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 said @ur momisugly May 30, 2013 at 12:04 am
You’re not on your Pat Malone, pal 🙂
Hint to those attempting to overturn the Received View of physics: try understanding what you are attempting to overturn first. There’s a very good reason that your “explanations” do not appear in the standard texts and it ain’t censorship!
ferdiegb,
there we go then, warmer,
I thought a great example of the experiment but mirroring the atmosphere, ,
same experiment with the lid off, has the stove surface got warmer.
sorry simple words.
so the stove surface is not actually increasing its heat beyond its energy input, it’s just the trapped surface heat is not being allowed to escape, So hotter than it would be without the barrier.
does this actually happen in the natural world apart from the earths interior.
what I mean is with the barrier ,I am not actually getting more energy.
or is the energy in the water even when cooler giving the stove extra oomph.
obviously man has taken what would be throw away energy and maximised it with boilers and then double condenser boilers.
Richard,
In the case of the pot on the stove, much depends of the conduction of the metal surfaces. As that is mostly much better than between the metal surface and air, the stove gets cooler at the place where the pot is put. If you introduce some distance between pot and surface, then the stove’s surface will get hotter, partly because of less convection, partly because of more reflection and no conduction…
Magic Turtle says:
May 29, 2013 at 7:15 pm
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The simplest experiment is to get 2 lumps of soapstone. One is heated to a temperature of say 60degC, the other to say 35degC Monitor the temperature of each.
Place the cooler soapstone say 3 inches from the warmer soapstone and see whether by introducing it to the close proximity of the warmer soapstone, the warmer soapstone warms up or not. See whether the cooler soapstone warms up, or not.
As variations, one can see.whether the rate of cooling of either soapstone is reduced by the presence of the other.
One may have to play around with the temperatures (both absolute and the differential) depending upon the accuracy of the temperature measuring device.. One may also have to play around with proximity (again depending upon the sensitivity of the temperature measuring equipment).
I would not use IR measuring equipment, but rather thermocouples (IR measuring could be used in addition)
Well, with all the argument flying about re whether or not the filament heated up from backradiation, I feel incredibly stupid for thinking that, since the filament is mounted on a glass pillar and directly connected to the base of the globe and the holder, which are all included in the whole thing that heats up through being insulated, the filament may get warmer anyway regardless of backradiation? Did I miss it or has this simple little detail not been mentioned? How on earth would one know why the filament heated up (marginally) unless it was conductively isolated from everything else? In my vast ignorance it seems as if this experiment, like the first one, proves or disproves nothing. A total crock and waste of time. And the convoluted arguments of some commentators.. too many to even list. Good for a laugh though. And PLEASE don’t tell me that yeah, but that was the original experiment proposed. Pointless to waste all this time on a poorly thought out first poser. No need to add insult to injury.
Of course, not even to mention the glass globe around the filament. Totally not conductively insulated… yet some suggest that the glass is heated through backradiation? How the heck would one know? Puleeez….
richard says:
May 30, 2013 at 1:00 am
does this actually happen in the natural world apart from the earths interior.
In theory, yes: physics tells us that hindering the escape of energy increases the temperature of an object until its energy out again is equal to energy in. It doesn’t matter if the hindrance is by isolation or reflecting part of the energy back to the source: in this case the surface of the earth at night. Which is highly felt between clear and overcast nights.
The problem is that near the earth’s surface not only radiation is at work, but also evaporation, conduction and convection. That makes that it is very difficult to separate these different influences and that the warming caused by the hindering of escaping IR near ground may be increased (as most “warmists” say) or decreased (as most skeptics say) by the other three influences. That is the real debate between the “mainstream” and the “skeptics”. The current discussion between slayers and non-slayers is just a distraction of the main discussion…
GabrielHBay says:
May 30, 2013 at 2:00 am
How on earth would one know why the filament heated up (marginally) unless it was conductively isolated from everything else?
The filament heats up by the current introduced to a few thousand °C. After a few minutes, everything is in equilibrium: the heat input from the filament equals the light/heat output via all parts by conduction, including the lamp foot, via convection (bulb glass/air) and via radiation.
If there is some backradiation to the filament, that gets hotter. That gives more heat loss to the foot and more light radiation (the light gets brighter). How can we measure that (the difference in brightness is hardly measurable)? If the filament gets hotter its resistance increases, which was measured by the small drop in current during the experiment.
@Ferdi: Missing the point here? I understand all that… The experiment tries to prove backradiation… in the the total complexity of all the heat flows and differences in insulation detail at work, I contend that it is not possible to pick up a backradiation signature. All so far has been pure speculation.
GabrielHBay said @ur momisugly May 30, 2013 at 2:55 am
No! Curt measured things. This is called empirical evidence. You are waving your arms about and believe this makes you look clever. It doesn’t.
LOL Did he measure heat flow via conduction through the base, for instance? A few simple temperature readings of the putcomes do not remotely cover the complexity of the situation or inform about the real reasons for the outcomes. So the experiment is pure handwaving, to use your own term….
Sorry, outcomes… ‘p’ too close to ‘o’…
Curt Wilson
Congratulations on a great experiment, and on doing an experiment at all, as developed further from the Anthony Watts original. The results seem to repudiate the Doomslayers ‘thought’ experiment, but do they do it well enough is my question?
As a keenly interested, but retired scientist, I have thought about your experiment and have come up with a couple of points that I think need elaboration, perhaps experimentally as well as mind exercising.
I am assuming that the rational behind all this is to demonstrate or repudiate the notion that radiative heat transfer (RHT) from a cooler black body (BB) can additionally warm and raise the surface temperature of the originating hotter BB source, the latter with its own (internal) energy source.
In which case I think that the Wilson Expt would only confirm this IF the 35W light bulb could be shown to be emittinga under BB conditions. I suspect that it could only do that if it were painted sufficiently opaquely matt black. As it stands, it obviously transmits most of the incandescent filament radiation at <1000℃, and not at the BB radiation, at the surface temp of ~100℃. Spectral analysis required?
Secondly, the experiment was done in a normal air environment which already contains water vapour and CO2 gas, and as these constituents are also linked to the ‘main question’, perhaps repeating the experiment ± (CO2 + water vapour) would also yield some useful pointers, perhaps in a dry nitrogen atmosphere?
From a debating point of view, I seem to remember that the Doomslayers original text on the ‘light bulb’ demonstration only mentioned that the light brightness wasn’t added to by the reflected radiation?
Thanks for listening, Imarcus.
GHB, you seem to have either not read the experiment, or not understood the details.
The crucial point is the change that occurs when replacing a glass surround with the reflective one. Conduction and convention will be largely unchanged, by the radition environment, and the bulb temperature, change dramatically. Q.E.D.
jimmi_the_dalek says:
May 29, 2013 at 6:34 pm
Myrrh, I think part of the problem with your posts is that you do not understand what “Thermal Radiation” is. You seem to think that it means “radiation which heats the target”. It does not mean that. It means radiation whose characteristics depend on the temperature of the source. The visible light from the sun is thermal radiation.
It’s an AGWScienceFiction fake fisics meme that claims it means “radiation whose characteristics depend on the temperature of the source” – it is part of the AGWSF package of sleights of hand, magicians’ tricks, to make you think visible light from the Sun is capable of heating matter. These are very clever, very subtle, magicians’s tricks tweaking real science terms which you will not spot unless you know what they are hiding, what they are distracting you from seeing.
Your pretend ‘electromagnetic waves’ are actually absent most of their characteristics as well as having been given erroneous properties, because the base sleight of hand tweak in this deliberate science fraud is the meme “all electromagnetic radiation is the same and all create heat on being absorbed”.
All electromagnetic radiation is not the same, that’s why in traditional physics they have been given different names…
Because they are different from each other.
These have completely different characteristics from each other firstly with regard to their wavelengths, their sizes, and because of this they have different effects from each other on meeting matter – compare the size of a gamma ray with a radio wave and what each does on meeting matter.
The visible light wave is much tinier than near infrared, which is itself much tinier than longwave infrared, neither are classed in with heat waves which are the longer waves of infrared, and which, because we have improved on Herschel’s measurements, we now know empirically these shortwaves are not thermal, are not waves “of heat”.
Thermal comes from the Greek and means “of heat”: longwave infrared is called thermal to distinguish it from waves which are not thermal and thermal infrared to distinguish it from shortwave infrared waves which are not thermal, which are not waves of heat.
I have already explained the fakery in the part of that base AGWSF meme “all create heat on being absorbed”, by showing that visible light’s energy does not convert to heat energy but to chemical and electrical energy in photosynthesis and sight, so does not physically heat up matter without which characteristic, property, we would have no weather, but what this meme also does is hide that heating matter is a powerful characteristic, it has effectively destroyed appreciation of how great a power it takes to heat matter and so destroyed appreciation of how great a power heat energy is in itself, great heat heats matter greatly. It was in understanding heat’s power that kick started the industrial revolution and so all the innovations of the modern world we enjoy today as our scientists continued to explore matter and energy.
Applying heat to something will physically heat that something by moving the whole molecules of it into vibration, which is heat. Heat is the thermal energy of something, which is movement of the whole molecules in vibration, the hotter something is the more it is vibrating, visible light works on the much tinier electonic transition level, it is only capable of moving the electrons of the molecule, it is not capable of moving the whole molecule, and unless the whole molecule is vibrated it does not heat up.
Traditional physics understands empirically the different properties and processes of electromagnetic waves, not only has it given them different names but it has put them into categories, into sets, where they share characteristics in different contexts. As I’ve given Heat/Light and Reflective/Thermal, so also we have Ionising/Non-Ionising. Look up what ionising gamma rays do to the body, and the difference between ionising and non-ionising uv.
Visible light from the Sun is benign, it does not damage the DNA which ionising uv does, for example, which is how we get “sunburn”. This is not uv heating the skin, it is uv scrambling the DNA protection from which damage our bodies produce more melanin to absorb the uv.
If our production of melanin can’t keep pace with the amount of ionising uv we are getting, we don’t get a tan, we get our DNA scrambled which destroys the skin – and we can’t feel this is happening because we can’t feel uv as heat because it isn’t thermal, so, we can be “burnt” by uv even on cloudy days because it does not have the power to heat us up so we don’t notice it. When the threat of too much uv is over, our tans fade.
However, in exploring this aspect be very aware that pharmaceutical/medical interests particularly have raised cancer fears about this to a scam level. There is now a growing incidence of rickets caused by lack of vitamin D, which had been practically eliminated after the world wars, because parents have been frightened into slathering greater and greater “sun protection” lotions onto their children even in lands where we adapted to the weak Sun by evolving white skin after leaving Africa.
Three links, the first showing research thoughts on how uv can be stronger on cloudy days, the second a typical scare page and the third the antidote to these scares:
http://zidbits.com/2011/04/on-cloudy-days-are-uv-rays-stronger/
http://www.besunsible.ca/outdoor-workers/myths-facts/
http://rense.com/general48/sunlight1.htm
Anyway, uv is even tinier than visible light, it does not penetrate beyond the first layer of the epidermis, three layers, and simply putting a shirt on will stop it.. cover up or acclimatise slowly to allow your melanin production to keep pace.
These shortwaves that AGWSF claim are heating energies from the Sun are not thermal energies, not only are they not thermal in that they are not hot, but they are not physically capable of heating matter because they work on a different much smaller electronic transition DNA level in reflection/scattering as in our blue sky and in conversion to chemical energies. They are not big enough to work on the whole molecule vibrational level which is the property of longwave infrared from the Sun and which AGWSF has taken out of its Greenhouse Effect Illusion’s energy budget.
Don’t get confused by those who would like to bamboozle you about the “meaning of heat”, hello Tim, heat is the thermal energy of the matter which is its particles in motion in vibration where they are in space, and it is the same thermal energy on the move in transfer, as it radiates out. The different contexts will use the word “heat” for one aspect, such as “in transfer”, so it is easier to describe what that discipline is talking about, what stage that thermal energy is in.
The thermal energy of the Sun which is the vibration of its matter is the same thermal energy in transfer by radiation, both are heat and are rightly called heat because they are the same thing, the same particles.. Photons are packets of particles. The heat of the Sun in transfer is the same hot matter of the Sun. Just as the heat of a campfire is the same heat you feel radiating out to you, you are feeling the heated matter move, the particles/photons. They move in electromagnetic waves.
The Sun’s millions of degrees heat is so great that its hot particles reach us at the speed of light through the 93 million miles of space between us. That’s one big campfire in the sky..
You will need to get a feel for heat energy and what it does to counter these sleights of hand from AGWSF fisics, thermal energy whether by conduction, convection or radiation, heats us up. It heats up our skin by making the molecules of our skin vibrate and the faster they vibrate the hotter they get, and it is readily absorbed by the water in us which heats us up internally.
And radiant heat from the Sun which is thermal infrared which is longwave infrared is particularly good at this as it penetrates several inches into our bodies which absorb it, rub your hands together, this is mechanical energy, friction, moving the whole molecules of your skin heating them up. That’s how powerful radiant Heat from the Sun heats us up. Light from the Sun cannot and does not do this.
http://tes.asu.edu/MARS_SURVEYOR/MGSTES/TIR_description.html “What is Thermal Infrared Energy?
“Light and Heat
“Thermal IR energy is more commonly known as “heat”. Everyone is familiar with heat because of our sense of touch. But what exactly is heat? Heat is a form of light invisible to our eyes, but detectable with our skin. Visible light is part of a large spectrum of energy that includes other familiar electromagnetic energy regions: microwaves, radio waves, ultraviolet, and X-rays all are forms of light that we can not see. The colors of a rainbow form a continuous spectrum of light in the visible wavelength region as does the “light” in the other regions. Infrared light occurs at wavelengths just below red light, hence the name, infra- (below) red. Near-infrared is the “color” of the heating coil on an electric stove just before it glows red. The thermal (or mid-) infrared colors are found at even longer wavelengths.”
http://science.howstuffworks.com/dictionary/physics-terms/heat-info4.htm
And note as it says here, http://answers.yahoo.com/question/index?qid=20070322131802AAsc6M7, that not only does heat energy, longwave infrared aka thermal infrared aka radiant heat, move the whole molecule into vibration, that molecule expands (because it is a real gas and not the AGWSF fictional massless ideal gas). Heat cause the molecule to become less dense, to expand in volume, the hotter it gets the faster and more it expands, and becoming lighter than air under gravity will rise in air, which is how we get our winds. When heated less dense lighter than air under gravity molecules of air rise taking the heat away from the surface (convection) in their combined volume, ‘packet’, they release their heat on reaching the colder heights and condense, become more dense again, and thus now heavier than the fluid gas air will sink displacing air. These are called convection currents, in the atmosphere they are known as winds and in the ocean simply currents.
Areas of high pressure are combined volumes of cold, heavy, individually condensed volumes of gas, areas of low pressure are hot, light, individually expanded volumes of gas. Hot air rises, cold air sinks, and, heat flows from high to low.
The hotter the combined volumes of gas and the greater the contrast between cold and hot volumes, the faster the air will flow from one to the other – the air heated intensely at the equator flows rapidly to the much colder poles becauuse the intense cold is absorbing it faster.
Hold an ice cube in your hand, it is not the ice cube making your skin colder by transferring cold to you, but your skin getting colder because it is losing its heat to the great power of cold to absorb heat.
This is what creates our great wind systems as the much colder so more condensed heavier air at the poles sinks and flows back to the hot low pressure area at the equator.
Where it first got heated by the powerful heat energy of the Sun directly heating the land and water, the Sun’s direct thermal energy in transfer, which is longwave infrared.
Which AGW has excised from its energy budget – so you have no heat at all from the Sun in your world.
Greg House says:
May 29, 2013 at 3:22 pm
“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“ ”
What is your problem? I gave the link with to the original. I quoted the section that Anthony had highlighted, which is specifically what he put to test, and proved wrong.
@steveta: Obviously we have different standards as to proof required for definitive scientific outcome. Convection and conduction patterns and values were not measured in any way and the contention that they remained largely unchanged is unsupported. The interpretation of outcomes hangs on this assumption. So, the interpretation is speculation, albeit perhaps an attempt at intelligent speculation. But speculation is speculation.
Greg House says:
May 29, 2013 at 3:04 pm
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.
Yes Greg you’ve told this lie several times before and been rebutted, it’s rather pathetic the way you cling to your belief system by refusing to read the evidence. You’re as bad as Myrhh who trots out his mistaken view of the universe and erroneous science and won’t listen to any rebuttal. His latest gem is that the sun’s surface temperature is millions of degrees (whereas it’s about 5800K). He does say that it emits near IR photons which is correct (but would actually emit hard X-rays if he were right about the temperature!)
Myrrh,
Instead of long meaningless posts, try just looking up the definition of “Thermal Radiation” in a standard physics text book – not a website, especially not a website concerned with the pros and cons of climate “science” , just an ordinary physics text book.
Faulty premise. There is no such thing as “backradiation”. It’s just radiation.
“to make you think visible light from the Sun is capable of heating matter.”
Myrrh, one of the favourite skeptical fringe themes is that LW IR cannot warm the ocean as it cannot penetrate water so can only increase evaporation (nonsense, but commonly stated).
In that case, since visible light that does penetrate a few 100 feet into the ocean is “incapable of heating matter”, exactly what is left that makes the top layers of the ocean so much warmer than the deeper layers?