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
Greg Mansion:
You say, “The term (Th^4 – Tc^4) refers to the heat transfer from the hot to the cold body and does not imply any “back radiation”. There is no room for any ‘exchange’ in this term, only transfer.”
I’m sorry, I’m used to dealing with people who are facile in math and science and do not need every tiny little step explained to them. So I will break it down into baby steps for you, using basic equations from standard heat transfer textbooks.
A body radiates a power flux density q (usually expressed in W/m^2) according to the formula:
q = ε σ T^4
where ε is the emissivity of the body (unitless, 0.0 to 1.0), σ is the universal Boltzmann constant ( in W / m^2 / K^4), and T is the absolute temperature of the body (in K).
So we have a Body 1 at temperature T1 radiating with power flux density:
q1 = ε1 σ T1^4
And we have a Body 2 at temperature T2 radiating with power flux density:
q2 = ε2 σ T2^4
These two bodies radiate toward each other through a common effective cross-sectional area Ac. So the radiative power flux from Body 1 to Body 2 is the flux density times the area:
Q(1->2) = q1 Ac = ε1 σ T1^4 Ac
The radiative power flux from Body 2 to Body 1 is:
Q(2->1) = q2 Ac = ε2 σ T2^4 Ac
The net radiative power transfer from Body 1 to Body 2 is:
Q12 = Q(1->2) – Q(2->1) = ε1 σ T1^4 Ac – ε2 σ T2^4 Ac
Now, σ is a universal constant, and Ac is by definition common to both terms. In most introductory problems, the two bodies have equal emissivities that do not vary with temperature, so to keep things simple, we will employ a common emissivity ε here. This allows us to simplify the equation to:
Q12 = ε σ (T1^4 – T2^4) Ac
Of course, if T1 is greater than T2, then we will have a net power transfer (usually manifesting itself as heat) from Body 1 to Body 2. So let’s set Th = T1 and Tc = T2 (I told you this would be baby steps), and we get:
Q12 = ε σ (Th^4 – Tc^4) Ac
This gets us to the equation I was discussing with Wikeroy. Now, when I said that the “Tc^4” term was the back radiation term, I assumed I was addressing an audience that understood the distributed property of mathematics that we all learned in second grade, which tells us that the term is really:
ε σ Tc^4 Ac
which is a power term representing the radiative power transmitted from Body 2 to Body 1 (i.e. the “back radiation”). But I guess I overestimated you.
You seem to accept the equation, but don’t understand its derivation or its implication.
Even if you don’t the term “back radiation”, it is a term in the equation. If you don’t think it is real, you still must acknowledge, if you accept the equation, that there is some other phenomenon that leads to the identical results predicted by the people who do think there is back radiation.
Greg Mansion says:
This whole argument is basically irrelevant since once you agree on the mathematical equation that you’ve written down (specifically, that the rate of heat flow away from an object depends on the temperature of the surroundings in addition to the temperature of the object), you have the greenhouse effect.
However, it happens that you are wrong about there being no radiative exchange. Here is what a few introductory physics textbooks have to say:
Young & Freedman, “University Physics”, 13th edition (2012) says:
Knight, Jones, & Field, “University Physics”, 2nd edition (2010) says:
So, there you have it from 2 textbooks. And, just in case you still think this is still a fluke, I’ll tell you that Walker, “Physics”; “Halliday, Resnick, & Walker (a different Walker), “Fundamentals of Physics”; Giancoli, “Physics”; Tipler, “Physics”; Serway & Faughn, “College Physics”; and Giambattista, Richardson, & Richardson, “College Physics” all also identify the second term in that equation as radiation that is absorbed by the body from its surroundings. I have not yet found an introductory physics textbook that doesn’t.
See also: Blackbody radiation. Or Planck’s Law.
It’s best to have done at least a little reading.
Crispin:
Or to put it another way, radiation heat energy exchange occurs by the exchange of photons between two objections. The photons being emitted by atoms of object A don’t know of the existence of object B when they are emitted, let alone what net kinetic energy the atom that absorbs it will have have when it gets absorbed.
The point is these photons from body A just get emitted based on a radiation pattern dictated by the properties of body with a spectral distribution dictated by the kinetic energy distribution of body A.
Even if you wanted to argue that it’s impossible for a photon to be absorbed by an atom in body B that has a lower mean kinetic energy in a particular inertial rest frame (there will be an infinite number of rest frames where this is not true of course), given the large number of atoms and molecules in bodies A and B, there will always be some overlap in their respective kinetic energy distributions, as long as we confine ourself to ordinary laboratory experiments, or the Earth’s surface in a realistic scenario for climate science.
Of course the statement “it’s impossible for a photon to be absorbed by an atom in body B that has a lower mean kinetic energy in a particular inertial rest frame” is simply not true, but even given that false premise, you will get photons emitted by body A and absorbed by body B, regardless of their classical thermodynamic temperatures.
Slartibartfast:
That gets in the way of belief.
Consider a satellite in orbit about the Earth. It is crossing the noon meridian. The surfaces facing the sun will receive a radiation flux equal to the solar constant. The surfaces facing the earth will receive a radiation flux equal to the Earth’s thermal blackbody radiation (plus a little of the reflected sunlight). Nobody doubts that the Earth has a far lower temperature than the sun. Is the satellite temperature determined by both heat inputs, or only by the sun’s? The answer is, of course, both, and we have to design the satellite to withstand that condition. When the satellite crosses the midnight meridian (in Earth shadow), it is heated by the Earth alone (neglecting minor thermal losses from battery currents). In both cases, it is cooled by its own radiation to intergalactic space (4 K background temperature). Satellites control their temperature by controlling their effective absoprtion/emission ratio.
If there was not a phenomenon of back-radiation, there would be no Thermos flasks. I hope the Slayers are not dismayed.
Curt says:
May 29, 2013 at 8:10 am
The last time I checked, the bulb’s glass surface was a physical object, and therefore subject to the laws of thermodynamics.
And radiation laws. The transparent glass is at 95- 170C and so radiating and absorbing much less than the filament at 3000C. ( even allowing for the larger surface area of the glass envelope ).
We are discussing a specific experiment proposed by Slayer Alan Siddons, performed and described recently by Anthony Watts and repeated by you. The claim is not about the temperature of the bulb surface it is about total radiation from the bulb ( does it shine brighter).
We are not discussing the earth’s surface or a different very universal>/i> claim. Changing the subject suggests you know you have lost the argument.
Your electrical measurements show that the bulb did not get brighter, no increase in power consumed ( W = V x A ) and emitted
Alex says:
May 29, 2013 at 8:45 am
‘A single counter-example would refute this universal claim. ‘
Sorry, proof is needed . Your experiment proved nothing. Try again. This time focus on anything connected with the filament and the slightest changes involved with that. That is the heart of it.
The point was to see if the source will increase in temperature if its own heat was reflected on it.
Many measurements of different situations were taken. Unfortunately we did not know what happened to the source. The closest measurement to the source was the surface of the globe filled with an inert gas. That’s a lot of ‘stuff’ between the source and the measurement.
The example I gave in the companion thread:
For example from GE:
“Standard incandescent and halogen lamps lose approximately 76% of the input energy by radiating heat, and convert only 8% into useful light. The Precise™ IR halogen capsule has multiple layers of very durable, thin, interference film which redirects heat, which would otherwise be wasted, back onto the lamp filament. This increases the filament temperature and allows it to give off more visible light for the same input power.
The increased burning efficiency provides the same light performance with a significantly reduced power input, alternatively allows a longer lamp operating life or a combination of both.”
Page 2 of
http://www.gelighting.com/LightingWeb/emea/images/Halogen_MR16_IR_Lamps_Data_sheet_EN_tcm181-12732.pdf
So these patented lamps which you can buy at a store demonstrate that the source does increase its temperature if its own heat is reflected back on it. Temperature measurements of the filament have been made and compared with bulbs without the coating as have measurements in the change of light emissions.
Slartibartfast.
“So, your theory is that this heat coming from the electric current being supplied to the bulb knows when there’s a reflector in place?”
So, YOUR theory is that this heat coming from the electric current being supplied to the bulb knows when there IS NOT a reflector in place?
HAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAH
Roger Clague says:
May 29, 2013 at 1:57 pm
Your electrical measurements show that the bulb did not get brighter, no increase in power consumed ( W = V x A ) and emitted
The small drop in current at near constant voltage showed that the filament got hotter, thus brighter with decreased power input. Which only is possible of some of the reflected radiation (light and/or heat) reaches the filament.
The experiment that I tried to perform was exactly that: counting the decrease in lifetime of the filament when all radiation is reflected back into the bulb, where at least some of it will reach the filament. Hotter filaments give brighter light, but shorter lifetimes. Unfortunately the bulb base got too hot to continue the experiment over a longer period. Measuring the power consumption with very accurate equipment in Curt’s experiment would give the definitve answer within minutes.
Cho_cacao,
“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!!!”
Except that is EXTREMELY oversimplifying!! The noble gases of our atmosphere do not radiate and are heated by conduction at the surface and conduction with GHG’s and then are convected away. Only conduction to GHG’s which then radiate the energy typically cool them away from the surface!!! At least that is what current science tells us. NASA did a news release on an exaggerated example of this with recent solar flares.
usurbrain says:
May 28, 2013 at 3:51 pm
to: Curt says: May 28, 2013 at 3:05 pm
Then I wasted all the money I spent on that high dollar 3M Sun Control Film that only blocks 15% of the visible light and 97% of the IR? (ratings from 3M) The Sample they gave me seems near invisible. If So, why is my “sun” room (windows on 3 sides) 20-30 degrees cooler in the summer? Also wouldn’t this make all Solar Film a scam?
Sadly, real world examples are blocked out of mind. I think it’s because of the enormity of this ‘indoctrination’, not something anyone in the educated and free West wants to think could happen, let alone that it has happened..
May be there will be a few who can face up to this, start the ball rolling, but it’s a monumental task to change this as it goes to PhD level and beyond.. . It seems practically inevitable, now that elementary physics has disappeared completely from the general education system, that there are none among the majority population capable of designing such technological innovations we take for granted in traditional physics.
As it stands, they have no idea why they have no heat from the Sun in their AGW world, nor any sound in their empty space atmosphere populated by massless ideal gas ‘molecules’. Which could explain why they can’t hear this..
Oh, and Cho,
the GHG’s also absorb INCOMING radiation preventing it from heating the surface in the first place.
If I said anything that led you to conclude that, please quote it back to me. I could see that there might be a communications error, particularly if English is not your first language.
That’s one of those “you’re going to have to show me” claims.
Carrick says:
May 29, 2013 at 9:22 am
Asking whether the filament warms is akin to asking if the GHG warms the center of the Earth too.
No, it’s asking if the GHG warms the Sun too.
Phil. says (May 29, 2013 at 2:30 pm): “For example from GE:
“Standard incandescent and halogen lamps lose approximately 76% of the input energy by radiating heat, and convert only 8% into useful light. The Precise™ IR halogen capsule has multiple layers of very durable, thin, interference film which redirects heat, which would otherwise be wasted, back onto the lamp filament. This increases the filament temperature and allows it to give off more visible light for the same input power.
So these patented lamps which you can buy at a store demonstrate that the source does increase its temperature if its own heat is reflected back on it.”
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Absolutely not.
What the manufacturer has written about back/reflected radiation warming the source has no basis in science. No experiment and no reference to any published experiment has been presented by the manufacturer. It is just a claim about their lamp being more efficient. Patent Offices do not check whether such claims are true or not, they do not conduct experiments.
Wayne:
You strenuously object to my sentence in the conclusion: “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.”
At best you have a semantic quibble. I stand by the correctness of my statement — it certainly “can” happen, as I demonstrated. I did not say “will” or “will always” or “will under all circumstances”. I don’t even consider it misleading in the context of the entire aritcle. (It could be misleading standing alone.) I carefully discussed the nature, magnitude, and constancy of the power source.
The experiment was done to test the claims of those (notably the Slayers) who state that this cannot, under any circumstances, occur, even if the warmer body has a power source. So they say “it cannot happen” and I conclude from my experiment that “it can”. The entire article provides the qualifications you want in that particular sentence.
@Slartibartfast says:
May 29, 2013 at 3:00 pm
The three most common molecules in dry air, N2, O2 & Ar, consist of one or two atoms, so don’t absorb IR radiation to the same degree (or at all) as more complex (though still simple) chemical compounds, such as CO2, CH4 or H2O, so don’t radiate as much, either.
So: if you heat them up, they stay heated? Do they have to contact a solid to lose their kinetic energy?
This is one area where my education has been sadly lacking.
Curt Wilson wrote in small part on using 40W lightbulbs:
“Also, I would not be working with contraband high-wattage devices ;-”
1: It is still legal in USA to manufacture and import 60W standard
incandescents. They get banned on 1/1/2014, same time as 40W ones.
Currently, 75W and 100W ones are banned.
2: The ban is on manufacturing and importing banned lightbulbs, not
on posession, sale or usage.
3: The ban has many exceptions.
I have a web page on the ban, its many exceptions, with links to the
legislation and a US government website explaining the ban:
http://donklipstein.com/incban.html
Ulric Lyons says (May 29, 2013 at 10:49 am): “I think this may have started with Derek Alker’s “Do moths make a light bulb burn brighter?” http://principia-scientific.org/publications/New_Concise_Experiment_on_Backradiation.pdf
though Siddons really puts his foot in it as Anthony highlights; “a light bulb facing a mirror does not heat up”: http://wattsupwiththat.files.wordpress.com/2013/05/psi_siddonscapture.png
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You quotation is inaccurate. You left out the essential part of the original statement: “from its own radiation coming back”.
This would be correct: “a light bulb facing a mirror does not heat up or shine brighter from its own radiation coming back“
[Dupe entry? Mod]
Ulric Lyons says (May 29, 2013 at 10:49 am): “I think this may have started with Derek Alker’s “Do moths make a light bulb burn brighter?” http://principia-scientific.org/publications/New_Concise_Experiment_on_Backradiation.pdf
though Sidons really puts his foot in it as Antony highlights; “a light bulb facing a mirror does not heat up”: http://wattsupwiththat.files.wordpress.com/2013/05/psi_siddonscapture.png
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You quotation is inaccurate. You left out the essential part of the original statement: “from its own radiation coming back”.
This would be correct: “a light bulb facing a mirror does not heat up or shine brighter from its own radiation coming back“
Just a test, sorry.
Slartibartfast says:
May 29, 2013 at 3:19 pm
When gaseous CO2 molecules in the air absorb IR photons of the right energy, their electrons gets excited & radiate photons in all directions. I hope this helps.
[snip – Greg House under a new fake name. Verified by network path. Mr. House has been shown the door but decided to come back as a fake persona preaching the Slayer/Principia meme. -Anthony]