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
Wow – I’ve just attempted to read the rest of the Myrrh posting – Joe Postma could pull holes in it. Total gibberish.
That’s just about enough nitwittery from you. You’ve been provided with the counterfactuals on this claim, and responded to them with nothing.
Visible light does heat. A ruby laser emits monochromatic visual light which can heat matter to the point of vaporization. A CO2 laser emits in the UV and THAT can cut metals.
These are points that directly contradict your claim. Do the right thing with those contradictions. Here‘s the scientific method, stripped down to something even a kid can understand. Use it.
GabrielHBay says:
May 30, 2013 at 4:44 am
I suppose that most of the energy input is radiated away, not conducted via the glass/isolation/foot of the bulb. The bulb glass gets hot because some part of the spectrum is absorbed and some of the inside gas conducts heat to the glass.
The setup of the experiment is to reduce the outside convection, so that that plays a similar role in all experiments.
The main difference in experiments is the difference in radiation balance. Now you can argue that the increase in temperature of the bulb is caused by less convection inside the cover, or less conduction through the cover, but that is more or less registered by measuring the temperature of the outside of the cover. You can compare the bulb temperature with the outside cover temperature to have an impression of the difference in convection/conduction/convection over the bulb-cover-outside path.
Take the two cases: glass cover only and aluminum cover only:
glass: bulb: 105.5°C, cover: 37.0°C
aluminum: bulb: 148.7°C, cover: 46.5°C
Conduction of aluminum is way better than glass and the higher temperature of aluminum at the outside of the cover should give more outside cooling convection. Despite that, the bulb and cover temperatures are way higher… The difference? Most of the visible/NIR light is going directly through the glass cover and less heat need to be conducted/convected through the glass than with the aluminum cover, where near all visible/NIR is kept within the cover and only conduction/convection can remove the inside generated energy. That it is near impossible to measure the foil temperature with an IR thermometer shows that the foil is a bad emitter in a broad range of wavelengths, including visible and NIR.
Anyway, the experiment proves qualitatively that reflecting light back to a bulb heats up the bulb (and even gives slightly more brightness). The question was if that was anyway possible, not to quantify that.
Myrrh’s claim that visible light cannot heat effectively has energy getting destroyed. A card painted with the best currently available black substance (used to be Martin black but that has been outdone, I hear) absorbs nearly all energy in the IR and visible bands. What happens to the energy in the visible bands, after being absorbed? Something magical, possibly.
Seriously, people: do a little thought experimentation. Einstein didn’t do those because he had a rich fantasy life; he did them to try out his theories and see what kind of consequences might fall out as a result.
FerdiEgb says (May 30, 2013 at 2:22 am): “physics tells us that hindering the escape of energy increases the temperature of an object until its energy out again is equal to energy in. It doesn’t matter if the hindrance is by isolation or reflecting part of the energy back to the source: in this case the surface of the earth at night.”
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No, not physics. It is warmists who tell us that and it has no basis in science.
Note the contradiction in your concept. You said “hindering the escape of energy increases the temperature of an object..”, but in fact in case of radiation energy does escape the object. Radiation always says “by-by object”. It is gone.
Back radiation can not compensate for this loss at all, no effect on the temperature of the source is possible.
For those interested in the interaction of EM radiation in water
In particular the transparency of Pure Water in the Visible Range.
There is a rather nice explanation and graph in this link
http://hyperphysics.phy-astr.gsu.edu/Hbase/chemical/watabs.html
These wavelengths have no physical link to electronic transition levels in pure water.
Any absorption of light is caused by particulates in the water for example sea water phytoplankton, seaweed and inorganic particulates.
The container of the water for example sea and pond bed may also absorb in the visible spectrum.
Ulric Lyons says (May 30, 2013 at 4:30 am): “What is your problem? I gave the link with to the original. I quoted the section that Anthony had highlighted, which is specifically what he put to test, and proved wrong.”
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I am surprised that you do not see a problem here. What was proven wrong was not the statement attributed to certain people. They did not say that. They said something different. What they really said was not proven wrong.
Eh, this experiment is too complex for slayers….
Slartibartfast says:
May 30, 2013 at 6:10 am
I have already explained the fakery in the part of that base AGWSF meme “all create heat on being absorbed”, by showing that visible light’s energy does not convert to heat energy
That’s just about enough nitwittery from you. You’ve been provided with the counterfactuals on this claim, and responded to them with nothing.
Visible light does heat. A ruby laser emits monochromatic visual light which can heat matter to the point of vaporization.
In my experiments with Laser Induced Incandescence I used green light (tripled YAG->532 nm)
to heat soot particles to ~5000K, something Myrrh claims is impossible!
I’d agree with that, but only because there is no such thing as “back radiation”.
Slartibartfast: I find myself oddly aligned with Phil.
“Beng” is correct. I think maybe slayers just aren’t capable of assimilating new information. Well, you had a good run.
After 411 comments, time to close the door on this nuttery, as it has devolved into nothing more than Principia Circusmania
Since Principia/Slayers has taken my advice and opened comments, you can continue the discussion of nuttery over there: http://principia-scientific.org/