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
richard says: May 29, 2013 at 4:08 am
now if the lake was the same temp as the house you would still need a geothermal heat pump to extract the difference in temp to make the house hotter, now put that lake in the sky same temps as down at the surface,, you would need a heat transfer pump to make the surface hotter.
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Its not about making it hotter it is about slowing the cooling.
If your house is surounded by -40°C air and your house internal temperature is 20°C it will be gaining energy from the -40°C air and losing even more to the air.
surround your house with a blanket of lake water at 4°C suitably contained and you will still be losing similar energy from the 20°C room but you would now be gaining energy from -4°C instead of 40°C i.e you will be getting extra joules from the COLD water and your room will cool slower. IT WILL NOT HEAT UP that is obvious and it is not what GHG theory suggest will happen.
“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”
ok then what happens. the warmer body gets hotter warming the colder body in turn warming the hotter body, when does it stop, sounds wonderful, sounds like free energy to me.
“And why would that be, do you think? Both filaments will be drawing the same power, have the same ability to lose energy via convection and/or conduction. Only radiation is being blocked.”
You have wrapped an insulator around the globe and also entrained air. No surprise that the globe overheats. That does not prove that the filament has increased in temperature. It just means that you did not allow the globe to cool as it was designed to.
When you cover an electric bulb, the bulb will heat up until it reaches an equilibrium becoming extremely hot, possibly melting the mains until it shorts out, as an engineer, I am not happy about this being an explanation of a “greenhouse effect”. as it is governed by Ohm’s law, and we know where the energy is and what it is doing, if the ‘Slayers’ invented this logical fallacy I 100% disagree with them. “The Green house effect” as I understand it; is like poring boiling hot water into a cup at room temperature and watching the cup heat the boiling hot water up. There fixed it for ya!
[snip – your language is the issue -mod]
This is an easily falsifiable assertion. CO2 lasers emit in the UV but are capable of welding or cutting materials.
Don’t tell me lasers are special. It’s true, but it has nothing to do with heating.
ok then what happens. the warmer body gets hotter warming the colder body in turn warming the hotter body, when does it stop, sounds wonderful, sounds like free energy to me.
Please, Richard, think. If you are cold, you get under a blanket, which is colder than you originally. Then your heat warms the blanket. As it does you are suddenly surrounded not by cold air but by a warmer blanket and your rate of heat loss slows, since you lose less heat when you are surrounded by warmth than when you are surrounded by cold. Given an essentially constant rate of biological heat production, your skin warms up. For a lossy blanket, eventually this warming slows. If you piled on too many blankets, you get hotter than you like and your body’s cooling mechanisms kick in and you sweat.
The point is that the warmer body has a constant supply of energy. It generates, or absorbs, or produces, power — energy per unit time. This energy has to go someplace, and the rate that the warmer body loses it by any channel but latent heat depends on the temperature difference of surface of the body and its surroundings.
So listen carefully — you can understand this. As the warm body warms the cooler body, it reduces the rate of heat loss from the warmer body. But the rate of heat production has not changed! The body accumulates energy, which increases its temperature, until the temperature difference once again permits the warmer body to lose heat at exactly the rate of production.
If you surrounded the heated body with a perfectly insulating wall, what physicists call an “adiabatic container” (one that does not pass heat) then the First Law of thermodynamics clearly predicts that as long as you continue adding energy through that wall (via, for example, electrical wires running to an electrical heater) the temperature of the interior of the cavity will increase. This isn’t “free energy”, it is energy conservation, rather the opposite.
What the slayers propose violates this principle. According to them, the temperature of the surroundings doesn’t matter. The interior heated body will remain at the same temperature no matter what because nothing can modulate the rate at which it loses energy radiatively, even when it is placed in a vacuum so that radiation is the ONLY way it can lose energy. Surround the vacuum with ice cold walls, it won’t matter. Surround the vacuum with walls only a tiny bit colder than the object, it doesn’t matter. Surround the vacuum with a perfect reflector of radiation (making a “thermos jar” and it doesn’t matter (to them). The rate of radiative heat loss is the same, and the temperature of the heated source does not change.
Which is absurd.
rgb
Whether the filament has increased in temperature is an interesting question, but unknowable and (says I) irrelevant.
The filament has to get hotter, because the hotter light bulb is radiating in both directions. How much hotter? I have no idea, and you probably don’t either.
The question of whether the bulb got brighter (asked by someone else) betrays some absence of knowledge. Of course it got brighter. Its emissivity is unchanged, and its temperature increased. It’s radiating more energy at all wavelengths.
Forget about the physics. Has anyone ever attempted to assess how much warming globally is derived from the heat generated from the myriad of electric lights worldwide? Or the TVs, dishwashers, washing machines etc., all of which are hot to touch? Is the UHI effect proportionate to the wealth of a city as well as its population?
“You have wrapped an insulator around the globe and also entrained air”
I assume you mean “entrapped air” – but only a minute amount, as the foil is directly wrapped arounf the glass.
And why do you think foil is an insulator? Would it be because it’s used in vacuum flasks, and insulation sheets, and space blankets, and all the other applications where it is there to reflect IR? It would have no effect on convection, and minimal effect on conduction, so what else is there?
‘The question of whether the bulb got brighter (asked by someone else) betrays some absence of knowledge. Of course it got brighter. Its emissivity is unchanged, and its temperature increased. It’s radiating more energy at all wavelengths.’
Huh?
emissivity and temperature are correlated.
Adding a layer of material over the bulb has the same effect as painting it a different color or even replacing the glass with a different material, which means you changed the emissivity, did you not?
Adding a mirror means one side of the bulb has reduced emissivity relative to the other side, does it not?
Are you sure the lack of knowledge is on my part?
Is the room lit up more, less, or the same amount in this experiment?
I was going to post something that in effect posited certain rules that matter adheres to as far as EM radiation is concerned, but I thought there were huge possibilities for quibbling over details, so I refrained. Contrary to my better judgement, though:
Can we agree that all objects that have nonzero absolute temperature and nonzero emissivity (this is basically all real objects) emit electromagnetic energy (i.e. heat, light, etc) in general accordance with Planck’s Law?
Can we agree that all matter either absorbs, reflects or transmits incoming radiation? And that whatever is not transmitted or reflected MUST be absorbed?
Can we agree that radiation absorbed will in general be at least partially converted into thermal energy? I didn’t say completely because there are a few special cases where some of the incoming energy is NOT converted to thermal, and I am acutely aware of the propensity for engineers and scientists to nitpick.
I am nowhere near an expert in these matters; by training I am an Electrical Engineer and I have but dabbled in matters pertaining to radiant energy. Nevertheless, I think that dabbling has been handy, and instructive.
No, that wasn’t done. The glass of the bulb still has the same emissivity.
Adding a mirror means one side of the bulb has reduced emissivity relative to the other side, does it not?
To the other side of what? Again: the bulb has the same emissivity.
This question is not relevant. No one is talking about whether the system got brighter.
I would say it’s more in the way of flawed thinking.
“Given an essentially constant rate of biological heat production, your skin warms up”
Much like a steam cooker without a safety valve open, it’s not the actual metal cooker adding heat.
you make it sound like I am increasing energy, which if possible would be amazing in industry, please illustrate where this happens if it does.
all that is happening is the heat is not being allowed to escape, so insulation.
richard, yes you can call it ‘the heat is not being allowed to escape’ or ‘reducing the cooling rate’. Like closing an open window in winter and thereby warming the room.
Igloos!!!!!
but in the real world of our atmosphere, it becomes irrelevant i guess. I believe that is what they are trying to address.
of course a saucepan of hotter water with a sealed lid will explode eventually – man made,
take the lid off and top up with cool water, tranquility – natural.
I have to say I really missed something here , I want to learn about the atmosphere and how it works and I am given man made products, insulation and how it could happen.
Curt says:
May 28, 2013 at 7:31 pm
But in the different cases, temperatures internal to the system (especially the bulb surface temperature) are radically different: 105C for the glass shell, 129C for the black metal shell, and 176C for the reflective metal shell.
Alan Siddons says radiation sent back to its source does not make it hotter. Your experiment is to test that claim. So you need only to measure the temperature of the the source, the filament inside the glass envelope. The bulb surface temperatures are not relevant to the purpose of your experiment.
The power output is a measure of the temperature of the filament.
The current flowing in the filament ( A ) is a measure of the temperature of the filament
Current changes by from 0.2894 to 0.2887.
A change of 0.0007. That is 0.0007/ 0.2894 x 100%
= 7/2894 x 100%
= 0.24%
This is within the accuracy of the meters used to measure V and A.
The temperature of the filament, the source of the radiation does not change.
You have conformed that radiation sent back to its source does not make it hotter
That is not what you set out to do. But as you are a good scientist you correctly included measurements ( current flowing in filament ) , made as a control, that you could have left out.
They negate your premise but should not be ignored..
My apologies to Anthony for saying those were his words, when I saw “Watts – part Two” I didn’t even check the author. Then I hope Curt Wilson doesn’t really believe the way his statement read. More on this later. I’m frustrated, both Anthony and Dr. Robert Brown who both carry much weight here are making statements that though correct partially (by not saying it all) are leaving the wrong impression to many and I can see those impressions then in following comments and this ‘partially incorrect as stated’ concepts needs to be cleared up. This thread just adds more to it.
Briefly, an gases radiation, specific lines, because it is isotropic, goes in any direction, can never oppose energy leaving a planet’s surface but by less than one half per complete opacity (per the optical depth) due to the curvature of the surface. Here on Earth that factor is close to 48.5% down and 51.5% toward and lost to space. The ratio of those is approx. 51.5/48.5 to be 106%. By subtracting what leaves unimpeded through the window frequencies you get this equation to closely estimate what the OLR should be:
window UWIR = 63 W/m²
OLR = window + (396-window) * 1.06/2 = 239.5 W/m²
… and you don’t have to go much further to understand this about DWIR (“back radiation”) influence. Doubling co2’s concentration is only going to affect this by a very, very tiny amount. Everyone should know why by now why… co2 lines are but a small portion of the spectrum, co2 already absorbs all in those lines near the surface, etc, etc, etc.
As for the mirrored image of a light bulb heating the filament heat source, it can only occur per the fraction of the 4pi steradians that the mirrored image projects backwards upon that tiny area of the filament itself… a really tiny area. So these such ‘experiments’ must be taken with a grain of salt if they imply some huge effect.
[snip – your language is the issue -mod]
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[snip . . mod]
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,
two ice cubes, one a lot colder than the other, if you place them next to each other what happens.
richard, there are people distracting and confusing the matter by claiming you cannot warm a body by reducing its cooling rate (and keeping its warming rate constant at the same time). The postulated radiative GHG effect is another matter.
does the warmer one melt quicker than if left standing by itself.
richard, go solve some problems from a Heat Transfer textbook and see if you can the correct results.