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
I continue to be amazed that this remains a controversial issue. 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. The result is that a lower filament current can achieve the same filament temperature, thanks to the mid infrared energy being reflected back to the filament. This results in a 15% – 20% increase in lumens/Watt.
Designing a filament that has low emissivity in the infrared and high emissivity in the visible could achieve similar improvements in lumens/Watt.
And part of the Gulf of Mexico flows up to Missouri, right?
Vince Causey says:
May 28, 2013 at 11:07 am
“How can the amount of energy leaving the system be less than the amount entering, if the system is at equilibrium?
The temperature of the bulb increases until equilibrium is reached, at which point energy leaving the system (at the surface of the foil box) equals energy entering.via the filament, which is assumed to have remained constant.
In fact, the energy leaving the system never changed, and neither did the energy entering. What changed was the size of the system, moving from the filament and bulb, to filament, bulb and foil box.”
I am talking about the total amount of energy (Joules) in the system, you are talking about the rate (Watts – Joules/second) the energy enters or leaves the system. The system is at equilibrium when the rate of energy entering the system is equal to the rate that the energy is leaving the system. The amount of energy in the system determines its temperature, as the temperature rises the rate of energy leaving the system also rises until it matches the rate of energy entering the system.
P Gosselin:
Given that macroscopic objects are composed of greater than 10^23 molecules and atoms, and further these molecules and atoms obey a distribution in kinetic energy, there will always be an overlap in kinetic energy distributions between molecules in the colder object and the warmer object.
That is on average some molecules on the colder object will be warmer than some molecules on the warmer object (actually it’s “very many”), so even if your pet theory were correct, what you are saying would still be wrong.
Beyond that, how does a photon once it is emitted by the colder object “know” that it came from a cold object so that it “knows” not to be absorbed by the warmer object? These are some smart photons in your universe, don’t you think?
Since Anthony snipped an earlier comment from the Slayer, he will visit other blogs that allow him a forum to espouse his belief and attempt to beat into submission those who disagree with him and/or do not believe has he believes. He will, along with the rest of his cadre. It is quite remarkable considering they do not allow an open forum on their site.
pgosselin:
Another physics fail from the failed physics group:
Radiative transfer of photons and macroscopic flow aren’t appropriate physical analogies.
John West says:
May 28, 2013 at 11:15 am
“Bottom line is that the bulb temperature increases in the presence of a colder outer layer, thus the colder outer layer warmed (or slowed the cooling of if you prefer) the inner layer. Sky-Dragon-Slayer premise busted!”
Well, I am not so sure.
If I have understood the discussion correctly, the Watt’s camp says ” thus the colder outer layer warmed the inner layer” , and the Slayer camp says “slowed the cooling” ( Or Insulated it, thereby making the temperature go up).
So, when you say “the colder outer layer warmed (or slowed the cooling of if you prefer) “, you basically embraced both camps……or??
Curt Wilson,
Simple and well thought and results far beyond reasonable doubt.
That is what real science is all about!
Ferdinand
Bryan says:
No…What is says is that a colder object will increase the temperature of a warmer object where that increase is measured relative to the situation where we have a MUCH colder object there instead. In other words, try putting a vat of liquid nitrogen in your living room and then replacing it by a block of ice instead. This replacement will cause an increase in the temperature of objects in the room (with the details depending on how close the object is to the vat or block, etc.) just as replacing the 2.7 K radiative temperature of outer space with the radiative temperature of the atmosphere causes the Earth’s surface to be warmer.
Julian Flood says:
What Julian is describing is a very nice experiment first carried out around 1800 by Pictet and discussed here in a 1985 American Journal of Physics article: http://www2.ups.edu/faculty/jcevans/Pictet%27s%20experiment.pdf It is particularly nice in that it really manages to get rid of any significant conductive or convective effects. See p. 741 of the article I linked to for Pictet’s own description of the experiment and result. Also note that there is a discussion of our modern understanding of what the experiment demonstrates on pp. 749-750 and then also a discussion of how to reproduce the experiment.
_Jim says:
May 28, 2013 at 11:02 am
Problem being, this varies dynamically over the applied AC sinusoid even; above is a post where I propose to measure the bulb filaments characteristic over an applied cycle in 500 us steps (using DAQ or data acquisition cards), then the instantaneously measured V / I performance of the filament could be used to calculate it’s instantaneous R, and infer the change in temperature.
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Please. None of this matters. You could use DC, and you’d get exactly the same result.
A light bulb that has vacuum in it (not a gas filled one) is an ideal object to demonstrate the principle of back radiation absorption. However a different setup than the above would be better. While glass passes short wave radiation well, it has high absorption and emissivity at much longer wavelengths. The best way to make the glass absorb all wavelengths is to paint the outside with black paint. Then coat the outside with a layer of insulation (not too thick or the inside will get too hot). Now use the fixed voltage power in, and an accurate ammeter to track current. As the bulb and insulation heat up inside, the only back radiation will be long wave thermal radiation, not reflected short wave light. Nevertheless, the resistance of the light bulb will increase as the filament heats up from back radiation. This will cause a drop in current. There is no other possible cause for the drop in current other than low temperature glass heating the much hotter filament. Convection and short wave reflection were removed from the test.
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)…”
_____________________________
Do you suppose that they paid for their investment in research and tooling?
The last GE plant making incandescent bulbs closed in 2010.
http://www.washingtonpost.com/wp-dyn/content/article/2010/09/07/AR2010090706933.html
My whole question posed about GE profiting from their bulbs is but a gnat’s poke at the world economy and politics. GE doesn’t pay taxes, while it funnels millions into the coffers of “Green” causes and politicians, who then divert funds from individuals making rational business decisions into the pockets of GE via tax subsidized wind generators, etc.
In my comment, the glass I refer to is the bulb itself, not an external container.
Hello Everyone,
Thanks for your interest and comments. Some clarifications in response to questions and comments:
Arthur4563, you say: “Offhand, I’d say the experiment makes an invalid claim : that the foil is a cooler body radiating anything. Clearly the foil is not the source of the radiation being reflected, nor could any foil body at its claimed temperature radiate any such powerful heat radiation. Explanation please.”
Sometimes when you are evaluating, um, “unusual” claims, you have to go down their road a bit. (When I watch Mythbusters with my kids, I occasionally find myself yelling at the screen, “Did you really need to test that?!”) In this case, I was testing the claim I have seen made several times that reflecting radiation back to a hot emitting body such as a light bulb could not increase the temperature of the bulb. I think I disproved that claim here. I was careful to distinguish this case from the other case with the black shell that absorbed the bulb shortwave radiation and re-emitted longwave radiation. This also caused the temperature of the bulb to increase.
philr1992, you say: “I’m not a slayer, but some of the effect you’re getting is the glass-greenhouse effect..where heat from the bulb is conducted to the surrounding gases which cannot convectively cool. The gases surrounding the enclosed bulb are thus warmer and direct heat loss via conduction is reduced.”
The whole point of the experimental design was to control for this “glass-greenhouse effect”. That’s why the temperatures of the metal shell cases were compared to that of the glass shell case and not the bulb in open air. I cannot think of any way in which the metal shells would inhibit conductive/convective losses more than the glass shell does, leaving changed radiative transfer as the mechanism for the difference.
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.
To all who missed the point about the filament resistance, please read the whole post. I did measure the current through the filament, and it did decrease slightly when the bulb surface temperature was hotter. This demonstrates two things. First, the higher temperature cannot be explained by greater electrical power input. Second, the decrease indicates higher electrical resistance of the metal filament, which indicates a higher temperature, even as it dissipates less power.
Mistaken impression by the public (I’m afraid); 2.4 GHz is not a magic number selected b/c ‘water’ is resonant there. Verily, industrial processors use 915 MHz equipment for the deep ‘reach’ (penetration) into foodstuffs. 2.4 GHz happens to have a very manageable wavelength compared to 915 MHz (or 27 MHz, the old ‘diathermy’ equipment used that!)
I have been using this experiment to heat my pump house for thirty five year , insulation is a wonderful thing .
Leonard Weinstein says:
May 28, 2013 at 11:46 am
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I actually did that futzing around when I was a kid. I painted a light bulb with dark red paint (don’t ask why). It got so hot it burned the bulb socket.
Cho_cacao says:
May 28, 2013 at 10:52 am
“OMG you finally understood how it works! The sun is providing a (so to say) constant energy input, the GHGs reduce the amount of energy being released to space… And voilà!!!”
Why, then, does the temperature not go up since 1998?
Does the energy enter the oceans? How, when it is delivered by IR photons? IR does not enter water beyond the skin layer.
Something is wrong in the models.
Assertion monkey is assertive.
Look, I am beginning to suspect that y’all are equating energy transfer with net energy transfer. Sure, the cooler object receives more energy from the warmer one than flows in the opposite direction. What of it? The warmer object is still getting more radiation than it would from e.g. intergalactic space.
wickeroy:
The PSI group has claimed the bulb won’t heat up, so it’s hardly embracing both camps.
“Slowing the rate of cooling” is a alternative description to the “colder outer layer warmed”. They mean the same thing. Some of us prefer one description over the other, but the point is the physics isn’t changed by choosing a different description of the physical laws.
AFTER all the detail I included? PLEASE, who is ignoring factors here? You may ‘lose’ your possibly small change in signal (current value) in the ‘noise’, as, for instance, today most mains line voltage has the appearance of being ‘clipped’ owing to the amount of … you tell me, since you’ve assumed expert status on this! When is the last time you viewed on an O-scope the AC mains? Ever?
Dear readers, use caution when seeking technical advice from this man …
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Carrick says:
May 28, 2013 at 11:54 am
“Some of us prefer one description over the other”.
Okay, I must be “old school” then, since I prefer “slowing the rate of cooling, then. “Backradiation” sound so….backwards.
DirkH:
It’s gone up actually, just slowed a bit.
You need to specify which wavelength of IR you’re talking about… but for IR photons form the sun, the average penetration depth is on the order of meters.
From there it gets distributed by the mechanisms of thermal conduction and convection.
Something is always wrong in the models. That’s why we call them models and still collect experimental data.
Before I would attempt to screw this whole thing up with AC analysis, I’d consider these things:
1) You’re going to have to estimate the thermal mass of the filament and model its thermal rise and decay…at 60 Hz.
1.5) Since it’s the temperature of the glass bulb that is being measured, you’re going to have to model its response to the filament radiant energy.
2) The experiment is still a decent exhibition of the author’s point even if the filament temperature is oscillating.
3) All of that said, I’d prefer to rectify to DC and redo the experiment than tailchase the AC aspect of things.
Not really an open system at all. The only time it was an open system for purposes of the experiment was when the bulb was all by itself.
I also used lost incorrectly. I meant absorbed into the glass, thus preventing that energy from being reflected.
My base argument is that of all the testing, the one thing that did not get tested was what is being argued, that gasses cause a greenhouse effect.
My understanding of the situation is the following.
Yes, there is a green house effect from some gasses.
That greenhouse effect is limited by the amount of energy available in the frequencies those gases work in.
That once the bandwidth is saturated no amount of additional gas will have any measurable, within the next 100 years (technological limit, not effect), effect on the climate of the Earth.
The sun and our position in the galaxy drives our climate through direct energy input and galactic cosmic rays effecting cloud cover.