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
The bulb only gets hotter because less heat gets away from the bulb when reflective foil surrounds the bulb ,energy flow does not change unless the flow is absorbed or more radiation is produced if no additional energy is added then if one part of the energy flow gets hotter then another part of the energy flow must get cooler in this case the outside gets cooler.
1 – No my dear Michael. The heat flowed downhill more slowly.
2 – Saying that the temperature of the filament is the only temperature that matters is like saying that the temperature of the Earth’s core is the only temperature that matters.
It was nice of the Slayers to suggest such a simple experiment. It was delightful that Anthony actually did the experiment. I would say though that Curt Wilson nailed it. The only thing missing is a couple of pages of equations. 😉 Simply elegant.
The comparison between the black cube and the foil cube was interesting. The black cube radiates both away from and toward the bulb. In that regard it is an analog of the case with greenhouse gasses.
The foil reflects energy back to the bulb. It is an analog with clouds. Unsurprisingly, all versions of the foil produced a hotter bulb than did the black cube.
This is easier than one would think. Take a 100 watt and a 60 watt incandescent bulb. Paint them black. Set them next to each other. Turn each one on independently and measure their surface temperatures facing the other bulb. Now turn them both on and notice that both temperatures will be higher.
There are numerous fundamental flaws with both of these arguments and experiments.
I am certain they do NOT actually demonstrate what is claimed.
1. It is confusing light with heat ! A light bulb is designed to generate light – NOT heat !
2. As other people have said before, in BOTH of these experiments the glass of the bulb is NOT the source of the radiant energy.
While the light is allowed to pass through the glass of the bulb the light energy is only contributing a minimal thermal response at the surface of the glass – the majority of the “heat” is coming from the heated filament being heated hot enough by the electrical current, so hot it emits visible light, by diffusion in the gases inside the bulb !!!
Remove the glass of the bulb and the filament would require significantly more energy to heat sufficiently to emit bright visible light.
If the light energy contributed significantly to the heat it would not be a good source of light.
The glass is being heated – not a source of heating – it is never itself hot enough to emit visible light on its own !!
It is being heated by the filament of the bulb which is far hotter than the glass of the bulb.
So the glass of the bulb is heated by the energy from the filament and is designed to transmit as much of that energy as possible – a useful feature for a light bulb.
The inverse square law actually proves this assertion – the filament, whilst not a point source of energy being all twisty, is still equivalent to one.
The energy at the internal surface of the glass of the bulb is already significantly less in terms of watts per square metre than that being emitted by the filament. I always thought temperature and watts per square metre had some relationship.
Exposed to the fee open air the glass is allowing the radiant energy to escape – its design purpose – and is also losing significant thermal energy by diffusion/convection in the free atmosphere.
Trap all that energy and convert the light to “heat” by absorption by various substances and of course the temperature will rise – allowed to escape freely the light contributes little to a thermal action.
The glass of the bulb is therefore only at equilibrium at a given circumstance and the fact that restricting it energy loss “capabilities” by restricting diffusion, convection and radiant energy losses simply prove that the bulb exposed to the atmosphere is NOT the TRUE equilibrium state for the TRUE heating element – the filament being heated by the electrical current.
The TRUE equilibrium state for the filament is when as much energy as it is capable of generating in response to the electrical current is truly accounted for not simply allowed to escape as it designed to do – it is a light bulb after all.
All that is demonstrated is that the glass is being heated by energy originating from a high energy source and trapping all THAT energy – that previously escaped because that is what light bulbs are designed for – will obviously cause a heating effect but I do not believe it can be claimed that the bulb was at a TRUE equilibrium originally because the glass of the bulb is not an active source of energy – it is completely passive.
Both of these experiments DO NOT demonstrate what is claimed.
It is converting the visible light that previously escaped into “heat” that causes the increases in temperature.
There is no possible way to determine what response is caused by the “reflected” thermal energy !!!
I believe the REAL physics has been misunderstood in this orgy of insult and revenge for name calling by all concerned.
Insuly leads to anger.
Neither is conducive to informed reasoning.
An incandescent lightbulb radiates around 5% visible light and 95% thermal infrared, aka longwave infrared. Visible light cannot heat matter, it is not a thermal energy. What were you measuring?
@BillMArsh,
I notice you didn’t actually answer my one question …
but you did try and make hay of my other statements …
1) There is no requirement that the objects be ‘similar’ for the experiment to work
there should be some sort of control … it is an “scientific” experiment …
2) The Aluminum foil is an ‘object’ just as the light bulb is an ‘object’.
my point is that your definition of object is pretty nebulous … which you verify here … the light bulb is hardly similar to the foil … several different materials for one …
3) Please elucidate on the several mechanisms you believe would slow the dissipation of the energy
the lack of air movement … I’ll think about some others … I’m sure you could “eludicate” some too … did you really need to ask that ? eludicate … cool word …
4) I think you’re mixing energy absorption and temperature, they aren’t the same thing.
no I’m not, I’m asking how the absorption of the radiation from the cooler object into the warmer object was measured … just because radiation is directed at an object doesn’t prove it was absorbed by said object … you are assuming that higher temp = energy absorption not me …
In general I find the discussions here interesting …. this thread appears to have broken that pattern …
Obviously I meant “Insult” – Insuly isn’t even a word.
@BillMArsh,
I notice you didn’t actually answer my one question …
but you did try and make hay of my other statements …
1) There is no requirement that the objects be ‘similar’ for the experiment to work
there should be some sort of control … it is an “scientific” experiment …
2) The Aluminum foil is an ‘object’ just as the light bulb is an ‘object’.
my point is that your definition of object is pretty nebulous … which you verify here … the light bulb is hardly similar to the foil … several different materials for one …
3) Please elucidate on the several mechanisms you believe would slow the dissipation of the energy
the lack of air movement … I’ll think about some others … I’m sure you could “eludicate” some too … did you really need to ask that ? eludicate … cool word …
4) I think you’re mixing energy absorption and temperature, they aren’t the same thing.
no I’m not, I’m asking how the absorption of the radiation from the cooler object into the warmer object was measured … just because radiation is directed at an object doesn’t prove it was absorbed by said object … you are assuming that higher temp = energy absorption not me …
In general I find the discussions here interesting …. this thread appears to have broken that pattern …
Bryan, you say, “Curt’s summary implies that the presence of a colder object will ALWAYS increase the temperature of a warmer one. It contradicts common sense. Try putting a large block of ice in your living room.”
You misinterpret my argument. The key follow-on question is, “Compared to what?” If I were worried about keeping warm, would I want a block of ice near me compared to objects at ambient temperatures of 20+C? No! But would I want a block of ice near me rather than having a direct path in that direction to a clear winter night sky? Absolutely! It’s why the Inuit build igloos. More to the point, it’s why I always carry an aluminized mylar “space blanket” when I am hiking in cold weather or skiing. If I get stuck in the cold and have to use it, that blanket will be the temperature of a block of ice, but my body temperature would be higher with it than without it.
I’ll turn the argument around on you. Take a thermostatically controlled room held at an air temperature, say, of 23C. I’m brought into the room suffering from hypothermia, which means that my body is generating as much heat as it can and my body temperature is still low. The Slayers would claim that it would make no difference to me whether the room was filled with large blocks of stone at 23C or large blocks of ice at 0C, because both are colder than my body temperature and could not possibly affect my higher body temperature. I say that is wrong, and I think that my experiment demonstrates it.
The foil shells, even though colder than the bulb, do increase the bulb’s temperature by radiating back to it compared to the control case where there is less radiation coming back to the bulb (from the room walls). Slayers have often claimed this was not possible.
This experiment, as described it, is completely laughable.
Anthony, if you are going to pretend to do a scientific experiment, you should know that you have to isolate your variables, which you utterly failed to do. For just ONE of the many flaws, nothing you are attempting to show here means anything in the presence of the conductive and convective conditions you describe.
Hell, even your initial assumptions are fatally flawed. While glass may be transparent to the near-infrared initially emitted by the bulb, anything warmed by your bulb by conduction or convection will be radiating strongly in the MID-infrared, which is almost completely reflected by normal glass, invalidating what you are trying to measure. THE GLASS ITSELF will be radiating strongly in the mid-infrared. While some of your observations about plain reflection may have some validity, they are going to be completely drowned out by the other effects.
If this is seriously intended to demonstrate your point, you fell flat on your face.
All other arguments aside, don’t even bother if you don’t have a vacuum chamber. You won’t prove anything.
I think people need to think about the real physics – the glass of the bulb is not the source of the energy which is designed to escape in the formof visible light.
The glass is never hot enough to emit visible light on its own and that simple fact plus the Inverse square law invalidate much of what is claimed.
Myrrh, you say, “Visible light cannot heat matter, it is not a thermal energy.”
Visible-light lasers, containing only a single wavelength well inside the visible spectrum, are in common use to melt steel. It is a mainstream industrial process.
Go outside on a sunny day wearing a white T-shirt. Note how warm you feel. Now change into a black T-shirt, and note how warm you feel. This is something a first-grader understands.
How much wasted time, experimental tools, and brain-electricity etc. for determining the obvious.
Just a question for all of you.
Why do you think that the inside of a thermos flask is painted mirror-silver?
Let see who knows the answer.
I have not read all the posts, but it seems to me that in the above experiment and in the GHE we are not seeing colder objects give off radiation that heats up hotter objects. We are simply seeing colder objects slow the dissipation of the energy from the hotter objects to the outside environment, causing the hotter objects to get hotter still. There is no violation in the laws of thermodynamics in such a system.
The cooler objects are not heating up the hotter objects. That would be impossible. If we turn off the energy supply to the hotter objects, the whole system begins to cool immediately. The cooler objects do not keep the hotter objects hot (or make them hotter) by radiating heat back to them.
This isn’t about one object heating another with E/M radiation. This is about one object altering the energy dissipation of another object by interfering with the radiation’s journey to the surrounding environment.
Soooo, the bulb does not become brighter after all?
It really is simple.
the energy escaping originally NEVER came from the glass of the bulb – it came from the filament.
A portion causes a thermal response in the glass both by radiant energy and diffusion/convection the rest escapes.
Stop that escape and cause the visible light to heat some object and of course the temperature will rise.
This easily explains a significant proportion of what is happening here.
There is no way to determine what contribution “reflected” thermal energy originating in the glass is contributing.
1. It is watts per square metre that is related to temperature.
2. All of the energy is originating in a point source – the filament.
3. The Inverse square law insists that intensity in watts per square metre DECREASES as you move away from that source.
4. The bulb is designed to be a source of light and allow it to escape – of course trapping the light by absorption will cause an increase it temperature !
5. Light is not the same as “heat” – light can cause “heat” by being absorbed.
5. As there is no measure of the power flux before and after there is no valid basis for any conclusions as drawn.
I don’t believe these experiments actually demonstrate what they claim when a true appraisal is applied.
A better experiment is to heat an object with external energy sources.
1. Use one at a certain power flux and record the temperature
2. Double its power and record the temperature.
3. At the original power flux add another equal and record the temperature both combine to generate.
That way you have real data to work with !
Jim Clarke says: May 28, 2013 at 3:11 pm
The cooler objects are not heating up the hotter objects. That would be impossible. If we turn off the energy supply to the hotter objects, the whole system begins to cool immediately. The cooler objects do not keep the hotter objects hot (or make them hotter) by radiating heat back to them.
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I have just completed an experiment here The results show that an object at 65°C will cause a rise in temperature of an object at 75°C. (0.3°C). I cannot explain this other than radiation from cool plate increases the temperature of the hot plate.
Feel free to criticise the experiment or make suggestions to improve the test.
Results here:
http://climateandstuff.blogspot.co.uk/2013/05/back-radiation-early-results-no-fan.html
setup here:
http://climateandstuff.blogspot.co.uk/2013/05/proposed-back-radiation-test-setup.html
chris y says (May 28, 2013 at 11:29 am): “GE R&D center developed an incandescent bulb back in the late 1980′s (and GE lighting commercialized it in the 1990′s) that placed a spherical clear glass shell around the filament. The shell was coated with a multilayer (anywhere from 15 – 30 separate layers) optical filter that reflects mid infrared back onto the filament, while allowing visible light to pass through.”
I remember chris y gave this example back in 2009, in the “Steel Greenhouse” thread.
http://wattsupwiththat.com/2009/11/17/the-steel-greenhouse/#comment-227006
It seems the Slayers/trolls at WUWT have been “in the dark” about this light for over three years. 🙂
Do the same experiment in a vacuum.
As far as I understand the matter, the major problem with thinking of people around the Principia web is that they think that when you have an object (e.g. a volume of gas) at thermal equilibrium at certain temperature, all its particles will have exactly the same energy. You can easily come to all their wrong conclusions if you take that as a fact. The only problem is that it is not true.
[+emphasis]
Does the “cooler” object have some sort of, say, ESP or precognition, to “know” that a nearby body is warmer than it, and therefore “withhold” its radiation in that direction. If you suddenly remove the “warmer” body, does the “cooler” body breathe a sigh of relief and start radiating again — assuming that it is above absolute zero?
“Bulb covered by inner reflective foil shell inside glass container 177C 68C (foil)
Bulb covered by inner reflective foil shell alone 176C 50C”
It seems that the bulb is being heated not by energy “emmitted” by the foil, but “reflected”.
You have effectively proven that a light bulb dissipates its heat mostly by radiative means. Well done.
Curt says (May 28, 2013 at 3:05 pm): ‘Myrrh, you say, “Visible light cannot heat matter, it is not a thermal energy.”’
Good examples, Curt. Myrrh is also trivially refuted by a backyard version of the Herschel Experiment:
http://coolcosmos.ipac.caltech.edu/cosmic_classroom/classroom_activities/herschel_example.html
Myrrh is so often wrong, he’s probably the only WUWT commenter I know who makes the Pink Unicorn Brigade look reasonable by comparison. 🙂
rgb,
you slightly overstate your case here:
“So LWIR heads up, encounters gas molecules, is absorbed (or not, depending on wavelength and chance) and if absorbed, is radiated back down at the surface just under 50% of the time.”
I don’t know the numbers, but quite often the “extra” energy that a photon gives to a CO2 molecule is transferred by collision to another nearby molecule. The mean free path in the atmosphere is of the order of 10^-7 m, and the particles are move with speeds on the order of 10^2 m/s, so they collide ~ every 10^-9 second. Depending on how long the CO2 molecule can stay in its excited state, there is a very good chance that a collision will occur before an emission.
Of course, *other* collisions will be giving the CO2 molecules extra energy from time to time, so eventually they should emit ~ 1/2 of the photons up and ~ 1/2 of the photons down.