curt Wilson how many layers of foil there are? And where are they placed — it looks like there are more layers on top where you say the heat mostly accumulates. It’s a critical issue for this experiment.
It’s just automatically assumed by the author that the difference between glass/board and foil (a difference which is rather small) is due to the reflection issue rather than the foil being a more effective blanket to convection and conduction.
Kurt, might not a better experiment be to have 2 light bulbs, say 100W & 25W. Turn on the 100W & find its equilibrium. Then position the 25W nearby, turn it on & measure the changes in the 100W bulb. I would also suggest that you should position the bulbs at least a few feet away from any other surface. This method would eliminate the greenhouse effect you cause by enclosing the bulb as you did in your experiment, & also minimise any other sources of reflected energy.
[Reply: actually, Anthony tells me he has an experiment already in the works like that – mod]
Myrrh says:
May 28, 2013 at 4:19 pm
Sigh. Back in the debate about {snip – I don’t want to drag that up again}
someone pointed out a NASA (I think) page that supported his (wrong) position.
I wrote the author of the page, and he agreed with me and got the page fixed.
In this case, NASA has already fixed it – the new page at http://missionscience.nasa.gov/ems/07_infraredwaves.html says things like “The region from 8 to 15 microns (µm) is referred to by Earth scientists as thermal infrared since these wavelengths are best for studying the longwave thermal energy radiating from our planet.” And “A typical television remote control uses infrared energy at a wavelength around 940 nanometers. While you cannot “see” the light emitting from a remote, some digital and cell phone cameras are sensitive to that wavelength of radiation.” (Gone is the misleading “Shorter, near infrared waves are not hot at all – in fact you cannot even feel them.”)
BTW, the reason an IR remote control doesn’t feel hot is because it emits something like 30 or 40 milliwatts of power. Not enough to feel on your skin. A flashlight with an incandescent bulb doesn’t get hot because it puts out a fraction of a watt.
On that T-shirt description – what happens when a photon is absorbed by a black T-shirt? Photons have energy, when they are absorbed they wind up leaving some heat which can show up as a warmer T-shirt. It’s really that simple. Just follow the photons. It doesn’t matter if they’re from a visible light laser, a LED, or hot tungsten – if there are photons a T-shirt can absorb, it will, and it will get warmer. Unless, of course, it’s radiating or conducting heat away faster than it’s coming in….