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.
=====================================
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
Curt says:
May 28, 2013 at 3:05 pm
Myrrh, you say, “Visible light cannot heat matter, it is not a thermal energy.”
Visible-light lasers, containing only a single wavelength well inside the visible spectrum, are in common use to melt steel. It is a mainstream industrial process.
The incandescent lightbulb is not a laser, neither is the Sun.
Otherwise there would be no life on Earth…
Visible light as radiated from the Sun and from an incandescnet lightbulb is not enhanced in any way. This works on the electronic transition level, that is, reacts with the electrons of matter, it does not work on the molecular vibrational level which is what it takes to heat matter, getting the whole molecule to virbrate – this is done by thermal infrared, aka longwave.
Go outside on a sunny day wearing a white T-shirt. Note how warm you feel. Now change into a black T-shirt, and note how warm you feel. This is something a first-grader understands.
Since visible light cannot be heating either t-shirt, just what do you think a first grader can understand here?
I’m sorry to have to tell you, that the general education system was corrupted some time ago to fit in with AGW agenda, if you want to know basic physics in this you need to find a traditional science teacher, or applied scientists in the field. For example, those building photovoltaic cells to convert shortwaves to electricity and building thermal panels to capture the longwave infrared direct from the Sun, aka thermal infrared, aka heat. Or, what might be easier to think about, consider the manufacture of window for hot countries designed to admit maximum light and reflect away heat, longwave infrared direct from the Sun, to save on air conditioning costs. If visible light heats matter, then they are maximising heating the room, not cooling..
AGWScienceFiction has replaced the direct heat we actually feel from the Sun which is longwave infrared, aka thermal infrard, aka radiant theat, with mainly visible light and shortwaves either side in order to create the illusion of “backradiation” – so now, whatever heat they measure down welling they can attribute this mythical “backradiation”.
You can continue arguing about it, but if you, generic, don’t know the difference between heat and light electromagnetic radiation then all your experiments are, quite frankly, meaningless, because you can’t even begin to think about what you are measuring.
You have no heat from the Sun at all in the AGW/CAGW world.
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?
Just to note the obvious, may have been said, but it’s worth adding again:
The bulb was heated by the power supplied to it, not back radiation, obviously.
The foil/aluminum (nice job btw, I had the same suggestion in the other thread, though I suggested controlling for convective changes by leaving a sheet of glass in between the bulb and another piece of glass/black surface/mirrored surface) simply reduced the ability of the bulb to radiate heat away into the room, it did not make the bulb brighter, you might have noticed that this was the way Joe phrased it (I never noticed the bit by Siddons about heating, silly oversight on his part) as “the bulb will not be made brighter”, and this experiment confirmed that.
It started out dissipating 35 W to the room, it wound up dissipating 35 W to the room.
So, congratulations, you’ve invented the lamp shade while busily confirming your own preconceptions, bravo! 😀
Baa Humbug at May 28, 2013 12:55 pm asks why the foil outside the glass, foil outside the glass and foil only have different results. The Analysis section of the original post already lays this out but let me try paraphrasing.
Remember that glass is not a perfect transmitter and foil is not a perfect reflector. Some energy is absorbed in both cases. When comparing the foil-outside-glass to foil-only, the glass is absorbing some small fraction of the energy as it’s radiated from the bulb out to the foil. It is also absorbing some of the reflected energy on it’s way from the foil inward. The energy is not lost but it is converted to heat in the glass which is then dissipated by convection and conduction and is no longer available to radiate back to the bulb. That’s why the bulb’s equilibrium temperature in foil alone is higher (148C) than the bulb’s equilibrium temperature in glass then foil (138C).
The foil-inside-glass is almost identical to foil-alone WHEN USING THE SAME SIZED FOIL (177C vs 176C). The difference between the two foil-only measures (148C when sized to fit outside the glass and 176C when sized to fit inside the glass) is due to 1) a minor effect from the smaller floor (the sixth side of the “box” which is more reflective than air but not as reflective as the foil) and 2) a more significant effect from the relative surface areas for dissipating the fraction of energy that is absorbed by the foil. The smaller shell has much less surface area (980 cm2 vs 1280 cm2). The tiny increase between the small-foil-only and foil-inside-glass scenarios may be measurement error or may be a tiny effect from the change that the addition of exterior glass makes to convection/conduction.
While the air gap at the bottom may have had some slight effect on the results, it appears to me that any effect would have been common to all versions of the experiment and thus do not affect the conclusions. The increasing temperature would increase air pressure slightly but that’s a one-way and temporary movement that stops when the temperature reaches equilibrium. A gap that small would not allow for true circulation with the outside.
It also appears everyone is ignoring the transmittance factor of the Glass box to the IR, VL, etc.
wavelengths. This will effect the temperature with the foil inside or outside the glass. Standard glass (not special low-E) is about the same for both but with just small amounts of impurities it drops for IR – the main reason a greenhouse works. Thicker glass, like the glass box also has lower values for IR.
What’s Up With This? Did I use a forbidden word?
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?
Gary Hladik says:
May 28, 2013 at 3:39 pm
Curt says (May 28, 2013 at 3:05 pm): ‘Myrrh, you say, “Visible light cannot heat matter, it is not a thermal energy.”’
Good examples, Curt. Myrrh is also trivially refuted by a backyard version of the Herschel Experiment:
http://coolcosmos.ipac.caltech.edu/cosmic_classroom/classroom_activities/herschel_example.html
Myrrh is so often wrong, he’s probably the only WUWT commenter I know who makes the Pink Unicorn Brigade look reasonable by comparison. 🙂
======
Gary, when Herschel did his experiment it was breakthrough of real genius, but, it was crude. He physically moved the prism with his hand at the edge of a table..
We now call longwave infrared, thermal infrared, to differentiate it from near infrared which is not thermal, i.e. longwave infrared is the electromagnetic wave “of heat”, which is what thermal means. Near infrared is classed with Light, not Heat, with Reflective, not Thermal.
We have improved our knowledge since then, the only reason AGWScienceFiction leads you to that experiment is because it cannot show the fine detail, not only that visible light isn’t hot, but nor is shortwave infrared.
This used to be taught in general junior level education, but that has now changed to conform with the AGW agenda. Traditional teachers still teach it, it hasn’t been completely lost…
Here’s how NASA used to teach it:
“Infrared light lies between the visible and microwave portions of the electromagnetic spectrum. Infrared light has a range of wavelengths, just like visible light has wavelengths that range from red light to violet. “Near infrared” light is closest in wavelength to visible light and “far infrared” is closer to the microwave region of the electromagnetic spectrum. The longer, far infrared wavelengths are about the size of a pin head and the shorter, near infrared ones are the size of cells, or are microscopic.
“Far infrared waves are thermal. In other words, we experience this type of infrared radiation every day in the form of heat! The heat that we feel from sunlight, a fire, a radiator or a warm sidewalk is infrared. The temperature-sensitive nerve endings in our skin can detect the difference between inside body temperature and outside skin temperature
“Shorter, near infrared waves are not hot at all – in fact you cannot even feel them. These shorter wavelengths are the ones used by your TV’s remote control.”
http://science.hq.nasa.gov/kids/imagers/ems/infrared.html
This page was originally taken down completely, then about a week later it reappeared, but it’s stand alone, doesn’t connect to the normal NASA science pages on the electomagnetic spectrum.
So, note the difference in size, visible light is much smaller than long near infrared and so very much smaller than thermal infrared – in the Herschel experiment he was measuring overlap.
Visible light is not hot just as near infrared is not hot, we cannot feel them as heat. This is important to understand..
..if you want to understand the difference between heat and light.
AGW/CAGW both say that no heat energy reaches us from the Sun, heat energy is thermal infrared, longwave infrared, and they have replaced this with the fake fisics meme that visible light and shortwaves from the Sun heat matter of the Earth’s surface, land and water. This is a science fraud.
However, there are two different versions of why the direct radiant heat from the Sun “doesn’t reach us”.
The first is the original CAGW claim that there is an “invisible barrier like the glass of a greenhouse at TOA preventing longwave infrared from entering” – this “invisible barrier” is unknown to traditional science. As you can see from the traditional teaching from the old NASA page – the heat we feel direct from the Sun is the same as the heat we feel from a fire, thermal infrared, longwave infrared.
The second version of why we “don’t get longwave infrared direct from the Sun, is because the Sun produces insignificant longwave infrared and we get insignificant of insignificant”, this is attributed to AGWs and they back calculate from the planckian curve on the 300 mile wide visible light atmosphere around the Sun, so say that the Sun is only 6,000°C.
The Sun is millions of degrees centigrade.. We can feel its great invisible thermal energy on the move to us transfered by radiation in around eight minutes, we’re talking about a Star here..
So, this is a far greater science fraud than that of temperature manipulations, it affects the education of the next generation in basic science. If you care about that, then you should investigate what I’m saying, dismissing it without understanding that we have gained a lot more knowledge than the real scientist Herschel on whose shoulders we stand isn’t helpful in a science debate.
“I mean, the filament is not producing any more heat when the cover is on than when the cover is off.” Right, but it is loosing less heat due to the radiation fed back from the reflecting surrounding bodies (foil). So it receives back some of the heat emitted, which leads to a higher equilibrium.
usurbrain:
No, that’s a quite small effect, but possibly economically significant still.
The main effect is because it prevents free-air convective heat loss.
Myrrh, let’s go back to first principles. All electromagnetic radiation is transmitted via photons. All photons have a frequency or wavelength (two different ways of measuring the same property). Photons that have a wavelength between 380 and 740 nanometers are considered “visible light”.
When a photon impacts matter, it can be transmitted, absorbed or reflected. If absorbed, it’s energy is added to the energy of the matter as heat – that is, the photon causes the molecules to vibrate slightly faster. It is a property of the molecule whether a photon of a given wavelength is likely to be transmitted, reflected or absorbed. Some materials easily absorb photons of a particular wavelength but transmit or reflect all others. Most materials display a mixed behavior. This is called the material’s absorption spectrum.
Materials which reflect most visible light are said to be “white”. Materials which absorb most visible light are said to be “black” (even if they are highly reflective in the wavelengths above 740 or below 380).
Photons in the infrared part of the spectrum are not “heat” – they are merely photons that have a longer wavelength. We sometimes think of them as “thermal radiation” because they are easily absorbed by the skin, raising the temperature of the skin but they are not otherwise perceptible without instruments. Regardless, they are still merely photons, different from visible light photons only by wavelength. If a photon in the visible light spectrum is absorbed by your skin, it will increase the temperature of your skin just as much as that infrared photon. (Actually, slightly more but that’s not important now.)
Here’s an experiment for you to prove it. Low-pressure sodium lamps emit light in a remarkably narrow spectrum – tightly concentrated around 600 nm and dead in the middle of the visible light range. They are often used as street lamps. Find one, turn it on and hold it in your hand. Come back to us after you have the burn on your hand bandaged up.
I have seen various posts that consider either actual experiments or thought experiments and all seem to end in confusion. I am pleased that it is not just me!
My concerns include:
o) the use of glass in two places while still expecting IR to pass;
o) the lack of consideration for any of the surfaces to absorb radiation and become warmer as a result resulting in conduction/convection;
o) in the real world, the gases warmed by conduction at the surface will emit increased amounts of IR as a result. So GH gases, 1% of total, will absorb and re-radiate, but all gases, 100%, will radiate because of their temperature. Should this be ignored?
Max:
Depends on what you mean by “brighter.” Being as it is enclosed in aluminum, I’d say the opposite is true.
If you mean “did the amount of radiation produced by the filament increase” we can see that this is true (indirectly) from the data, because the current drops, indicating a higher internal temperature, and since the temperature is related to its brightness by the Stefan-Boltzman equation, that definitely means it got brighter (internal to the cage) as well as hotter.
Regarding this:
Gee last I looked back-radiation referred to radiation emitted in response to an external stimulus. So no *** Sherlock.
That’s the condition for equilibrium. If your net rate of heat loss were 35 W to the room, and you put a blanket on, after you came to a new equilibrium, the rate of heat loss through the blanket would still be 35 W, even though … this should be obvious even to Sherlock … you are now warmer under the blanket.
Jim — You asked about the current sensor. I don’t think it’s really relevant to the test, as I was really looking at only possible small changes, but you can see from the picture that it is an Extech 380942 “True RMS AC/DC Mini Clamp Meter”. It is advertised as providing real RMS measurements even when there is significant distortion. I’m sure it is better than the el-cheapo meters that just look at the peak and infer a sinusoid.
With a 120 Hz thermal excitation (i.e. 8.33msec period), would the filament really cool down that much at the zero crossings? What do you estimate the thermal time constant of the filament in gas as?
Stop saying this, a greenhouse works by stopping convection, the experiment above stopped convection, whatever material was used it would lead to a temperature increase without invoking any ridiculous nonsense about “the selective transmittance of IR vs visible is why greenhouses get hot”, thanks much though.
This is why I suggested using a sheet of glass between the mirror/black surface and the bulb rather than cutting off convection entirely.
Max, you are mistaken.
There is an effect from blocking IR, though it’s small under most nocturnal conditions.
Google “IR blocking greenhouse plastic”.
Carrick,
” In all cases, there is about a 35-watt (120V x 0.289A) electrical input to the system,” The writer may have contradicted himself, and in his summary corrected it.l
Someone needs to teach you some manners. I volunteer for this task.
Bryan says:
I don’t think anybody disagrees with that. However, that is a very different situation than the atmospheric greenhouse effect:
(1) You don’t have an external or internal source around (like the sun).
(2) You are not doing a comparison between two cases: The steady-state temperature in the case where there is a cold object around vs. the steady-state temperature in the case when there is a less cold (but still colder than the warm object) around.
mkelly says:
It squares almost exactly with what I described. The one difference is that it involves reflected rather than absorbed and subsequently emitted radiation. But, that doesn’t fundamentally change any of the accounting.
Carrick says: May 28, 2013 at 4:27 pm – The main effect is because it prevents free-air convective heat loss.
BINGO – that is why the box whether it be glass, tinfoil, black, etc. makes the bulb hotter.
Max says:
And, the Earth was heated by the power from the sun supplied to it, not back radiation. However, the amount of back-radiation still changes the temperature of both objects because the steady-state temperature of an object is determined by having to balance the energy emitted and the energy absorbed.
Michael Moon, you’ve repeated made the same erroneous statement when the body of the article contains this text, and you’ve been repeatedly told multiple times by multiple people:
So how is this not a reading comprehension issue on your part again?
Explain that, then we can move on to other topics of your choice.
ururbrain:
Yep. And if that were the primary effect, the temperature would be nearly the same “whether it be glass, tinfoil, black”. The fact that the reflectivity of the material affects the temperature (and is the primary predictor of the magnitude of the effect) tells us that blocking free-air convective heat loss is not the primary mechanism by which the interior of the enclosed volume became hotter.
Moon? Mann? I think somebody’s trying to do a funny.
No I won’t, actually.
I had recently been receiving emails from John O’Sullivan which, although appeared to disagree with CAGW, were seriously lambasting Monckton, Anthony and Roy Spencer for supposed “mistakes” in a “competition” of put up or shut up.
I have never seen Monckton et al ever make a serious scientific mistake, or at least nothing more than a grammatical typo, So I started to furrow my brow a bit.
I can’t remember IF I ever subscribed to those emails, but once I saw the language colour it put me off reading the rants entirely. I never paid much attention to them, except that something in the science content was lacking.
I’ve now witnessed the “science” at PSI embarrassingly eviscerated for all to see at WUWT. No thanks Mr. Postma, I shall NOT be keeping an eye at PSI. Regardless your supposed intentions, your science is not as solid as you dream.
Yeah, you know what people use that plastic for generally?
Keeping their greenhouse warm in the winter.
A greenhouse works just fine with glass that transmits IR, but it won’t work if you vent the top, if you think it does, by all means produce a greenhouse which doesn’t restrict convection yet still heats up and put it on the market, you’ll be rich before you know it!
Myrrh says (May 28, 2013 at 4:19 pm): “Gary, when Herschel did his experiment it was breakthrough of real genius, but, it was crude.”
Yet sophisticated enough to prove you wrong, as grade school students today can and do routinely. In my experience, most people who spout nonsense at least pick nonsense that can’t be so easily disproved.
They’re not quite as entertaining, though, I’ll give you that. 🙂