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
Carrick,
“Once again you are confusing electrical power (which actually isn’t a constant, it decreases slightly as the filament heats up) with temperature, which is a measure of thermal energy.”
Once again? When was the first time? Power is Joules/second. Joules are energy. Heat is a measure of the average kinetic energy of the molecules in the object whose temperature is being measured. Power from a resistance heater such as the filament of this light bulb is voltage times amperage. The wire’s resistance changes with temperature. The writer states that the current measured to three decimal places did not change, hence the filament’s resistance did not change, hence its temperature DID NOT CHANGE.
Confusion does not justify your insulting comment, let’s all be polite to each other here.
I looked up the GE lamp. They did patent such a thing. This does not mean it worked. The vacuum inside a light bulb is not perfect. Warming the gas inside a bulb would reduce conductive losses from the filament to the gas, could give a benefit I suppose. Try it yourself, shine a flashlight directly into a mirror, touch the lens directly to the mirror, see if the light gets brighter. It will not not not…
I tire of this.
Which tool box – the radiative flux toolbox?
This doesn’t work for “clouds and the ground” on an overcast night … does it?
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If nothing else, this thread proves Mosher’s theorem that skeptics aren’t a tribe, but a confederation of tribes.
Climate skeptics are the Jews of the science world; if you get two in a room, you’ll have three opinions.
Good demonstration on INSULATION. This is a very good demonstration of GHE (heat entrapment), insulation (dead air space) and reflective insulation – nothing more. The glass cover is the GHE/dead airspace/heat entrapment, The tinfoil is reflective insulation, The black box is again “reflective” (just not as good) insulation. The piping of most Electrical Power Plants have “reflective” insulation around all (most) of the piping and all major components TO PREVENT HEAT LOSS! The boiler sure isn’t warmed by the thin layer of shiny tin-plate/aluminum.
Have you never seen the Barbie Doll Ovens of the 50’s? They used a light bulb and got up over 200 degrees F.
mkelly says:
No…It doesn’t. The experiment and what we are saying is completely compatible with this equation: The rate of heat flow away from an object depends on both the temperature of the (warmer) object and the temperature of the (cooler) surroundings. [And, what that equation doesn’t specifically tell you, but nearly any introductory physics textbook will, is that our understanding of the equation is that the first term represents radiation emitted by the object to the surroundings and the second term represents radiation absorbed by the object from the surroundings.]
Hence, if the temperature of the surroundings is higher, then the heat flow (net macroscopic energy flow) will be less. When the heat flow away from the object decreases while the inputs from other sources (such as the sun or an internal source converting some form of energy into thermal energy) stay the same, the object will find itself emitting less heat than it is absorbing and will have to increase its temperature until the balance between absorption and emission is restored. (Since, we have been trying to explain these simple ideas to you for years with no success, I don’t expect a different result now, but at least others will understand where your reasoning goes wrong.)
I have some questions that better minds than mine can hopefully answer.
The author states…
In which case, whether we place the foil on the outside of the glass or on the inside of the glass should make no difference to the result (radiative transfer restricted in both cases) but there is a substantial difference i.e. bulb T = 138DegC with foil on the outside and 176DegC with foil on the inside. How do you explain the difference?
Furthermore, the instance of foil cover alone should return the same result as foil on the inside or outside of the glass, yet it returns 148DegC. Again, how do you explain the difference.
I also note that all covers – glass, foil or a combination – are sitting on wires of varying thicknesses. What steps did you take to stop air circulation due to possible gaps between the cover and the table top? I imagine the increase in air pressure inside the covers would force some air circulation between the very warm interior and relatively cool exterior.
My jury is out at mo.
wikeroy says:
”So, when you say “the colder outer layer warmed (or slowed the cooling of if you prefer) “, you basically embraced both camps……or??
Or I’m talking about reducing radiant net energy emitted from the bulb via radiant energy from the outer layer, something the Sky-Dragon-Slayers think is impossible.
In order to calculate the net radiant emission of the bulb one would have to subtract the radiant emission from the outer layer being received by the bulb from the Stephan-Boltzmann emission of the bulb.
Full Sarc on: Photon control to photon 123423564978563p389364j262533, you are not cleared for emission along pathway G34523433445j388 due to an object 0.000001 degrees hotter than us a mere 15000 light years along this path. Please redirect emission path to S63535226522k363. /sarc
Michael Moon says:
May 28, 2013 at 8:52 am
The Temperature of the Filament is the only temperature that matters here. Did the Filament warm from back-radiation?
In his description of the proposed experiment Siddons asks does the light shine brighter?
It is the filament that is shining not the surface of the glass bulb.
Did the filament warm?
The temperature ( T ) of the filament is proportional to the power loss
power loss ( W ) = V x A.
The temperature of the filament is proportional to current flowing in it.
Bulb surface exposed T 96C current 0.289A
Bulb surface cover in foil T 156.6C current 0.289A
Covering the bulb with foil has no effect on the filament temperature. The concept of heating by back-radiation is not supported
You were called on this once, twice, thrice, maybe more.
Can you point out what paragraph that states ‘three decimal places’? He shows a LOT of data measurements (including voltage and current measurements) with FOUR significant digits. You don’t see that?
All that can be found is which could be construed to support your contentions: “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.”
Then there are these reported current values:
“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.”
Nope. Doesn’t support your contentions on multiple fronts.
Either you have comprehension issues, are acting dishonestly in debate, maybe think we’re all really stupid, or I really, really missed something in the head post.
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Macro behavior in solids is not transferable to micro behavior in individual, three atom gas molecules. The Aluminum foil is at least several thousand molecules thick, the glass even more. Radiative and convective flux are both restricted by both materials based on mass, specific heat and in this case, the sealed cubic configuration. In a free gas environment there are limited restrictions to convective and radiative flows, and a three atom gas molecule, at virtual rest compared to a photons speed, is not capable of stopping or redirecting this high velocity OLR force for more than milliseconds. Atmospheric CO2 is a feather in the path of a howitzer round.
Or, you could choose between learning some actual quantum theory and learning something about the optical cross-section of CO_2, compute or measure the mean free path of LWIR photons in the CO_2 absorption bands, and conclude that this “feather” is utterly optically opaque and indeed saturated.
The weakness of the GHE in response to increases in CO_2 concentrations isn’t because there is no effect, it is because getting an LWIR photon in the strong absorption part of the spectrum from the surface out to space is ALREADY as unlikely as being able to throw a tennis ball straight through a crowd of groupies next to the stage at a rock concert without hitting one. Adding twice as many just means that the tennis balls go an even shorter distance into the crowd before bouncing.
I mean, do you just make this stuff up as you go along, Joe, or do you even TRY to look up facts and numbers before spouting it off? Do you even understand that all mass is not only at “virtual rest” compared to a photon’s speed, it is actually at precise rest compared to a photon’s speed? Do you have any idea that you are conflating force, momentum, energy, and that the time scale of milliseconds you cite is itself unintelligible and unfounded. The time of flight of a photon traveling directly from the surface to the top of the troposphere at 9 km is 9 x 10^3/3 x 10^8 = 3 x 10^-5 seconds, that is, 3 hundredths of one millisecond, and we won’t even bother looking at the radiative lifetime of excited CO_2 as I’m sure it is order 10^-8 seconds or thereabouts, too short to care about. ONE millisecond is hence tens to hundreds of times the unobstructed time of flight to where the atmosphere is no longer optically opaque.
Why not look up, or compute, the mean free path of an IR photon in the strongly absorptive CO_2 bands before asserting that this “feather” cannot possibly be important? Or just look at TOA IR spectrographs to see just how much surface radiation is empirically blocked by this “feather”.
rgb
Semiconductor materials is my physics so quantum mechanics was not my ultimate subject to comprehend but simplistically : energy in the form of radiation is Hµ (nu). Freedom of movement and therefore bands of absorption for CO² there are 2. Energy quantum levels available to absorp electromagnetic energy are down to the electron binding energy between the O and C nuclii. If a molecule is in its’ ground state then energy of both µ (wavelength) can be absorped. However, the emitter does have to have at least one electron above the ground state, so 0K the colder object cannot warm the warmer object or more accurately a molecule at ground state cannot emit energy to an object above ground state. There is no energy to transfer. If a molecule is 2 quantum states above ground and another molecule is 1 quantum state above ground then energy can be transfered between the 2 but only higher energy to lower energy if both systems are two level systems. If an EM wave packet of energy greater than 2 quantum energy levels should enter a 2 level system at ground state then the energy will not be absorped. It must have a quantum value hµ exactly the same as the quantum level(s). If these are 3 level systems then energy can move from cold to warm and warm to cold.
The problem, for me, with all of these heat experiments is the elimination of all other forms of energy transfer. Heat transfer from the filament to the glass can only be said to be radiative if there is no gas inside the bulb otherwise heat transfer from the filament will be a combination of kinetic and radiative (convective would be infinitessimally small).
Now, When a molecule is energised above ground state it’s only objective is to divest itself of this excess energy as quickly as possible and it will do so as quickly as possible and it will do so in one direction. The sense of that direction will be a probability function that I simply cannot remember. It may or may not be totally random, but the walk towards the upper atmosphere will be but within the convection pattern prevailing and the available molecules. At 400 ppm the absorption rate is going to be very small and the vast majority of ejected EM energy is likely to head straight into space. As Niels Boor discovered, even when you know where the atom is that you wish to hit, hitting it is a very big problem.
I just hate the term “back radiation”. It grates.
OK, I’M READY. FIRE AT WILL
First, a quantum atom or molecule in the ground state is perfectly happy first absorbing a photon, then re-emitting it in a new direction. A cold gas will scatter optical energy just fine. It does need to be approximately resonant with the photon being scattered or the scattering process is second order and strongly suppressed.
Second, a quantum atom in any known quantum state is at zero temperature. Or, if you prefer, its temperature is undefined and irrelevant, but really a definite state has zero entropy and zero entropy is zero temperature. Quantum statistical mechanics requires an ensemble of atoms and a fraction of them must “probably” not be in the ground state to have a temperature. Typically such a thermal state is described by a thermally weighted “mixed state” trace of a density matrix, effectively a semiclassical state. A better treatment is to start with a density matrix and go through the process that leads to e.g. the Nakajima-Zwanzig equation for an open quantum system in contact with a thermal bath and then equilibrate it (possibly numerically via a langevin or fokker-planck sort of approach in monte carlo) — this lets you look at internal correlations in the thermally equilibrated state.
As for calling back-radiation back-radiation — in some sense I don’t like it either. But like or or not, all you have to do is step outside at night and point an IR spectrometer up and there it is, coming down, and the source of it is definitely not outer space. Nor is it just the heat already in the atmosphere radiating down. It is at least partly ground radiation being actively absorbed by the atmosphere and re-radiated back down. Back radiation is as good a term as any, but I’m open to alternatives.
Just FYI.
rgb
And now, time to stop making the world safe for sky-dragons and removing the insult to the revered name of Newton for a bit and go fishing for a while. It looks like the physics of the experiment above is well-defended by knowledgeable people so that no attempt to defend the indefensible by PSI slayers is likely to get any traction at all. Not that this should be necessary.
Joel Shore commenting on my post above
Where I say that 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.
Its unfortunate that Anthony snipped the post that this refers to and you unfortunately have got the ‘wrong end of the stick’, so to speak
I agree with your comment.
See what you and RJB think of this situation
A hollow sphere at a higher temperature and black body emissivity has a colder object placed at its centre.
Lets say that there is a vacuum between the cold and hotter sphere to eliminate conduction and convection.
Now the cold object is a new source of radiation.
Its radiation falls on the spheres inner surface and is absorbed.
Yet the sphere will not increase in temperature.
Quite the opposite in fact.
Good demonstration on INSULATION. This is a very good demonstration of GHE (heat entrapment), insulation (dead air space) and reflective insulation – nothing more. The glass cover is the GHE/dead airspace/heat entrapment, The tinfoil is reflective insulation, The black box is again “reflective” (just not as good) insulation. The piping of most Electrical Power Plants have “reflective” insulation around all (most) of the piping and all major components TO PREVENT HEAT LOSS! The boiler sure isn’t warmed by the thin layer of shiny tin-plate/aluminum.
Again, you could try actually drawing a nifty little graphic picture with arrows and everything and try using some actual equations instead of just saying “I think you’re wrong because pink unicorns tell me so”.
At least three or four pieces of evidence above indicate that the “boiler” was in fact warmed by the thin layer of shiny foil. The measured current in the filament dropped as the filament got hotter. The temperature of the bulb got warmer with any sort of optical barrier than it did with just a convective barrier. The conductivity of the foil was MUCH less than the conductivity of the glass, yet it produced the greatest warming even without the glass.
And besides, read your own words. You speak of “reflective insulation” helping out with the GHE “heat entrapment”. The conductive insulation capability of aluminum foil is zilch — try using a piece as a potholder if you doubt that. The convective trapping essentially didn’t change when switching from glass to glass surrounded by foil but the temperature went up further, so we can ignore the convective increase, there is more than just that.
You have just conceded that the foil reflects heat back into the cavity, and that this nice cool foil thereby causes the contents of the heated cavity to become warmer still. This is a nearly precise definition of the radiative atmospheric effect, a.k.a. the GHE. The sun heats the Earth. The atmosphere reflects some of the heat back at the Earth. The Earth gets warmer than it would be without the foil.
This is exactly what the slayers have repeatedly claimed cannot happen “because a colder body cannot heat a warmer one, even if the warmer one is actively being heated”. Obviously, it can. Obviously, it does. Thank you for conceding this. I will even forgive your minor error that the foil doesn’t raise the temperature of the “boiler”, or in this case the bulb filament. Insulation of a heated space does, in fact, cause its temperature to rise, all the way back to the heater, because the heater has to lose the energy you add to it against a thermal gradient. As you heat the space, you reduce the gradient. The heater gets hotter until it is power balance once again.
As the evidence above rather graphically proves for the specific case of a heated light bulb.
rgb
“Skydragon Slayers”
Say are they selling refrigerators, because if they are, my electrical meter will run backwards. A downside is everything inside will freeze, even when I jam the light switch so the light stays on while the door is closed….
Slartibartfast: 3) All of that said, I’d prefer to rectify to DC and redo the experiment than tailchase the AC aspect of things.
For precision work I’d use something like this.
“snip – if you want to point out errors, do so now. No other discussion from you is of any value at this point – Anthony”
I find it strange that you post an article refuting the “slayers” and then you delete their comments/objections. This behavior reminds me of the CAGW extremists. Not pretty.
REPLY: You are right, when somebody says you are wrong, but refuses to tell you WHY (after being repeatedly asked on another thread) it isn’t pretty when they make accusations without substance. He’s had dozens of opportunities. – Anthony
Roger Clague:
Actually, we are able to infer from the data that the filament temperature increased. This is because the measured current dropped from 289.4 mA to 288.7 mA, and the resistance of a tungsten filament increases with its temperature.
Based on the data, you are wrong.
Michael Moon:
As the report stated, the measured current dropped from 289.4 mA to 288.7 mA. The measured current did change in a way consistent with an increase in temperature for a tungsten filament.
A reading comprehension course might be needed before you take that physics course you need so desperately.
“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.”
joeldshore says:
May 28, 2013 at 12:53 pm
Joel the above quote is from the write up of the experiment. This does not square with your explanation. By the way no need to be condescending I understand the idea of insulation, but we are talking radiation exchange.
Jim:
This depends on the signal to noise of what you’re measuring.
_Jim says:
May 28, 2013 at 1:03 pm
Then there are these reported current values:
“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.”
Jim is correct. These are the important results. The part of the bulb that uses the electrical energy supplied and radiates it is the filament not the glass envelope.
To less than !% the energy radiated by the filament does not change. So we know that the temperature of the filament does not change when the bulb is covered by reflecting foil.
This depends on the signal to noise of what you’re measuring.
And more generally the quantity you’re trying to measure. The average temperature of the Earth isn’t going to be easy to measure to a part in 10^6 for example.
Roger Clague: So we know that the temperature of the filament does not change when the bulb is covered by reflecting foil.
So having quoted results that show the temperature change, you conclude the temperature does not change?
Not very interesting.
_Jim says:
May 28, 2013 at 12:16 pm
I am pretty familiar with the improved incandescent work at GE, because I worked there when the effort was going on. A genius chemist had worked out a process for depositing the multilayer coatings in a batch process for about 10 cents per bulb. That in itself was a miraculous engineering achievement.
Down the hall from my lab, a group was working on an electrodeless gas discharge lamp technology that was routinely achieving more than 200 Lumens/Watt for thousands of hours. GE Lighting poured millions of R&D dollars into that technology. In the end they couldn’t get the Lumen maintenance and color stability to meet customer demand (it was targeted for commercial and industrial lighting with a 250W bulb that was about a 1 inch diameter hollow quartz sphere).
I note that LED’s are finally starting to approach this efficiency, almost 25 years later.