Slaying the 'Slayers' with Watts – part 2

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.

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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.)

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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.

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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.

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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.

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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).

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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.

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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.

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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.

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With the bulb covered by a foil shell alone and a surface temperature of 158.6C, the bulb drew slightly less, 288.7 milliamperes.

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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

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Joseph E Postma
May 28, 2013 8:01 am

This experiment makes the same mistakes than Anthony made in Part 1. It actually makes worse mistakes. Keep an eye at PSI, or possibly here if we are allowed to post a link/reply, for the analysis.
REPLY: Then point out the ‘mistakes’, or please do shut the hell up on this thread.
Principia might get a link here if they open up the web site to comments with the article, so people can go there and point out the ‘pink unicorns’ blocking the back radiation. Something tells me that O’Sullivan doesn’t have the integrity to allow that, though. – Anthony

Julian Flood
May 28, 2013 8:05 am

A neat illustration of the process is to use two hyperbolic reflectors, placed facing each other. At the focus of the first we place a thermoneter and at the second a cup of boiling water. The thermometer will show an increase in temperature as the heat from the water is focussed on it. Now remove the water and a new temperature, room temperature will show.
Now is the clever bit. Put a block of ice at the second focus. The temperature indicated by the thermometer will fall.
So, we’ve focussed the cold? No, all we have done is alter the state of equilibrium of the thermometer: essentially we have demonstated that the thermometer is recording its state of balance with the environment around it, and that the balance alters when we alter the environment.
I still like the ‘focussing the cold’ idea though, the stuff of fifties SF. Even Campbell might have fallen for that.
JF

Joseph E Postma
May 28, 2013 8:17 am

Of course, I’ll point out again that these table-top experiments, beside being wrongly interpreted etc, are beside the point of our empirical real-world data which has already proven that there is no GHE.
REPLY: So, let me get this straight.
1. Principia/Slayers cites a table top experiment in your essay.
2. Two people (soon to be 3) replicate the experiment in different ways, showing the Principia tabletop experiment is flawed, and the premise you state is ridiculous, as is clear by the data.
3. Slayers go on record in comments saying “we don’t need to replicate the experiment”.
4. Ignoring the fact that you’ve never done the experiment in the first place, you say others have misinterpreted it.
5. Slayers specifically in comments claim absurd easily disprovable things, such as passive element microbolomters emit a signal and temperature is measured by the “shift” in the returned signal
Remote read IR thermometers are also used to ‘explain’ this back-radiation warming effect. These instruments work be sending out an IR signal and measuring the shift in the returned signal.
…and when shown to be wrong, slayers refuse to admit you’ve confused the laser aiming with the sensing system.
6. You state the table top experiment you promoted has errors, but refuse to divulge what the supposed errors are.
7. You plan a rebuttal on a website that doesn’t allow open comments to the article, while asking me to carry your rebuttal.
Do you listen to yourselves when you say these things and take these positions? Do you realize that you’ve made a fool of yourself and killed any scrap of integrity you had with these claims?
I think the Ron House pink Unicorn analogy to describe your membership is right on:

We have here a really strange phenomenon: a truly fantastic claim (that our understanding of radiation is dramatically wrong) is being put forward, without any evidence, as the reason why our understanding of something derivative (namely the greenhouse effect) is mistaken. Surely it is obvious why that is so very perverse a process? It differs in no substantive way from, for example, claiming that powered flight is impossible because invisible pink unicorns block the takeoff of aircraft.

-Anthony

richard
May 28, 2013 8:23 am

not sure if this makes sense. jut thinking how one body can cause the other to heat up.
how does it work with a fridge, two bodies of air , the outside and the inside- both the same temp. the only way to get the heat from the air inside to the outside is to switch the heat pump on.

LamontT
May 28, 2013 8:31 am

Joseph you again claim that the experiment makes critical errors but you don’t explain what the errors are. Nor do you do your own experiment demonstrating the correct way and I might add proving your point all in one neat and easy experiment. Or is it that you can’t craft an experiment that would prove your point?

May 28, 2013 8:32 am

Offhand, I’d say the experiment makes an invalid claim : that the foil is a cooler body radiating anything. Clearly the foil is not the source of the radiation being reflected, nor could any foil body at its claimed temperature radiate any such powerful heat radiation. Explanation please.

Joseph E Postma
May 28, 2013 8:40 am

[ snip – if you want to point out errors, do so now. No other discussion from you is of any value at this point – Anthony]

JeffC
May 28, 2013 8:43 am

I am curious about the assumption that an object at a higher energy state (the light bulb) can absorb the energy being radiated by the objects at a lower energy state (the air and the glass) ? are you sure it can actually absorb the radiated energy ? I don’t see how you measure this ? I can think of several mechanisms that would slow the dissipation of the energy of the bulb that have nothing to do with the bulb actually absorbing energy … any slowing of the energy dissipation would raise the temperature of the bulb …
Plus you aren’t using 2 similar objects like 2 light bulbs side by side … you have the bulb and a 360 degree enclosure around that bulb … not exactly a hot and cold object assuming that by object you mean similar items …

intrepid_wanders
May 28, 2013 8:44 am

Awesome experiment Curt! While this experiment is quite satisfactory for the majority of us, a compliant that the ‘slayers’ will have is the convection was included. If you have the time, you can do a quick check by sampling 2-3 enclosed elements (glass, glass foil and foil), but just flip the enclosure to allow the the convection to escape and just have the radiative effects.
I know, nothing will satisfy the ‘slayers’, but the silly lightbulb-mirror figure is an open convection system…

May 28, 2013 8:52 am

The Temperature of the Filament is the only temperature that matters here. Did the Filament warm from back-radiation? Did heat actually “flow uphill?”
Did you measure the temperature of the filament? Could you, please, and end this ludicrous “debate?” You only mention the current twice. If the filament warmed its resistance would increase, and with a constant voltage the current would drop a little. A direct measurement of filament temperature would be better but obviously difficult to achieve.
“In all cases, there is about a 35-watt (120V x 0.289A) electrical input to the system” OK, I answered my own question, the filament did NOT warm, and the Second Law reigns supreme as always!!!

wsbriggs
May 28, 2013 8:53 am

Well done!
Just like the other group of D*ists, the music is on and the tap dancing starts. Unspecified mistakes, uncommentable blogs (I won’t say unspeakable), and no real science just arguments from “theory.”
In the meantime it appears to me that a number of recent papers UofH among others, show someone is paying attention to Bob T. and Willis E. and trying to at least explain (I caught myself using ex-pain) the observations. It’s still modeling all the way down, but at least other phenomena are being considered.
OT but pertinent to “consensus” science, there has be a major breakthrough in understanding part of how viruses are attacked by the body. A theory, roundly poo pooed by “experts,” has been found to describe a completely new mechanism of viral identification and destruction – can you say TRIM21, I knew you could.
http://online.wsj.com/article/SB10001424127887323582904578489743949572994.html
Hey, that science was settled too. Sheesh!

Nylo
May 28, 2013 8:54 am

Absolutely flawless and very well explained. JEP’s response is typical of a troll.

Bill Marsh
May 28, 2013 8:59 am

@jeffC,
1) There is no requirement that the objects be ‘similar’ for the experiment to work
2) The Aluminum foil is an ‘object’ just as the light bulb is an ‘object’.
3) Please elucidate on the several mechanisms you believe would slow the dissipation of the energy
4) I think you’re mixing energy absorption and temperature, they aren’t the same thing.

Darren Potter
May 28, 2013 9:00 am

Joseph E Postma says: “It actually makes worse mistakes. Keep an eye at PSI”
That is all you got “makes worse mistakes”, and you want people to hang around PSI waiting on PSI to fabricate some response?
Joseph E Postma says: “This experiment makes the same mistakes …”
Yes, both experiments made mistake of assuming each would be considered by people with enough forethought to know they lacked basic scientific knowledge to comment on experiments.

OldWeirdHarold
May 28, 2013 9:00 am

Thus proving what Max Planck knew over 100 years ago.

May 28, 2013 9:01 am

Thanks for taking the trouble Curt. Now sit back and be amazed/amused/frustrated by the misinterpretations which will shower down on your demonstration.

Nylo
May 28, 2013 9:04 am

Michael Moon, “The Temperature of the Filament is the only temperature that matters here”
Obviously, both the Earth and the bulb glass surface have a constant source of energy which allow their temperature to be constant despite radiative losses (the sun in one case, the filament in the other), and that source of energy is necesarily hotter. But dragonslayers’ claim is that GHGs cannot warm the EARTH, not the Sun. So the correct comparison is with the bulb’s GLASS, not the filament.

OldWeirdHarold
May 28, 2013 9:05 am

Michael Moon says:
May 28, 2013 at 8:52 am
The Temperature of the Filament is the only temperature that matters here. Did the Filament warm from back-radiation? Did heat actually “flow uphill?”
=====
That actually could be measured by a very precise measurement of current through the bulb. As the filament warms (and yes, it will warm), the resistance will go up, and the current will drop. By how much, I don’t know, but it should be calculable from the known resistivity properties of tungsten.
Go for it. Do it. And when the current drops (and it will, but possibly by a very small amount), tell me why that happened. Show all work.

Bill Marsh
May 28, 2013 9:07 am

@Michael Moon
““In all cases, there is about a 35-watt (120V x 0.289A) electrical input to the system” OK, I answered my own question, the filament did NOT warm, and the Second Law reigns supreme as always!!!”
You seem to have missed the significance of ‘about’ in that quote. It indicates that Anthony did not make exact measurements of the electrical current, therefore you can’t draw the conclusion you have drawn.

May 28, 2013 9:11 am

@ Curt Wilson, well-done. This is basic heat transfer ‘101’.
Millions upon millions of fired furnaces operate world-wide on the textbook principles.
If the textbook analysis were wrong, as Slayers insist, then the furnaces would not work. One suspects that we would have noticed, by now.

OldWeirdHarold
May 28, 2013 9:13 am

Matter of fact, based on this:
http://hypertextbook.com/facts/2004/DeannaStewart.shtml
Current measurement should be pretty robust. If someone can place a clamp-on ammeter on the wire to the bulb, and take a current measurement, I would expect to see a significant drop in current when the cover is placed over the bulb, and then when the cover is removed, the current should rise to the original value. Thus proving that the filament itself became hotter.

MattN
May 28, 2013 9:15 am

I vote Joseph gets no more posts unless his very next one lists the “mistakes”…

May 28, 2013 9:17 am

Nice!
I would expect current would drop a bit as the bulb got warmer.
And I wonder if the difference in temps between the large and small foils are due to the 4th power law and some of the energy is not reflecting back to the bulb.

Kevin Kilty
May 28, 2013 9:19 am

Michael Moon says:
May 28, 2013 at 8:52 am
The Temperature of the Filament is the only temperature that matters here. Did the Filament warm from back-radiation? Did heat actually “flow uphill?”
Did you measure the temperature of the filament? Could you, please, and end this ludicrous “debate?” You only mention the current twice. If the filament warmed its resistance would increase, and with a constant voltage the current would drop a little. A direct measurement of filament temperature would be better but obviously difficult to achieve.
“In all cases, there is about a 35-watt (120V x 0.289A) electrical input to the system” OK, I answered my own question, the filament did NOT warm, and the Second Law reigns supreme as always!!!

The filament did indeed become warmer as one can tell from the decline in current. As the author points out, the equation of state of tungsten is one of increasing resistance with increasing temperature. Use voltage and current to determine the resistance of the tungsten filament and obtain its absolute temperature in turn. But your assertion that the filament would become warmer from heat flowing “uphill” does not make sense. The filament temperature derives from an energy balance–input from the mains and output through radiation, convection, conduction, etc which are all functions of temperature. The radiation baffle impairs one branch of the energy output stream leading to a higher filament equilibrium temperature–there is no need to have heat flow uphill in this view.
However, radiation has an interesting characteristic in that one cannot tell if a photon came from a cool body or a hot one. So, in a microscopic view, photons emitted from a cold body can land on a hot one and transfer energy “uphill”. Cold bodies produce a less intense stream of photons than do hot bodies, so, statistically, net heat flows from hot to cold.

Nano Pope
May 28, 2013 9:20 am

I’m not sure you’re disproving the correct point. You’ve proved that radiation can be reflected and affect solids, but isn’t the point about energy transfer in gases under various pressures?

Peter Shaw
May 28, 2013 9:21 am

Your data may include something simpler and more remarkable:
Consider the filament as a tungsten resistance thermometer. Your “foil shell” change of 0.7 mA represents a 3-4C increase.
This is an impressive demonstration of “uphill” heat flow.
Measurement is a challenge, but this has the elegance of simplicity.

May 28, 2013 9:21 am

Michael Moon, you make the same fundamental mistake the Slayers do…equating energy input to temperature. Temperature is NOT determined by the rate of energy input alone.

Gary Hladik
May 28, 2013 9:26 am

Julian Flood says (May 28, 2013 at 8:05 am): [snip]
Yes, the Pictet Experiment. Joel Shore linked this article in a comment at Dr. Spencer’s site:
http://www2.ups.edu/faculty/jcevans/Pictet%27s%20experiment.pdf
Not surprisingly, the Slayers managed to misiterpret this one, too. 🙂

Ben Of Houston
May 28, 2013 9:27 am

To all those contrarians who are claiming this is wrong. Anyone with a cursory knowledge of heat transfer knows exactly how radiation works (it is by far the simplest mode of heat transfer, with conduction more complex and convection downright headache-inducing). If you think that it violates the second law then you have no idea how entropy works, do you? It’s simple. At every step, total entropy increases, from the absorbance of the infrared to the re-emittance, and then the re-absorbance when it hits the light.
Or you could try and explain how the re-emitted photons magically avoid striking the fillament or even more magically, strike it and are absorbed but do not warm it.
Seriously, grow up and take a freshman-level thermodynamics course before trying to proclaim yourselves the lord and master of physics.

Eustace Cranch
May 28, 2013 9:27 am

A better way to do this experiment is to heat a length of nichrome resistance wire. Its temp could be measured directly without the complications of the light bulb envelope. Wattage would be easily calculated from Ohm’s Law. Could a wire dissipating, say, 10W cause an adjacent wire dissipating 20W to rise in temperature? Hmmm….

Anton Eagle
May 28, 2013 9:29 am

If you’re going to get the science wrong… then at least be internally consistent. You acknowledge that the current and voltage don’t go up, and assert that the filament temperature doesn’t go up. So then what is the source of the increased temperature on the surface of the bulb?
Also, why would the temperature of the filament be the only one that “counts”? The surface of the bulb is a great analog the surface of the earth… a much better analog than the filament would be.
Your supposed rebuttal does nothing to address the observed temperature change on the surface of the bulb. It can’t be the filament, as its energy output is strictly a function of the current and voltage. It has to be back-radiation, and this experiment neatly demonstrates that it is so.
If you’re going to offer an alternative explanation, then you need to do more than assert that something else should have been measured… you need to account for the observed temperature increase. If you can’t, then you really have no valid argument, and should just keep quiet. Better to be quiet and be thought a fool, than to open your mouth and remove all doubt.
Lastly, when someone proves you wrong, just step up and admit you were wrong, learn from it, and move on; continuing to fight an argument that you already lost is just foolish.

Anton Eagle
May 28, 2013 9:33 am

Oops… my reply was directed to Michael Moon… somehow his quote didn’t end up in my post like I expected.

Konrad
May 28, 2013 9:35 am

Meh.
Yesterdays news.
If you couldn’t work out that PSI were not sceptics, how can you combat AGW promoters?

May 28, 2013 9:36 am

I find this entire discussion set absolutely hilarious but tedious.
As the captain said to Luke
    “what we have here, is a failure to communicate!”
Both sides seem to be wandering in the weeds, looking at details, focusing on mundane and arguing about nits and bits without conceptualizing.
(Neat prototypical science experiment, Curt, should be mandatory at elementary schools)
Key points that are muddled:
1. the “hotter body” has a source of energy:
    => if it doesn’t radiate/lose/use energy it will increase in temperature until doomsday.
2. the enclosing “cooler body” reduces the rate of energy flow away
Yes, the details are thrilling, but any object that has energy added, needs to
    use it or lose it!
The trivial analogy, that hopefully everyone likely understands, is your house:
-> the “cold” insulation” keeps the house warmer than what it would otherwise
        if your house is heated
    or “cooler than it would be otherwise” if your house is A/C’d
and, when the power dies, people get upset at the cold/heat.
The earth’s air, the glass in a greenhouse, windows in your house, etc
all let in energy and slow the rate of its departure,
=> like the insulation in the walls which slow the departure of the heat generated by your furnace. (the windows, not so much)
It’s the energy flow stu…
If you want to argue about the nits and bits of how fiber glass insulation
(or the earth’s air) slow the departure of thermal energy or how glass and air allow the influx of energy in some bands (when present), go for it.
It is usually entertaining, but mostly tedious.

Feeling is believing
May 28, 2013 9:36 am

My house is not very well insulated. That is true of many houses in New Zealand the same age as mine. Sadly, limited access to the ceiling cavity has made this problem difficult to completely rectify. Fortunately it seldom gets too far below freezing where I live. Often however when I get up in the morning at this time of year (winter) the temperature in my bathroom is unpleasantly cold.
I like to start my day with a nice hot shower. Before I start my shower however I always detach the shower head and hose down all four walls (two of them glass) of the shower box with hot water to take the chill off them. That way I can have a comfortable shower and not one where the parts of me not directly under the hot water feel cold. Taking the chill off the walls and windows of the shower box stops me losing so much heat to them by radiation and makes an amazing and quite perceptible difference to the comfort of my shower.
I’d be fascinated to see how a sky dragon slayer might explain this phenomenon.

Eustace Cranch
May 28, 2013 9:36 am

Another way to think about this: consider a double star system in space. One star will almost certainly be cooler that its mate. Could it possibly heat the hotter star without losing its own heat? Is the total radiation of the pair greater than the sum of each separately? I think not.

dp
May 28, 2013 9:39 am

These lamps are nearly constant current devices and this characteristic can be used to directly measure the thermal response of the lamp system. No thermisters needed. Use a precision voltage source or a precision current source, put the lamp in a vacuum, warm it by placing any other radiating object near by. Watch the voltage or current change, depending on which of the two you choose to regulate.
But none of this is necessary if one can explain what becomes of the radiated energy from the cooler object that strikes the warmer object. In fact that energy does to the warmer object what any energy does that strikes another object. It increases that object’s energy level. Oddly enough some of that energy is radiated back to the cooler object, warming it faster which creates greater emission back to the warmer object. Back and forth like this until equilibrium is achieved.
Any boy scout can tell you two burning logs adjacent will reach equilibrium and very high heat at the place of least distance even of only one log is initially lit. Neither log alone will reach the temperature of the pair, nor heat as quickly as the pair.
Objects that radiate have no awareness of what is around them. They are equivalent isotropic radiators if geometry permits, and anything that is illuminated by emissions from such an object is energized by that illumination. The sun is very bright – bright as it is we can still strike it with a powerful laser. That laser energy becomes part of the total energy of the sun. That we have no prayer of measuring it does not mean that laser energy has evaporated on the way to the sun.
The slayers are too ignorant to bother with further and it drags down the quality of the science to humor them.

May 28, 2013 9:40 am

[snip – we aren’t getting off topic on this – we are discussing this experiment, not some pie in the sky theoretical one – Anthony]

May 28, 2013 9:47 am

Macro behavior in solids is not transferable to micro behavior in individual, three atom gas molecules. The Aluminum foil is at least several thousand molecules thick, the glass even more. Radiative and convective flux are both restricted by both materials based on mass, specific heat and in this case, the sealed cubic configuration. In a free gas environment there are limited restrictions to convective and radiative flows, and a three atom gas molecule, at virtual rest compared to a photons speed, is not capable of stopping or redirecting this high velocity OLR force for more than milliseconds. Atmospheric CO2 is a feather in the path of a howitzer round.

OldWeirdHarold
May 28, 2013 9:51 am

Years ago, in a statistical mechanics lesson in P-chem, the prof, after wading through an insufferable derivation, drew an interesting conclusion. In a hurricane, over 40% of the molecules are moving against the direction of the wind. If they were all moving the the same direction, they’d be moving at the speed of sound.
The conceptual error that the Skydragons are making is failing to distinguish between individual dynamics and population dynamics. Just as a large percentage of the molecules in a hurricane move against the wind, a large percentage of the photons can and do move counter to the net heat flow, which implies moving counter to the thermal gradient.
The Second Law is an emergent phenomenon that applies to populations. It doesn’t apply to individual particles. Skydragons don’t seem to get that concept.

May 28, 2013 9:58 am

I’m not a slayer, but some of the effect you’re getting is the glass-greenhouse effect..where heat from the bulb is conducted to the surrounding gases which cannot convectively cool. The gases surrounding the enclosed bulb are thus warmer and direct heat loss via conduction is reduced.
It is very hard to measure the direct effect of LWR. With the mirror you get reflected SW which at the very least promotes a faster equilibrium.

PeterH
May 28, 2013 10:13 am

Hypothesis,
Clearly described experimental procedure,
Clearly presented data,
Analysis of how the data relates to experimental conditions and theory,
Debate and input how the experiment might be done differently or better.
This is SCIENCE!

May 28, 2013 10:16 am

Joseph E Postma says May 28, 2013 at 8:01 am

This is the same guy who conflates ‘night vision’ (visible light amplified by light intensifier tubes and the like) with thermal LWIR ‘FLIR” imagery equipment with near wavelength IR viewers requiring an external source of that near IR ‘light’ for the viewer?
He conflates an awful lot I’d say … and more, probably, than just visual/thermal/near-IR imagery subjects …
.

wikeroy
May 28, 2013 10:16 am

Susan Corwin says:
May 28, 2013 at 9:36 am
“I find this entire discussion set absolutely hilarious but tedious.”
That was what came to my mind too !
For many, many years I have noticed that the insulation in the walls of my house has kept it warmer than without. But the word “backradiation” never came up.
And I did the basic course in thermodynamics 30 years ago. I went to the addict and looked through the books. Couldnt find the word. Probably some new definition.

May 28, 2013 10:17 am

Anthony
My proposed experiment was addressing this experiment in particular the final comment
“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.”
This implies that the presence of a colder object will always increase the temperature of a warmer one.
It contradicts common sense.
Try putting a large block of ice in your living room.
This is more acceptable and perhaps is what Curt was really meaning.
‘It is not compatible with the idea that cooler bodies cannot transfer any power by radiative means to warmer bodies and SOMETIMES cause an increase in temperature of the warmer body.’

Slartibartfast
May 28, 2013 10:22 am

These guys are in effect claiming that some radiation is effective while other radiation is not. In other words, radiation only works if it’s going from hot to cold.
It’s bunk. It’s tantamount to saying that some photons are more privileged than others. Blackbodies don’t care where the photos came from; they just care that the photons came at all.

May 28, 2013 10:23 am

Michael Moon says May 28, 2013 at 8:52 am
The Temperature of the Filament is the only temperature that matters here. Did the Filament warm from back-radiation? Did heat actually “flow uphill?”
Did you measure the temperature of the filament? Could you, please, and end this ludicrous “debate?”

OK, I answered my own question, the filament did NOT warm, and the Second Law reigns supreme as always!!!

Premature jubilation; did you perchance read this tidbit:

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.

Pls join-up with Greg for some remedial education …
.

Slartibartfast
May 28, 2013 10:23 am

s/photos/photons

Editor
May 28, 2013 10:31 am

Nano Pope says:
May 28, 2013 at 9:20 am

I’m not sure you’re disproving the correct point. You’ve proved that radiation can be reflected and affect solids, but isn’t the point about energy transfer in gases under various pressures?

That’s what the point should be, but what it really is about is realizing an ill-designed experiment created by Dr. Siddons.
We don’t have a failure to communicate, we have a failure to comprehend. We have better things to do, but the Slayers have been so vocal and pig-headed that we have to spend an excessive amount of time to explain to everyone else why their claims are wrong.
“Follow the money” is often heard here. In this case, “Follow the photons” applies.

Carrick
May 28, 2013 10:31 am

OldWeirdHarold:

That actually could be measured by a very precise measurement of current through the bulb. As the filament warms (and yes, it will warm), the resistance will go up, and the current will drop. By how much, I don’t know, but it should be calculable from the known resistivity properties of tungsten.

That’s the thought I had too. The electrical properties of the filament allows it to be directly used as a temperature sensor.
It’d be interesting if Curt Wilson (or somebody else) were to calibrate the tungsten filament temperature as a function of power drawn (use a clear glass bulb so you can image the filament, vary the power applied using a variac), then repeat the experiment, tracking the resistance of the filament under the different experimental conditions.
What these experiments prove is how resilient the PSI group are to actual facts in addition to how very limited their own understanding of how science works is.
I nominate this for their official icon.

Slartibartfast
May 28, 2013 10:35 am

Another way to think about this: consider a double star system in space. One star will almost certainly be cooler that its mate. Could it possibly heat the hotter star without losing its own heat? Is the total radiation of the pair greater than the sum of each separately? I think not.

You really need to go look into what makes stars do what they do, and then consider that each star is in fact in the process of irradiating the other star. I can’t imagine what you could possibly mean by “losing its own heat”, because both stars are constantly in the process of losing their heat. They can’t NOT lose their own heat.
I don’t know that statistics of close binaries have been kept, but I would be shocked if they didn’t live less long than single stars of the same type, because they are in fact heating each other.
Look, only a hypothetical perfect reflector can fail to gain energy when exposed to a source of radiation. Are you saying one or the other of those stars is a perfect reflector?

OldWeirdHarold
May 28, 2013 10:36 am

Carrick says:
May 28, 2013 at 10:31 am
——–
If you think about it, what’s likely to happen if the filament temperature rises and current drops is the temperature of the filament will self-regulate within bounds. The power will drop slightly, which is probably the only thing that keeps the filament from overheating.

May 28, 2013 10:42 am

Thermodynamics requires that a direct transfer of heat by conduction go from the higher temperature to the lower.
But in the case of two bodies, if both are radiating, of course they are going to transfer energy. Imagine two bodies in space, separated by one foot that are radiating infrared. If you have Body A putting out 1kW radiatively and Body B putting out 500W radiatively, Body B is going to transfer X amount of joules to Body A, which would raise the temperature. (I could calculate how much, but it’s been years since I took Thermo…) Body A is also going to transfer Y amount of joules to Body B.

Michael Tremblay
May 28, 2013 10:44 am

Your conclusion is fundamentally flawed. The statement: “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.” is false. What you have is solid experimental evidence that reducing the amount of energy (heat) leaving a system while maintaining the amount of energy (heat) entering the system will result in a temperature rise in the system. The temperature rise is due to the increasing amount of heat in the system and this heat is coming from the electric current being supplied to the bulb, not from the back radiation.

Slartibartfast
May 28, 2013 10:45 am

If you think about it, what’s likely to happen if the filament temperature rises and current drops is the temperature of the filament will self-regulate within bounds. The power will drop slightly, which is probably the only thing that keeps the filament from overheating.

1) temperature rises
2) resistance increases
3) P=V^2/R decreases
4) temperature rate of increase slows
5) Goto 1)

William Sears
May 28, 2013 10:45 am

Seeing that the position of the PSI is that the proximity of a cooler object cannot cause a hotter one to get even hotter, which is something that I did not realize before due to its absurdity, I wonder why they have restricted it to thermal radiation. Surely, to be consistent, they must also apply this principle to conduction and convection as well, since they are all methods of thermal transfer. This being the case, how does invoking convective restriction as the explanation of the present experimental result help their case. When we accept this logical extension of the PSI theory most reasonable people should see the flaw since we are all very familiar with the benefits of blankets on a cold winters night. But since thermal radiation is a more abstract concept there may be either a tendency to ignore it or to descend into magical thinking aided by an endless stream of non sequiturs. There may also be an equating of conservation of energy with conservation of temperature. The latter does not. Temperature is a statistical parameter (one definition).

Slartibartfast
May 28, 2013 10:48 am

The temperature rise is due to the increasing amount of heat in the system and this heat is coming from the electric current being supplied to the bulb, not from the back radiation.

So, your theory is that this heat coming from the electric current being supplied to the bulb knows when there’s a reflector in place?
If not, what?

May 28, 2013 10:50 am

Author (Curt), needed now is a list of test equipment used (manf, and model number.) Some of us would like to look up the specs on such items as the clip-on ammeter and the method the AC voltage is calculated internally by the DVM (e.g. an ‘average’ value corrected to read RMS, or a true RMS reading/calculation technique). Thank you for this in advance, and also your work on this subject here today as well.
I assume the assembled multitude here is also cognizant of the dynamic change in filament R (resistance) over a complete cycle of applied AC, where the peak is on the order of 1.414 times the nominal 120 V RMS value, resulting in a figure around 170 Volts. This results in a peak current nearly coincident with the peak of the applied AV voltage, and falls off thereafter in a nonlinear manner (no longer observes the simple E = I/R relationship, but one with time as a parameter.)
This nonlinear response is MOST noticeable if one ‘powers’ a bulb during only one half (1/2) of the applied AV mains sinusoid (using a simple series diode), one can see a really lop-sided current ‘draw’ on the scope trace which is set to monitor the *dynamic’ current drawn by the bulb.
THAT is why I mentioned the Kill-A-Watt appliance measuring ‘instrument’ which calculates RMS power (Root of the Mean of the sum of Squares of a time series of measured values) consumed, as opposed to making use of ‘apparent power’ (simple I times E) calculation usually using the simple ‘peak’ reading obtained from a digital Volt or current meter (DMM or DVM).
(Better yet, utilize nowadays an NI DAQ (analog data acquistion) card and digitize the I and V values out to 12 or 14 bits over a complete waveform (at say, 500 us intervals) and analyze the ‘slope change’ of the I draw and calculate the temp change in the Tungsten filaments that way.)
Experiments in this area were undertaken in the early eighties looking at different means to extend Incandescent bulb life. A Tektronix 564B Mod121N O-scope was used in conjunction current ‘shunts’ to facilitate observations of bulb current-draw behavior over time under different applied AC waveforms.
.

Slartibartfast
May 28, 2013 10:50 am

I mean, the filament is not producing any more heat when the cover is on than when the cover is off.
Sincerely baffled, here.

Cho_cacao
May 28, 2013 10:52 am

TO Michael Tremblay
“What you have is solid experimental evidence that reducing the amount of energy (heat) leaving a system while maintaining the amount of energy (heat) entering the system will result in a temperature rise in the system.”
OMG you finally understood how it works! The sun is providing a (so to say) constant energy input, the GHGs reduce the amount of energy being released to space… And voilà!!!

son of mulder
May 28, 2013 11:02 am

Here’s another conceptually simple experiment that slays the slayers. Take 2 identical steel spheres, put two diametrically opposite embedded thermometers on each of them. Heat one of the spheres to 2,000 deg C and leave the other at room temperature. Then hang each of them closely together, but not touching in a vacuum, with one thermometer on each of them as close as possible to the nearest thermometer on the other, and the other 2 thermometers as far apart as possible.
Then measure how the temperature changes at each of the 4 thermometers as the spheres are allowed to approach thermal equilibrium. For the first part of the experiment the thermometer on the hot sphere closest to the cool sphere will cool more slowly than the thermometer on the hot sphere furthest from the cold sphere. The thermometer on the cold sphere closest to the warm sphere will warm more quickly than the thermometer on the cold sphere furthest from the hot sphere. As time passes there will come a points in time where the each of the thermometers on the cold sphere will start to cool after warming. Eventually both spheres will be in thermal equilibrium with each other and all thermometers will read the same.
The reason that the closest thermometer on the hot sphere cools more slowly is because some back radiation comes from the warming cold sphere. This slays the slayers.
There is no convection in the experiment, there is no conduction between spheres just within each sphere. There is no ongoing energy supply so the simplest form of the laws of radiative thermodynamics will apply. The experiment is symmetric. The Boltzmann profile of each sphere is identical.
If someone were to posit that conduction in the hot sphere were to keep both its thermometers cooling at the same rate then as the 2 thermometers would be at the same temperature there would be no net conduction but extra photons are arriving on the side closest to the cooler sphere so it must cool more slowly.

May 28, 2013 11:02 am

OldWeirdHarold:
That actually could be measured by a very precise measurement of current through the bulb. As the filament warms (and yes, it will warm), the resistance will go up, and the current will drop. By how much, I don’t know, but it should be calculable from the known resistivity properties of tungsten.
Carrick says May 28, 2013 at 10:31 am:
That’s the thought I had too. The electrical properties of the filament allows it to be directly used as a temperature sensor.

Problem being, this varies dynamically over the applied AC sinusoid even; above is a post where I propose to measure the bulb filaments characteristic over an applied cycle in 500 us steps (using DAQ or data acquisition cards), then the instantaneously measured V / I performance of the filament could be used to calculate it’s instantaneous R, and infer the change in temperature.
In lieu of such difficult measurement setup and/or apparatus, simply power the bulb from a DC power supply and do away with the dynamic thermal performance of the tungsten filament over the course of an applied AC sinewave … the values then read by simple instrumentation (simple DC volt and amp meters) will need no correction as the filament measurement conditions will be ‘static’ as opposed to ‘dynamic’ at a 60 Hz rate.
.

May 28, 2013 11:06 am

Anton Eagle says:
May 28, 2013 at 9:29 am
If you’re going to get the science wrong… then at least be internally consistent.
Anton and Gentlemen,
Our writer says “in every case” and states the current to three decimal places. Clearly he is telling us the filament temperature does not change. The bulb temperature is affected by a complex flux, of course it will change. Those who state that the bulb temperature is a better analog to the Earth’s surface have not even entered the debate. The only question of Physics here is, can a heated filament increase its own temperature by being exposed to its reflected IR? If it could we would certainly be able to lower our electric bills. Just think, an Einsteinian thought experiment: Say the filament temperature DID go up. Would it not consequently, in the most Stefan-Boltzmann’s Law way, radiate more heat and light, which would be reflected more strongly, increasing its temperature still more in a runaway cycle? Will the Perpetual Motion patent holder please tell us how he did it?
Photons of course have a frequency, and a “wavelength,” please do not forget this dear readers. Otherwise CO2 could absorb all photons, instead of just those at the 15-micron and its other resonant wavelengths.
Somebody certainly did get the science wrong…

astonerii
May 28, 2013 11:07 am

So, you are comparing a closed system to an open system and making conclusions about proof.
Interesting that everything was tested except the one item you are trying to prove has this certain quality, a greenhouse gas.
Lets see if we get this correct… Glass retains the heat and temperature goes up, aluminum foil reflects directly the exact same energy power as well as retaining the heat and the temperature goes up faster, putting glass between the reflective foil causes some of that energy to be lost as it goes through the absorbent glass twice.
This is science? It seems more like a vengeance attack against your enemy that is more or less making you look more the fool than your target.

Carrick
May 28, 2013 11:07 am

Michael Tremblay:

The temperature rise is due to the increasing amount of heat in the system and this heat is coming from the electric current being supplied to the bulb, not from the back radiation

It helps if you start by using the correct language to describe the problem:
Technically the electric current is providing a fixed rate of power that is providing a (nearly) fixed rate of thermal energy per unit time. For the temperature to increase, the net amount of thermal energy stored in the system would have to increase. “Heat” is a thermodynamic term relating to a process variable, namely it refers to the amount of heat energy exchanged between two bodies during one cycle of a thermodynamic process.
The fact that the temperature of the system increases when you manipulate the system by e.g. putting a reflective aluminum shroud around it tells you that something has changed to allow more thermal energy to be stored in an equilibrium state. That is something is impeding the outwards flow of thermal energy per unit time from the heat energy source.
As we see from the measurements, the amount of thermal energy stored actually increased while the amount of available electrical power decreased (slightly). The explanation cannot be simply that there is a power source internal to the system, because that fails to explain why the heat energy is being affected by the different experimental configurations.
It must be that there is a physical mechanism associated with the aluminum foil that is causing the temperature to increase more than with e.g. the black-anodized metal container.
Since this science has already been well tested experimentally, the most plausible explanation is that radiative physics is involved,
The only thing missing from this experiment is the control needed to actually calculate the amount of warming observed under the different experimental conditions. A quantitative comparison to a physics-based model, would make this publishable in e.g. American Journal of Physics, since they welcome physics demonstrations that are suitable for classroom environments.

MattN
May 28, 2013 11:07 am

Roy Spencer said: “Michael Moon, you make the same fundamental mistake the Slayers do…equating energy input to temperature. Temperature is NOT determined by the rate of energy input alone.”
I was going to point this out in the last thready but didn’t. It appears these guys do not know the difference between “heat” and “temperature”. They are absolutely NOT the same thing…

Vince Causey
May 28, 2013 11:07 am

Michael Tremblay says:
May 28, 2013 at 10:44 am
“Your conclusion is fundamentally flawed. The statement: “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.” is false. What you have is solid experimental evidence that reducing the amount of energy (heat) leaving a system while maintaining the amount of energy (heat) entering the system will result in a temperature rise in the system.”
How can the amount of energy leaving the system be less than the amount entering, if the system is at equilibrium?
The temperature of the bulb increases until equilibrium is reached, at which point energy leaving the system (at the surface of the foil box) equals energy entering.via the filament, which is assumed to have remained constant.
In fact, the energy leaving the system never changed, and neither did the energy entering. What changed was the size of the system, moving from the filament and bulb, to filament, bulb and foil box.

May 28, 2013 11:12 am

Slartibartfast says:
May 28, 2013 at 10:22 am
Heh, your comment made me wonder whether Slayers eat microwave dinners, or whether their food is too exclusive for photons at the wavelength of water.

Eustace Cranch
May 28, 2013 11:12 am

Another thought experiment. Two stars, 4 light-years apart. Measure the radiated energy of each, and sum them together. Now move them closer together by half the distance. The energy they impinge on each other increases by 4 (inverse square law). Does the sum of the radiated energies for the pair *as a system* increase? What about 1 light-year? Half a light year?

Carrick
May 28, 2013 11:13 am

astonerii:

So, you are comparing a closed system to an open system and making conclusions about proof.

Heat energy is being lost in all cases, so it’s alway an open system.

Duster
May 28, 2013 11:13 am

Might consider redoing this using a clear bulb. That would eliminate the effects of the interior coating of the bulb. Otherwise, very interesting indeed.

Carrick
May 28, 2013 11:15 am

Vince:

In fact, the energy leaving the system never changed, and neither did the energy entering. What changed was the size of the system, moving from the filament and bulb, to filament, bulb and foil box.

I think you mean in equilibrium, right? Clearly when the interior is heating up the amount of heat energy per unit time leaving would be less than at equilibrium.

John West
May 28, 2013 11:15 am

This whole filament/bulb argument is misdirection. The bulb is a grey body heated by a source; whether that source is the filament or the sun makes no difference. Bottom line is that the bulb temperature increases in the presence of a colder outer layer, thus the colder outer layer warmed (or slowed the cooling of if you prefer) the inner layer. Sky-Dragon-Slayer premise busted!

Carrick
May 28, 2013 11:26 am

Micheal Moon:

Our writer says “in every case” and states the current to three decimal places. Clearly he is telling us the filament temperature does not change

Clearly you don’t have any physics training.
Once again you are confusing electrical power (which actually isn’t a constant, it decreases slightly as the filament heats up) with temperature, which is a measure of thermal energy.

P Gosselin
May 28, 2013 11:28 am

“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.”
“Radiation from a cooler object”? That statement ought to dismiss anything else written in the essay. A cooler object doesn’t radiate energy to a warmer one.

chris y
May 28, 2013 11:29 am

I continue to be amazed that this remains a controversial issue. GE R&D center developed an incandescent bulb back in the late 1980’s (and GE lighting commercialized it in the 1990’s) that placed a spherical clear glass shell around the filament. The shell was coated with a multilayer (anywhere from 15 – 30 separate layers) optical filter that reflects mid infrared back onto the filament, while allowing visible light to pass through. The result is that a lower filament current can achieve the same filament temperature, thanks to the mid infrared energy being reflected back to the filament. This results in a 15% – 20% increase in lumens/Watt.
Designing a filament that has low emissivity in the infrared and high emissivity in the visible could achieve similar improvements in lumens/Watt.

May 28, 2013 11:31 am

And part of the Gulf of Mexico flows up to Missouri, right?

Michael Tremblay
May 28, 2013 11:36 am

Vince Causey says:
May 28, 2013 at 11:07 am
“How can the amount of energy leaving the system be less than the amount entering, if the system is at equilibrium?
The temperature of the bulb increases until equilibrium is reached, at which point energy leaving the system (at the surface of the foil box) equals energy entering.via the filament, which is assumed to have remained constant.
In fact, the energy leaving the system never changed, and neither did the energy entering. What changed was the size of the system, moving from the filament and bulb, to filament, bulb and foil box.”
I am talking about the total amount of energy (Joules) in the system, you are talking about the rate (Watts – Joules/second) the energy enters or leaves the system. The system is at equilibrium when the rate of energy entering the system is equal to the rate that the energy is leaving the system. The amount of energy in the system determines its temperature, as the temperature rises the rate of energy leaving the system also rises until it matches the rate of energy entering the system.

Carrick
May 28, 2013 11:40 am

P Gosselin:

“Radiation from a cooler object”? That statement ought to dismiss anything else written in the essay. A cooler object doesn’t radiate energy to a warmer one.

Given that macroscopic objects are composed of greater than 10^23 molecules and atoms, and further these molecules and atoms obey a distribution in kinetic energy, there will always be an overlap in kinetic energy distributions between molecules in the colder object and the warmer object.
That is on average some molecules on the colder object will be warmer than some molecules on the warmer object (actually it’s “very many”), so even if your pet theory were correct, what you are saying would still be wrong.
Beyond that, how does a photon once it is emitted by the colder object “know” that it came from a cold object so that it “knows” not to be absorbed by the warmer object? These are some smart photons in your universe, don’t you think?

Brad
May 28, 2013 11:41 am

Since Anthony snipped an earlier comment from the Slayer, he will visit other blogs that allow him a forum to espouse his belief and attempt to beat into submission those who disagree with him and/or do not believe has he believes. He will, along with the rest of his cadre. It is quite remarkable considering they do not allow an open forum on their site.

Carrick
May 28, 2013 11:42 am

pgosselin:

And part of the Gulf of Mexico flows up to Missouri, right?

Another physics fail from the failed physics group:
Radiative transfer of photons and macroscopic flow aren’t appropriate physical analogies.

wikeroy
May 28, 2013 11:43 am

John West says:
May 28, 2013 at 11:15 am
“Bottom line is that the bulb temperature increases in the presence of a colder outer layer, thus the colder outer layer warmed (or slowed the cooling of if you prefer) the inner layer. Sky-Dragon-Slayer premise busted!”
Well, I am not so sure.
If I have understood the discussion correctly, the Watt’s camp says ” thus the colder outer layer warmed the inner layer” , and the Slayer camp says “slowed the cooling” ( Or Insulated it, thereby making the temperature go up).
So, when you say “the colder outer layer warmed (or slowed the cooling of if you prefer) “, you basically embraced both camps……or??

FerdiEgb
May 28, 2013 11:44 am

Curt Wilson,
Simple and well thought and results far beyond reasonable doubt.
That is what real science is all about!
Ferdinand

Joel Shore
May 28, 2013 11:45 am

Bryan says:

This implies that the presence of a colder object will always increase the temperature of a warmer one.
It contradicts common sense.
Try putting a large block of ice in your living room.

No…What is says is that a colder object will increase the temperature of a warmer object where that increase is measured relative to the situation where we have a MUCH colder object there instead. In other words, try putting a vat of liquid nitrogen in your living room and then replacing it by a block of ice instead. This replacement will cause an increase in the temperature of objects in the room (with the details depending on how close the object is to the vat or block, etc.) just as replacing the 2.7 K radiative temperature of outer space with the radiative temperature of the atmosphere causes the Earth’s surface to be warmer.
Julian Flood says:

A neat illustration of the process is to use two hyperbolic reflectors, placed facing each other…

What Julian is describing is a very nice experiment first carried out around 1800 by Pictet and discussed here in a 1985 American Journal of Physics article: http://www2.ups.edu/faculty/jcevans/Pictet%27s%20experiment.pdf It is particularly nice in that it really manages to get rid of any significant conductive or convective effects. See p. 741 of the article I linked to for Pictet’s own description of the experiment and result. Also note that there is a discussion of our modern understanding of what the experiment demonstrates on pp. 749-750 and then also a discussion of how to reproduce the experiment.

OldWeirdHarold
May 28, 2013 11:46 am

_Jim says:
May 28, 2013 at 11:02 am
Problem being, this varies dynamically over the applied AC sinusoid even; above is a post where I propose to measure the bulb filaments characteristic over an applied cycle in 500 us steps (using DAQ or data acquisition cards), then the instantaneously measured V / I performance of the filament could be used to calculate it’s instantaneous R, and infer the change in temperature.
=====
Please. None of this matters. You could use DC, and you’d get exactly the same result.

Leonard Weinstein
May 28, 2013 11:46 am

A light bulb that has vacuum in it (not a gas filled one) is an ideal object to demonstrate the principle of back radiation absorption. However a different setup than the above would be better. While glass passes short wave radiation well, it has high absorption and emissivity at much longer wavelengths. The best way to make the glass absorb all wavelengths is to paint the outside with black paint. Then coat the outside with a layer of insulation (not too thick or the inside will get too hot). Now use the fixed voltage power in, and an accurate ammeter to track current. As the bulb and insulation heat up inside, the only back radiation will be long wave thermal radiation, not reflected short wave light. Nevertheless, the resistance of the light bulb will increase as the filament heats up from back radiation. This will cause a drop in current. There is no other possible cause for the drop in current other than low temperature glass heating the much hotter filament. Convection and short wave reflection were removed from the test.

Luther Wu
May 28, 2013 11:48 am

chris y says:
May 28, 2013 at 11:29 am
“…GE R&D center developed an incandescent bulb back in the late 1980′s (and GE lighting commercialized it in the 1990′s)…”
_____________________________
Do you suppose that they paid for their investment in research and tooling?
The last GE plant making incandescent bulbs closed in 2010.
http://www.washingtonpost.com/wp-dyn/content/article/2010/09/07/AR2010090706933.html
My whole question posed about GE profiting from their bulbs is but a gnat’s poke at the world economy and politics. GE doesn’t pay taxes, while it funnels millions into the coffers of “Green” causes and politicians, who then divert funds from individuals making rational business decisions into the pockets of GE via tax subsidized wind generators, etc.

Leonard Weinstein
May 28, 2013 11:48 am

In my comment, the glass I refer to is the bulb itself, not an external container.

Curt
May 28, 2013 11:49 am

Hello Everyone,
Thanks for your interest and comments. Some clarifications in response to questions and comments:
Arthur4563, you say: “Offhand, I’d say the experiment makes an invalid claim : that the foil is a cooler body radiating anything. Clearly the foil is not the source of the radiation being reflected, nor could any foil body at its claimed temperature radiate any such powerful heat radiation. Explanation please.”
Sometimes when you are evaluating, um, “unusual” claims, you have to go down their road a bit. (When I watch Mythbusters with my kids, I occasionally find myself yelling at the screen, “Did you really need to test that?!”) In this case, I was testing the claim I have seen made several times that reflecting radiation back to a hot emitting body such as a light bulb could not increase the temperature of the bulb. I think I disproved that claim here. I was careful to distinguish this case from the other case with the black shell that absorbed the bulb shortwave radiation and re-emitted longwave radiation. This also caused the temperature of the bulb to increase.
philr1992, you say: “I’m not a slayer, but some of the effect you’re getting is the glass-greenhouse effect..where heat from the bulb is conducted to the surrounding gases which cannot convectively cool. The gases surrounding the enclosed bulb are thus warmer and direct heat loss via conduction is reduced.”
The whole point of the experimental design was to control for this “glass-greenhouse effect”. That’s why the temperatures of the metal shell cases were compared to that of the glass shell case and not the bulb in open air. I cannot think of any way in which the metal shells would inhibit conductive/convective losses more than the glass shell does, leaving changed radiative transfer as the mechanism for the difference.
wikeroy, you say: “I did the basic course in thermodynamics 30 years ago. I went to the [attic] and looked through the books. Couldnt find the word [backradiation]. Probably some new definition.”
In engineering heat transfer texts, it’s usually referred to with a term like “radiative exchange” between bodies. Right near the beginning of the chapter on radiative heat transfer in any of these texts, you will see an equation for the radiative heat transfer between two bodies with some constants multiplying the term (Th^4 – Tc^4), where Th is the absolute temperature of the hotter body and Tc^4 is the temperature of the colder body. The Tc^4 term is the “back radiation”, whatever you choose to call it.
To all who missed the point about the filament resistance, please read the whole post. I did measure the current through the filament, and it did decrease slightly when the bulb surface temperature was hotter. This demonstrates two things. First, the higher temperature cannot be explained by greater electrical power input. Second, the decrease indicates higher electrical resistance of the metal filament, which indicates a higher temperature, even as it dissipates less power.

May 28, 2013 11:49 am

andywest2012 says May 28, 2013 at 11:12 am
… whether their food is too exclusive for photons at the wavelength of water.

Mistaken impression by the public (I’m afraid); 2.4 GHz is not a magic number selected b/c ‘water’ is resonant there. Verily, industrial processors use 915 MHz equipment for the deep ‘reach’ (penetration) into foodstuffs. 2.4 GHz happens to have a very manageable wavelength compared to 915 MHz (or 27 MHz, the old ‘diathermy’ equipment used that!)

Rob
May 28, 2013 11:50 am

I have been using this experiment to heat my pump house for thirty five year , insulation is a wonderful thing .

OldWeirdHarold
May 28, 2013 11:50 am

Leonard Weinstein says:
May 28, 2013 at 11:46 am
=====
I actually did that futzing around when I was a kid. I painted a light bulb with dark red paint (don’t ask why). It got so hot it burned the bulb socket.

DirkH
May 28, 2013 11:52 am

Cho_cacao says:
May 28, 2013 at 10:52 am
“OMG you finally understood how it works! The sun is providing a (so to say) constant energy input, the GHGs reduce the amount of energy being released to space… And voilà!!!”
Why, then, does the temperature not go up since 1998?
Does the energy enter the oceans? How, when it is delivered by IR photons? IR does not enter water beyond the skin layer.
Something is wrong in the models.

Slartibartfast
May 28, 2013 11:54 am

A cooler object doesn’t radiate energy to a warmer one.

Assertion monkey is assertive.
Look, I am beginning to suspect that y’all are equating energy transfer with net energy transfer. Sure, the cooler object receives more energy from the warmer one than flows in the opposite direction. What of it? The warmer object is still getting more radiation than it would from e.g. intergalactic space.

Carrick
May 28, 2013 11:54 am

wickeroy:

So, when you say “the colder outer layer warmed (or slowed the cooling of if you prefer) “, you basically embraced both camps……or??

The PSI group has claimed the bulb won’t heat up, so it’s hardly embracing both camps.
“Slowing the rate of cooling” is a alternative description to the “colder outer layer warmed”. They mean the same thing. Some of us prefer one description over the other, but the point is the physics isn’t changed by choosing a different description of the physical laws.

May 28, 2013 11:56 am

OldWeirdHarold says May 28, 2013 at 11:46 am

Please. None of this matters. You could use DC, and you’d get exactly the same result.

AFTER all the detail I included? PLEASE, who is ignoring factors here? You may ‘lose’ your possibly small change in signal (current value) in the ‘noise’, as, for instance, today most mains line voltage has the appearance of being ‘clipped’ owing to the amount of … you tell me, since you’ve assumed expert status on this! When is the last time you viewed on an O-scope the AC mains? Ever?
Dear readers, use caution when seeking technical advice from this man …
.

wikeroy
May 28, 2013 11:58 am

Carrick says:
May 28, 2013 at 11:54 am
“Some of us prefer one description over the other”.
Okay, I must be “old school” then, since I prefer “slowing the rate of cooling, then. “Backradiation” sound so….backwards.

Carrick
May 28, 2013 12:00 pm

DirkH:

Why, then, does the temperature not go up since 1998?

It’s gone up actually, just slowed a bit.

Does the energy enter the oceans? How, when it is delivered by IR photons? IR does not enter water beyond the skin layer.

You need to specify which wavelength of IR you’re talking about… but for IR photons form the sun, the average penetration depth is on the order of meters.
From there it gets distributed by the mechanisms of thermal conduction and convection.

Something is wrong in the models.

Something is always wrong in the models. That’s why we call them models and still collect experimental data.

Slartibartfast
May 28, 2013 12:01 pm

Before I would attempt to screw this whole thing up with AC analysis, I’d consider these things:
1) You’re going to have to estimate the thermal mass of the filament and model its thermal rise and decay…at 60 Hz.
1.5) Since it’s the temperature of the glass bulb that is being measured, you’re going to have to model its response to the filament radiant energy.
2) The experiment is still a decent exhibition of the author’s point even if the filament temperature is oscillating.
3) All of that said, I’d prefer to rectify to DC and redo the experiment than tailchase the AC aspect of things.

astonerii
May 28, 2013 12:03 pm

Not really an open system at all. The only time it was an open system for purposes of the experiment was when the bulb was all by itself.
I also used lost incorrectly. I meant absorbed into the glass, thus preventing that energy from being reflected.
My base argument is that of all the testing, the one thing that did not get tested was what is being argued, that gasses cause a greenhouse effect.
My understanding of the situation is the following.
Yes, there is a green house effect from some gasses.
That greenhouse effect is limited by the amount of energy available in the frequencies those gases work in.
That once the bandwidth is saturated no amount of additional gas will have any measurable, within the next 100 years (technological limit, not effect), effect on the climate of the Earth.
The sun and our position in the galaxy drives our climate through direct energy input and galactic cosmic rays effecting cloud cover.

Leonard Weinstein
May 28, 2013 12:06 pm

wikeroy says:
May 28, 2013 at 11:43 am
I have communicated many times with the slayers. You are correct that some of them agree with the process described as “slowed the cooling” ( Or Insulated it, thereby making the temperature go up). Radiation absorption by absorbing gases and radiation by radiating gases to the ground does exactly the same thing, and thus causes a warmer ground than otherwise, and this is called the atmospheric greenhouse effect. Many of the slayers deny this effect altogether. The issue of CO2 increases being significant is a separate issue, as negative feedback and natural variation probably overwhelm that small effect. This is exactly what Roy, Anthony, and I say, and is the point of the present writeup. However, be clear, it slows the cooling by absorbing back radiation, and thus decreasing net radiation heat transfer. Thus, not all slayers are wrong on all issues, but those that deny back radiation from a cooler source to a warmer source can cause increased temperature in the presence of constant supplied energy are totally wrong. Slayers such as Joseph Postma and some other totally deny the possibility of absorbing back radiation from a cooler source.

higley7
May 28, 2013 12:07 pm

Unless the light intensity is also added to this consideration, the energy flux and temperature are not meaningful. Did light intensity, which represents energy outward alter with changes in current or temperature of the filament? This experiment is far from being as simple as it first appears.
If you think about the atmosphere and the “GHGs,” IR radiation from the surface heads upward, encounters gas molecules and most often is immediately re-emitted in its original direction. Some molecules absorb the energy, and while energized, they collide with another molecule and, instead of the energy being re-emitted as IR, it becomes heat energy. Statistically this latter event is rare and thus there is little warming of the gases. Of course, the IPCC brains took the thermodynamic constant for CO2 and multiplied it by 12 to increase the effect, but even then it was still tiny. So, they called upon water vapor to be enslaved by CO2 and multiply the effect of CO2 by another order of magnitude. Of course, they also entirely deny the existence of the water cycle which is a huge heat engine which carries energy to altitude, moving perhaps as much as 85% of the energy budget, with the remainder as radiation (this is why Trenberth has so much trouble with “missing heat,” there being no water cycle in his world).
Now, any IR re-emitted by GHGs downward will encounter a warm surface whose equivalent energy levels are already filled, and thus, the RI will be absorbed as IR is emitted, being a wash, or simply reflected back upward. The result is no heating occurs. Furthermore, making this scenario even less likely to be at all detectable is the fact that the IPCC claims that this IR is redirected downward by the upper troposphere which is rather thin, at 0.25 atm, and has fairly little CO2 in terms of density per cubic meter. Sure, this is where Raleigh Scattering may occur and can be re-directed downward, but that is only one of 6 major directions and thus little IR is redirected downward. And, to make matters worse, the upper troposphere, which the IPCC says has to be warming faster than the surface, has been cooling a bit rather than warming and, in fact, the hotspot the IPCC MUST HAVE is entirely missing. This one missing feature of their model kills the whole model. It’s a must have!!!!!!!! Why do people feel free to ignore this glaring problem?

Anymoose
May 28, 2013 12:09 pm

I think what we have here is a confirmation that Ben Franklin knew what he was talking about, 250 years ago. Depending on radiation to heat air is a fool’s task.
Ben knew that standing in front of a fireplace would warm his backside to the point of discomfort, while leaving his belly cold. By placing the same fire (heat source) in a stove would heat the room to some level of comfort. Ben didn’t know about molecules and that sort of thing, but understood that the top, bottom and sides of the Franklin stove transferred heat to the air more efficiently. We understand now that those molecules bouncing off the stove, and off of each other, were being heated by contact and being carried off by convection, to eventually bring the room temperature to a more-or-less equilibrium temperature.
This also explains how the re-radiated heat energy from the earth’s surface can pass through the air molecules of the troposphere, without raising their temperature much
All of this coming from Ben Franklin’s highly calibrated backside.

Slartibartfast
May 28, 2013 12:10 pm

Not really an open system at all.

The only real open systems are hypothetical.

was what is being argued, that gasses cause a greenhouse effect

The argument here is not that gasses cause a greenhouse effect; the argument (as I understand) is that a cooler object can warm up a warm object. Which as I see it is intuitively obvious to the casual observer.

Carrick
May 28, 2013 12:12 pm

Jim, I do look at the line noise from AC line in. If you want to collect measurements with an accuracy of parts in 10^6 (I often do), then these things perhaps will matter. Even for high-end audio applications, filtering the AC matters (using balanced lines helps more though).
But for a physics demonstration where you’re just looking at qualitative changes, I don’t expect this to matter very much. Still replacing the AC line with a good DC digital power supply (where I can monitor the current draw directly) would be a nice improvement.

son of mulder
May 28, 2013 12:13 pm

” OldWeirdHarold says:
May 28, 2013 at 11:50 am
I actually did that futzing around when I was a kid. I painted a light bulb with dark red paint (don’t ask why). It got so hot it burned the bulb socket.”
I’m amazed it didn’t explode. I’ve had lightbulbs explode because they were too high a power for the light shade.

rgbatduke
May 28, 2013 12:13 pm

Go for it. Do it. And when the current drops (and it will, but possibly by a very small amount), tell me why that happened. Show all work.
Or, you could read TFA

Slartibartfast
May 28, 2013 12:15 pm

If you define line noise to include amplitude and phase variation on top of high-frequency noise, I’d say that yes, those will probably dominate here.
But if you take those away, you’re going to see apprximately the same result.

Rob
May 28, 2013 12:16 pm

Seems all one would have to do is call a bulb manufacturer , their R&D should have all the data as they manufacture a wide variety of color’s and inclosure’s .

rgbatduke
May 28, 2013 12:16 pm

where it is shown that the current does indeed drop with the foil shell and warmer temperatures, proving that the filament does, in fact, heat as its resistance increases a small amount (given the warming in degrees absolute and T^4 radiative cooling).
Sorry about the split post — mouse clicked in the middle of typing…

May 28, 2013 12:16 pm

chris y says May 28, 2013 at 11:29 am
I continue to be amazed that this remains a controversial issue.
GE R&D center developed an incandescent bulb back in the late 1980′s (and GE lighting commercialized it in the 1990′s) that placed a spherical clear glass shell around the filament.
The shell was coated with a multilayer (anywhere from 15 – 30 separate layers) optical filter that reflects mid infrared back onto the filament, while allowing visible light to pass through.
The result is that a lower filament current can achieve the same filament temperature, thanks to the mid infrared energy being reflected back to the filament. This results in a 15% – 20% increase in lumens/Watt.

Excellent post. Bears repeating. Thank you.
It appears GE suspended some of that work, according to press at the time (2010):
GE Suspends Development Of High-Efficiency Incandescent Bulbs
http://www.environmentalleader.com/2008/12/01/ge-suspends-development-of-high-efficiency-incandescent-bulbs/
A little more detail on the technology:

There were two publicly-known technologies they were working on at the time that, if improved, could raise the efficacy of incandescent lamps to the 50 to 60 lm/W range.
The first is IR reflecting films, a technology that is already in commercial use. Considering that 90% to 95% of the energy generated by an incandescent filament is radiated away as IR (depending upon where you define the long wavelength end of the visible spectrum), using IR films to raise the efficacy of incandescent lamps by a factor of 3 or even 4 is possible. Low-voltage IR-halogen filament tubes may already meet the initial goal of 30 lm/W.

From: http://sci.engr.lighting.narkive.com/Zjpr8gDa/new-ge-incandescent-lamp-technology
.

FerdiEgb
May 28, 2013 12:20 pm

Duster says:
May 28, 2013 at 11:13 am
Might consider redoing this using a clear bulb. That would eliminate the effects of the interior coating of the bulb. Otherwise, very interesting indeed.
I think the bulb used was an internally etched one, not coated. That only spreads the light in all directions, but doesn’t absorb it.

higley7
May 28, 2013 12:21 pm

Another point of view: There is no doubt that the Earth’s SURFACE is cooler for there being an atmosphere as it blocks some incoming radiation and also carries heat away by conduction and convection. BUT, and a big BUT, having an atmosphere makes the space ABOVE the surface much warmer, as the vacuum of space would be in the single digits of Kelvin. The ability of GHS to convert IR to heat in the atmosphere is pathetic and undetectable in the real world

Cho_cacao
May 28, 2013 12:23 pm

DirkH,
the discussion was about the heating bulb. Believe or not, but the planet is a slightly more complex system! 😉

rgbatduke
May 28, 2013 12:23 pm

And incidentally — well done! Too bad Joe won’t just concede that all of the times he has asserted to me personally in both WUWT threads and email threads that reflection does not increase the temperature of a heated source (while refusing to actually do the experiment properly himself or acknowledge the technology that is based on the principle) he has been just plain wrong.
“Back to the drawing board” wrong. “Start over, this time using the laws of physics” wrong. “Hard to be wronger than that” wrong. “Watch the movie” wrong. “Read the hard data points above in an easily reproducible experiment that controls for convection and conduction quite nicely” wrong. Really, truly, wrong.
Come on Joe. Time to man up. Acknowledge that you were WRONG so that we can go on to the next thing you are wrong about, then the next thing, until there is nothing left of the PSI assertions but hot air and a bad smell.
Or, you can shorten the process considerably and abandon PSI altogether, right now. Because they/you are also wrong about the Lindzen model, wrong about the supposed violation of the second law, and, in fact, wrong when they assert that there is no effect on surface temperature from clearly measurable atmospheric back radiation. All of which are perfectly obvious and empirically proven, but it’s time to acknowledge this and move on.
rgb
rgb

Just an engineer
May 28, 2013 12:26 pm

pgosselin says:
May 28, 2013 at 11:31 am
And part of the Gulf of Mexico flows up to Missouri, right?
————————————————————————————-
Actually the Gulf of Mexico IS the predominate source of Missouri rain.
So you are correct. This time.

Maxbert
May 28, 2013 12:30 pm

This experiment seems like going to a lot of trouble only to demonstrate that while you can slow the rate of net transfer, depending on the point of equilibrium, you cannot reverse it.

rgbatduke
May 28, 2013 12:31 pm

If you think about the atmosphere and the “GHGs,” IR radiation from the surface heads upward, encounters gas molecules and most often is immediately re-emitted in its original direction.
Actually, that turns out not to be the case. Not just “most often” but almost always, the gas is re-emitted in a dipolar distribution that is azimuthally symmetric around the polarization axis. For unpolarized light, this axis is itself essentially randomly distributed in the plane perpendicular to the original direction and hence reradiation is essentially spherically symmetric.
So LWIR heads up, encounters gas molecules, is absorbed (or not, depending on wavelength and chance) and if absorbed, is radiated back down at the surface just under 50% of the time.
The “just under” comes from transfer, stimulated emission while excited, and the slightly smaller solid angle of the sphere versus space at some height above the surface, usually negligible.
This is distinct from albedo. You can read Grant Perry’s book if you want to work through the details.
rgb

rgbatduke
May 28, 2013 12:35 pm

A cooler object doesn’t radiate energy to a warmer one.
I’ve got a blue-light laser here that says you are wrong. I will cheerfully direct it at your hand while holding a thermocouple against the laser. Any takers?
Any more obviously false idiocy?
rgb

Michael J. Dunn
May 28, 2013 12:36 pm

A horrendously long thread; I didn’t have time to read it all. And such confusion! Let me point out one simple matter: If either of two bodies are maintaining a surface temperature on the basis of an internal heat source being relieved by thermal radiation….there is no conservation of energy (duh!…internal heat source).
As for two bodies next to one another, they are both “hot” relative to absolute zero (the only “cold” that counts). Either will raise the other’s temperature relative to what it might have been if the other had been somewhere else alone (no conservation of energy, remember?), and more heat will be transferred to the less “hot” body than to the more “hot” body (net heat transfer from “hot” to “cold”), as we all expect from simple thermodynamics.
It is easy to show that the Earth’s temperature is essentially a function of the ratio of its integrated radiation absorption coefficient to its integrated emission coefficient. Absolute surface temperature will vary in proportion to the fourth root of this ratio. This has meant the difference between tropical (carboniferous) and frozen ages. We know our albedo to maybe a 10% error. For a nominal surface temperature of 300 K, that works out to a temperature variation of 7 C or 13 F. It doesn’t take much to result in really big shifts. The miracle is not that the surface temperature changes, but that it really doesn’t change much at all. There are powerful negative feedbacks at work, thank God!

mkelly
May 28, 2013 12:36 pm

q = ε σ (Th4 – Tc4) Ac
In view of the experiment what do we do with this from engineering tool box. It clearly says that once two temperatures are equal no heat transfer takes place. But the experiment and many say that the above equation is wrong. Where is the feed back loop in the equation that allows for Tc to add to Th? rgb according to this equation a reflection cannot heat the source since Th = Tc in reflection so where to we go? What equations or series of equations do we use?

May 28, 2013 12:41 pm

Carrick says May 28, 2013 at 12:12 pm

But for a physics demonstration where …

Ahem … I’d like to resolve the current draw issue (although I think it is now moot), I have a request standing for the model number of the clip-on style meter seen used in the above demo pics … I’m interested in those numbers and not so much in editorializing. Thank you.
I also addressed issues from an instrumentation and ‘dynamics’ standpoint for the more astute reader who has curiosity to that depth (I know I have).
BTW, accuracy in parts in 10^6 are really easy today, but if that is your ‘limit’ then you can improve on that by two and three orders of magnitude easy (for frequency anyway, which is what I am most often concerned with). I maintain a pair of IC-756ProII at around 10^7 (short term), and this is by “zero beat” (using human ear) with the NIST over-the-air SW reference station called “WWV”.
Replacing the AC with DC allows for low-cost instrumentation to be used and negates any ‘dynamics’ effects that can cloud (pardon the pun) or mask issues. This should remind many here who might have stumbled onto one or two (maybe more) YouTube videos put out by the ‘free energy’ crowd who use ‘static’ instrumentation (which measures/records/reports peak values but displays reports RMS-corresponding (normally for AC sinusoids) or ignores peaks altogether (most DMMs)) to observe obvious pulse phenomenon as they attempt to ‘extricate’ energy from magnets and pulsed coils/inductors, ignoring the ‘flyback’ effect as the built-up mag field collapses when the device’s ‘switch’ removes the periodicly voltage or current source. The so-called “Rosemary Ainslie circuit” being an example with widespread coverage and publicity …
.

Barry Cullen
May 28, 2013 12:42 pm

Haven’t read all the comments but consider this; one of the “bulb” manufacturers, don’t remember which one, had been producing tungsten/halide bulbs for many years that have a coating of IR reflector (multi-layer interference type) on the clear, high silica envelope. The W filament is supported in the center line of the bulb. The result is higher lumens/watt (~32-33) than without the coating (<30).
So, the cooler envelope coating IS heating the filament with IR that normally would escape. The increase in temperature is tiny (~53°K at 3000°K) for the 10% increased output because the radiation increases by T^4 and that bulb envelope temp is certainly way, way, way below 3000°K!
BTW – the T^4 relationship is likely why, in addition to or conjunction with Willis' tropical thunderstorm governor mechanism the temperature of the Earth is basically bi-modal, as determined by the long term EM input, like Momma and Baby bears porridge's but never Papa bears.
BC

joeldshore
May 28, 2013 12:44 pm

Slartibartfast says:

Look, I am beginning to suspect that y’all are equating energy transfer with net energy transfer. Sure, the cooler object receives more energy from the warmer one than flows in the opposite direction. What of it? The warmer object is still getting more radiation than it would from e.g. intergalactic space.

Exactly. What people need to understand is that at the time that the 2nd Law of Thermodynamics was formulated, the only thing that could really be considered was the net macroscopic energy flow. So, that is what “heat” is. We now understand this fundamentally time-irreversible behavior at the macroscopic level as arising from fundamentally time-reversible processes at the microscopic level, as described by statistical mechanics. (Thing of “time reversibility” as referring to whether a movie of the situation makes sense when run in reverse. For example, the collision of two molecules would be time-reversible: A movie in reverse represents a perfectly reasonable collision. However, the spontaneous flow of heat from hot to cold or the slowing down of a block sliding along a table due to friction is not time-reversible: We would immediately recognize the movie in reverse as showing something that we do not observe in nature.)
What “Slayers” are doing is applying macroscopic laws to microscopic behavior. I would say that they were stuck in the middle of the 19th century in their thinking, but this is unfair to the scientists of that time period because these scientists didn’t make this fundamental mistake. They merely describes the only world that they had the capability to observe at that time, which was the macroscopic world.
The whole point of the modern understanding of thermodynamics is that the 2nd Law does not arise because of arbitrary and capricious laws like saying that colder objects don’t radiate toward warmer objects (or the warmer objects absorb the radiation from colder objects). Rather, it is a consequence of simple facts such as that the amount of radiation from a warmer object that is absorbed by a colder object is always greater than the amount of radiation from a colder object that is absorbed by a warmer object, a consequence guaranteed by the basic laws governing such radiation (i.e., that radiation emitted is an increasing function of temperature and Kirchhoff’s Radiation Law relating the emissivity and absorptivity of an object at a given wavelength).

May 28, 2013 12:45 pm

It clearly says that once two temperatures are equal no heat transfer takes place.

No, once to items are at the same temperature, the *net* heat transfer between them drops to 0. If they are not isolated systems, the equilibrium will be the product of all the heat flows between them canceling out to 0.

May 28, 2013 12:45 pm

Carrick,
“Once again you are confusing electrical power (which actually isn’t a constant, it decreases slightly as the filament heats up) with temperature, which is a measure of thermal energy.”
Once again? When was the first time? Power is Joules/second. Joules are energy. Heat is a measure of the average kinetic energy of the molecules in the object whose temperature is being measured. Power from a resistance heater such as the filament of this light bulb is voltage times amperage. The wire’s resistance changes with temperature. The writer states that the current measured to three decimal places did not change, hence the filament’s resistance did not change, hence its temperature DID NOT CHANGE.
Confusion does not justify your insulting comment, let’s all be polite to each other here.
I looked up the GE lamp. They did patent such a thing. This does not mean it worked. The vacuum inside a light bulb is not perfect. Warming the gas inside a bulb would reduce conductive losses from the filament to the gas, could give a benefit I suppose. Try it yourself, shine a flashlight directly into a mirror, touch the lens directly to the mirror, see if the light gets brighter. It will not not not…
I tire of this.

May 28, 2013 12:47 pm

mkelly says May 28, 2013 at 12:36 pm
q = ε σ (Th4 – Tc4) Ac
In view of the experiment what do we do with this from engineering tool box. It clearly says that once two temperatures are equal no heat transfer takes place.

Which tool box – the radiative flux toolbox?
This doesn’t work for “clouds and the ground” on an overcast night … does it?
.

OldWeirdHarold
May 28, 2013 12:50 pm

If nothing else, this thread proves Mosher’s theorem that skeptics aren’t a tribe, but a confederation of tribes.
Climate skeptics are the Jews of the science world; if you get two in a room, you’ll have three opinions.

May 28, 2013 12:53 pm

Good demonstration on INSULATION. This is a very good demonstration of GHE (heat entrapment), insulation (dead air space) and reflective insulation – nothing more. The glass cover is the GHE/dead airspace/heat entrapment, The tinfoil is reflective insulation, The black box is again “reflective” (just not as good) insulation. The piping of most Electrical Power Plants have “reflective” insulation around all (most) of the piping and all major components TO PREVENT HEAT LOSS! The boiler sure isn’t warmed by the thin layer of shiny tin-plate/aluminum.
Have you never seen the Barbie Doll Ovens of the 50’s? They used a light bulb and got up over 200 degrees F.

joeldshore
May 28, 2013 12:53 pm

mkelly says:

q = ε σ (Th4 – Tc4) Ac

But the experiment and many say that the above equation is wrong. Where is the feed back loop in the equation that allows for Tc to add to Th?

No…It doesn’t. The experiment and what we are saying is completely compatible with this equation: The rate of heat flow away from an object depends on both the temperature of the (warmer) object and the temperature of the (cooler) surroundings. [And, what that equation doesn’t specifically tell you, but nearly any introductory physics textbook will, is that our understanding of the equation is that the first term represents radiation emitted by the object to the surroundings and the second term represents radiation absorbed by the object from the surroundings.]
Hence, if the temperature of the surroundings is higher, then the heat flow (net macroscopic energy flow) will be less. When the heat flow away from the object decreases while the inputs from other sources (such as the sun or an internal source converting some form of energy into thermal energy) stay the same, the object will find itself emitting less heat than it is absorbing and will have to increase its temperature until the balance between absorption and emission is restored. (Since, we have been trying to explain these simple ideas to you for years with no success, I don’t expect a different result now, but at least others will understand where your reasoning goes wrong.)

May 28, 2013 12:55 pm

I have some questions that better minds than mine can hopefully answer.
The author states…

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).

In which case, whether we place the foil on the outside of the glass or on the inside of the glass should make no difference to the result (radiative transfer restricted in both cases) but there is a substantial difference i.e. bulb T = 138DegC with foil on the outside and 176DegC with foil on the inside. How do you explain the difference?
Furthermore, the instance of foil cover alone should return the same result as foil on the inside or outside of the glass, yet it returns 148DegC. Again, how do you explain the difference.
I also note that all covers – glass, foil or a combination – are sitting on wires of varying thicknesses. What steps did you take to stop air circulation due to possible gaps between the cover and the table top? I imagine the increase in air pressure inside the covers would force some air circulation between the very warm interior and relatively cool exterior.

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.

My jury is out at mo.

John West
May 28, 2013 12:57 pm

wikeroy says:
”So, when you say “the colder outer layer warmed (or slowed the cooling of if you prefer) “, you basically embraced both camps……or??
Or I’m talking about reducing radiant net energy emitted from the bulb via radiant energy from the outer layer, something the Sky-Dragon-Slayers think is impossible.
In order to calculate the net radiant emission of the bulb one would have to subtract the radiant emission from the outer layer being received by the bulb from the Stephan-Boltzmann emission of the bulb.
Full Sarc on: Photon control to photon 123423564978563p389364j262533, you are not cleared for emission along pathway G34523433445j388 due to an object 0.000001 degrees hotter than us a mere 15000 light years along this path. Please redirect emission path to S63535226522k363. /sarc

Roger Clague
May 28, 2013 12:59 pm

Michael Moon says:
May 28, 2013 at 8:52 am
The Temperature of the Filament is the only temperature that matters here. Did the Filament warm from back-radiation?
In his description of the proposed experiment Siddons asks does the light shine brighter?
It is the filament that is shining not the surface of the glass bulb.
Did the filament warm?
The temperature ( T ) of the filament is proportional to the power loss
power loss ( W ) = V x A.
The temperature of the filament is proportional to current flowing in it.
Bulb surface exposed T 96C current 0.289A
Bulb surface cover in foil T 156.6C current 0.289A
Covering the bulb with foil has no effect on the filament temperature. The concept of heating by back-radiation is not supported

May 28, 2013 1:03 pm

Michael Moon says May 28, 2013 at 12:45 pm

The writer states that the current measured to three decimal places did not change, hence the filament’s resistance did not change, hence its temperature DID NOT CHANGE.

You were called on this once, twice, thrice, maybe more.
Can you point out what paragraph that states ‘three decimal places’? He shows a LOT of data measurements (including voltage and current measurements) with FOUR significant digits. You don’t see that?
All that can be found is which could be construed to support your contentions: “The current measurements shown above demonstrate that the current draw of the bulb was no higher when the bulb temperature was higher, and was in fact slightly lower.
Then there are these reported current values:
With the bulb in open air and a surface temperature of 96.0C, the bulb used 289.4 milli-amperes of current.
With the bulb covered by a foil shell alone and a surface temperature of 158.6C, the bulb drew slightly less, 288.7 milliamperes.
Nope. Doesn’t support your contentions on multiple fronts.
Either you have comprehension issues, are acting dishonestly in debate, maybe think we’re all really stupid, or I really, really missed something in the head post.
.

rgbatduke
May 28, 2013 1:07 pm

Macro behavior in solids is not transferable to micro behavior in individual, three atom gas molecules. The Aluminum foil is at least several thousand molecules thick, the glass even more. Radiative and convective flux are both restricted by both materials based on mass, specific heat and in this case, the sealed cubic configuration. In a free gas environment there are limited restrictions to convective and radiative flows, and a three atom gas molecule, at virtual rest compared to a photons speed, is not capable of stopping or redirecting this high velocity OLR force for more than milliseconds. Atmospheric CO2 is a feather in the path of a howitzer round.
Or, you could choose between learning some actual quantum theory and learning something about the optical cross-section of CO_2, compute or measure the mean free path of LWIR photons in the CO_2 absorption bands, and conclude that this “feather” is utterly optically opaque and indeed saturated.
The weakness of the GHE in response to increases in CO_2 concentrations isn’t because there is no effect, it is because getting an LWIR photon in the strong absorption part of the spectrum from the surface out to space is ALREADY as unlikely as being able to throw a tennis ball straight through a crowd of groupies next to the stage at a rock concert without hitting one. Adding twice as many just means that the tennis balls go an even shorter distance into the crowd before bouncing.
I mean, do you just make this stuff up as you go along, Joe, or do you even TRY to look up facts and numbers before spouting it off? Do you even understand that all mass is not only at “virtual rest” compared to a photon’s speed, it is actually at precise rest compared to a photon’s speed? Do you have any idea that you are conflating force, momentum, energy, and that the time scale of milliseconds you cite is itself unintelligible and unfounded. The time of flight of a photon traveling directly from the surface to the top of the troposphere at 9 km is 9 x 10^3/3 x 10^8 = 3 x 10^-5 seconds, that is, 3 hundredths of one millisecond, and we won’t even bother looking at the radiative lifetime of excited CO_2 as I’m sure it is order 10^-8 seconds or thereabouts, too short to care about. ONE millisecond is hence tens to hundreds of times the unobstructed time of flight to where the atmosphere is no longer optically opaque.
Why not look up, or compute, the mean free path of an IR photon in the strongly absorptive CO_2 bands before asserting that this “feather” cannot possibly be important? Or just look at TOA IR spectrographs to see just how much surface radiation is empirically blocked by this “feather”.
rgb

Stephen Richards
May 28, 2013 1:08 pm

Semiconductor materials is my physics so quantum mechanics was not my ultimate subject to comprehend but simplistically : energy in the form of radiation is Hµ (nu). Freedom of movement and therefore bands of absorption for CO² there are 2. Energy quantum levels available to absorp electromagnetic energy are down to the electron binding energy between the O and C nuclii. If a molecule is in its’ ground state then energy of both µ (wavelength) can be absorped. However, the emitter does have to have at least one electron above the ground state, so 0K the colder object cannot warm the warmer object or more accurately a molecule at ground state cannot emit energy to an object above ground state. There is no energy to transfer. If a molecule is 2 quantum states above ground and another molecule is 1 quantum state above ground then energy can be transfered between the 2 but only higher energy to lower energy if both systems are two level systems. If an EM wave packet of energy greater than 2 quantum energy levels should enter a 2 level system at ground state then the energy will not be absorped. It must have a quantum value hµ exactly the same as the quantum level(s). If these are 3 level systems then energy can move from cold to warm and warm to cold.
The problem, for me, with all of these heat experiments is the elimination of all other forms of energy transfer. Heat transfer from the filament to the glass can only be said to be radiative if there is no gas inside the bulb otherwise heat transfer from the filament will be a combination of kinetic and radiative (convective would be infinitessimally small).
Now, When a molecule is energised above ground state it’s only objective is to divest itself of this excess energy as quickly as possible and it will do so as quickly as possible and it will do so in one direction. The sense of that direction will be a probability function that I simply cannot remember. It may or may not be totally random, but the walk towards the upper atmosphere will be but within the convection pattern prevailing and the available molecules. At 400 ppm the absorption rate is going to be very small and the vast majority of ejected EM energy is likely to head straight into space. As Niels Boor discovered, even when you know where the atom is that you wish to hit, hitting it is a very big problem.
I just hate the term “back radiation”. It grates.
OK, I’M READY. FIRE AT WILL

rgbatduke
May 28, 2013 1:27 pm

First, a quantum atom or molecule in the ground state is perfectly happy first absorbing a photon, then re-emitting it in a new direction. A cold gas will scatter optical energy just fine. It does need to be approximately resonant with the photon being scattered or the scattering process is second order and strongly suppressed.
Second, a quantum atom in any known quantum state is at zero temperature. Or, if you prefer, its temperature is undefined and irrelevant, but really a definite state has zero entropy and zero entropy is zero temperature. Quantum statistical mechanics requires an ensemble of atoms and a fraction of them must “probably” not be in the ground state to have a temperature. Typically such a thermal state is described by a thermally weighted “mixed state” trace of a density matrix, effectively a semiclassical state. A better treatment is to start with a density matrix and go through the process that leads to e.g. the Nakajima-Zwanzig equation for an open quantum system in contact with a thermal bath and then equilibrate it (possibly numerically via a langevin or fokker-planck sort of approach in monte carlo) — this lets you look at internal correlations in the thermally equilibrated state.
As for calling back-radiation back-radiation — in some sense I don’t like it either. But like or or not, all you have to do is step outside at night and point an IR spectrometer up and there it is, coming down, and the source of it is definitely not outer space. Nor is it just the heat already in the atmosphere radiating down. It is at least partly ground radiation being actively absorbed by the atmosphere and re-radiated back down. Back radiation is as good a term as any, but I’m open to alternatives.
Just FYI.
rgb
And now, time to stop making the world safe for sky-dragons and removing the insult to the revered name of Newton for a bit and go fishing for a while. It looks like the physics of the experiment above is well-defended by knowledgeable people so that no attempt to defend the indefensible by PSI slayers is likely to get any traction at all. Not that this should be necessary.

May 28, 2013 1:35 pm

Joel Shore commenting on my post above
Where I say that Curt’s summary implies that the presence of a colder object will ALWAYS increase the temperature of a warmer one.
It contradicts common sense.
Try putting a large block of ice in your living room.
Its unfortunate that Anthony snipped the post that this refers to and you unfortunately have got the ‘wrong end of the stick’, so to speak
I agree with your comment.
See what you and RJB think of this situation
A hollow sphere at a higher temperature and black body emissivity has a colder object placed at its centre.
Lets say that there is a vacuum between the cold and hotter sphere to eliminate conduction and convection.
Now the cold object is a new source of radiation.
Its radiation falls on the spheres inner surface and is absorbed.
Yet the sphere will not increase in temperature.
Quite the opposite in fact.

rgbatduke
May 28, 2013 1:40 pm

Good demonstration on INSULATION. This is a very good demonstration of GHE (heat entrapment), insulation (dead air space) and reflective insulation – nothing more. The glass cover is the GHE/dead airspace/heat entrapment, The tinfoil is reflective insulation, The black box is again “reflective” (just not as good) insulation. The piping of most Electrical Power Plants have “reflective” insulation around all (most) of the piping and all major components TO PREVENT HEAT LOSS! The boiler sure isn’t warmed by the thin layer of shiny tin-plate/aluminum.
Again, you could try actually drawing a nifty little graphic picture with arrows and everything and try using some actual equations instead of just saying “I think you’re wrong because pink unicorns tell me so”.
At least three or four pieces of evidence above indicate that the “boiler” was in fact warmed by the thin layer of shiny foil. The measured current in the filament dropped as the filament got hotter. The temperature of the bulb got warmer with any sort of optical barrier than it did with just a convective barrier. The conductivity of the foil was MUCH less than the conductivity of the glass, yet it produced the greatest warming even without the glass.
And besides, read your own words. You speak of “reflective insulation” helping out with the GHE “heat entrapment”. The conductive insulation capability of aluminum foil is zilch — try using a piece as a potholder if you doubt that. The convective trapping essentially didn’t change when switching from glass to glass surrounded by foil but the temperature went up further, so we can ignore the convective increase, there is more than just that.
You have just conceded that the foil reflects heat back into the cavity, and that this nice cool foil thereby causes the contents of the heated cavity to become warmer still. This is a nearly precise definition of the radiative atmospheric effect, a.k.a. the GHE. The sun heats the Earth. The atmosphere reflects some of the heat back at the Earth. The Earth gets warmer than it would be without the foil.
This is exactly what the slayers have repeatedly claimed cannot happen “because a colder body cannot heat a warmer one, even if the warmer one is actively being heated”. Obviously, it can. Obviously, it does. Thank you for conceding this. I will even forgive your minor error that the foil doesn’t raise the temperature of the “boiler”, or in this case the bulb filament. Insulation of a heated space does, in fact, cause its temperature to rise, all the way back to the heater, because the heater has to lose the energy you add to it against a thermal gradient. As you heat the space, you reduce the gradient. The heater gets hotter until it is power balance once again.
As the evidence above rather graphically proves for the specific case of a heated light bulb.
rgb

upcountrywater
May 28, 2013 1:41 pm

“Skydragon Slayers”
Say are they selling refrigerators, because if they are, my electrical meter will run backwards. A downside is everything inside will freeze, even when I jam the light switch so the light stays on while the door is closed….

Carrick
May 28, 2013 1:44 pm

Slartibartfast: 3) All of that said, I’d prefer to rectify to DC and redo the experiment than tailchase the AC aspect of things.
For precision work I’d use something like this.

Johan i Kanada
May 28, 2013 1:48 pm

“snip – if you want to point out errors, do so now. No other discussion from you is of any value at this point – Anthony”
I find it strange that you post an article refuting the “slayers” and then you delete their comments/objections. This behavior reminds me of the CAGW extremists. Not pretty.
REPLY: You are right, when somebody says you are wrong, but refuses to tell you WHY (after being repeatedly asked on another thread) it isn’t pretty when they make accusations without substance. He’s had dozens of opportunities. – Anthony

Carrick
May 28, 2013 1:51 pm

Roger Clague:

Covering the bulb with foil has no effect on the filament temperature. The concept of heating by back-radiation is not supported

Actually, we are able to infer from the data that the filament temperature increased. This is because the measured current dropped from 289.4 mA to 288.7 mA, and the resistance of a tungsten filament increases with its temperature.
Based on the data, you are wrong.

Carrick
May 28, 2013 1:55 pm

Michael Moon:

The writer states that the current measured to three decimal places did not change, hence the filament’s resistance did not change, hence its temperature DID NOT CHANGE

As the report stated, the measured current dropped from 289.4 mA to 288.7 mA. The measured current did change in a way consistent with an increase in temperature for a tungsten filament.
A reading comprehension course might be needed before you take that physics course you need so desperately.

mkelly
May 28, 2013 1:56 pm

“But now the bulb is receiving radiant power from the metal shell, whether reflected in one case, or absorbed and re-radiated back at longer wavelengths in the other. Now the power into the bulb exceeds the power out, so the temperature starts to increase.”
joeldshore says:
May 28, 2013 at 12:53 pm
Joel the above quote is from the write up of the experiment. This does not square with your explanation. By the way no need to be condescending I understand the idea of insulation, but we are talking radiation exchange.

Carrick
May 28, 2013 1:59 pm

Jim:

BTW, accuracy in parts in 10^6 are really easy today

This depends on the signal to noise of what you’re measuring.

Roger Clague
May 28, 2013 2:03 pm

_Jim says:
May 28, 2013 at 1:03 pm
Then there are these reported current values:
“With the bulb in open air and a surface temperature of 96.0C, the bulb used 289.4 milli-amperes of current.”
“With the bulb covered by a foil shell alone and a surface temperature of 158.6C, the bulb drew slightly less, 288.7 milliamperes.”

Jim is correct. These are the important results. The part of the bulb that uses the electrical energy supplied and radiates it is the filament not the glass envelope.
To less than !% the energy radiated by the filament does not change. So we know that the temperature of the filament does not change when the bulb is covered by reflecting foil.

Carrick
May 28, 2013 2:06 pm

This depends on the signal to noise of what you’re measuring.
And more generally the quantity you’re trying to measure. The average temperature of the Earth isn’t going to be easy to measure to a part in 10^6 for example.

Carrick
May 28, 2013 2:07 pm

Roger Clague: So we know that the temperature of the filament does not change when the bulb is covered by reflecting foil.
So having quoted results that show the temperature change, you conclude the temperature does not change?
Not very interesting.

chris y
May 28, 2013 2:09 pm

_Jim says:
May 28, 2013 at 12:16 pm
I am pretty familiar with the improved incandescent work at GE, because I worked there when the effort was going on. A genius chemist had worked out a process for depositing the multilayer coatings in a batch process for about 10 cents per bulb. That in itself was a miraculous engineering achievement.
Down the hall from my lab, a group was working on an electrodeless gas discharge lamp technology that was routinely achieving more than 200 Lumens/Watt for thousands of hours. GE Lighting poured millions of R&D dollars into that technology. In the end they couldn’t get the Lumen maintenance and color stability to meet customer demand (it was targeted for commercial and industrial lighting with a 250W bulb that was about a 1 inch diameter hollow quartz sphere).
I note that LED’s are finally starting to approach this efficiency, almost 25 years later.

donald penman
May 28, 2013 2:14 pm

The bulb only gets hotter because less heat gets away from the bulb when reflective foil surrounds the bulb ,energy flow does not change unless the flow is absorbed or more radiation is produced if no additional energy is added then if one part of the energy flow gets hotter then another part of the energy flow must get cooler in this case the outside gets cooler.

commieBob
May 28, 2013 2:24 pm

Michael Moon says:
May 28, 2013 at 8:52 am
The Temperature of the Filament is the only temperature that matters here. Did the Filament warm from back-radiation? Did heat actually “flow uphill?”

1 – No my dear Michael. The heat flowed downhill more slowly.
2 – Saying that the temperature of the filament is the only temperature that matters is like saying that the temperature of the Earth’s core is the only temperature that matters.

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)

It was nice of the Slayers to suggest such a simple experiment. It was delightful that Anthony actually did the experiment. I would say though that Curt Wilson nailed it. The only thing missing is a couple of pages of equations. 😉 Simply elegant.
The comparison between the black cube and the foil cube was interesting. The black cube radiates both away from and toward the bulb. In that regard it is an analog of the case with greenhouse gasses.
The foil reflects energy back to the bulb. It is an analog with clouds. Unsurprisingly, all versions of the foil produced a hotter bulb than did the black cube.

ScottR
May 28, 2013 2:28 pm

This is easier than one would think. Take a 100 watt and a 60 watt incandescent bulb. Paint them black. Set them next to each other. Turn each one on independently and measure their surface temperatures facing the other bulb. Now turn them both on and notice that both temperatures will be higher.

Rosco
May 28, 2013 2:41 pm

There are numerous fundamental flaws with both of these arguments and experiments.
I am certain they do NOT actually demonstrate what is claimed.
1. It is confusing light with heat ! A light bulb is designed to generate light – NOT heat !
2. As other people have said before, in BOTH of these experiments the glass of the bulb is NOT the source of the radiant energy.
While the light is allowed to pass through the glass of the bulb the light energy is only contributing a minimal thermal response at the surface of the glass – the majority of the “heat” is coming from the heated filament being heated hot enough by the electrical current, so hot it emits visible light, by diffusion in the gases inside the bulb !!!
Remove the glass of the bulb and the filament would require significantly more energy to heat sufficiently to emit bright visible light.
If the light energy contributed significantly to the heat it would not be a good source of light.
The glass is being heated – not a source of heating – it is never itself hot enough to emit visible light on its own !!
It is being heated by the filament of the bulb which is far hotter than the glass of the bulb.
So the glass of the bulb is heated by the energy from the filament and is designed to transmit as much of that energy as possible – a useful feature for a light bulb.
The inverse square law actually proves this assertion – the filament, whilst not a point source of energy being all twisty, is still equivalent to one.
The energy at the internal surface of the glass of the bulb is already significantly less in terms of watts per square metre than that being emitted by the filament. I always thought temperature and watts per square metre had some relationship.
Exposed to the fee open air the glass is allowing the radiant energy to escape – its design purpose – and is also losing significant thermal energy by diffusion/convection in the free atmosphere.
Trap all that energy and convert the light to “heat” by absorption by various substances and of course the temperature will rise – allowed to escape freely the light contributes little to a thermal action.
The glass of the bulb is therefore only at equilibrium at a given circumstance and the fact that restricting it energy loss “capabilities” by restricting diffusion, convection and radiant energy losses simply prove that the bulb exposed to the atmosphere is NOT the TRUE equilibrium state for the TRUE heating element – the filament being heated by the electrical current.
The TRUE equilibrium state for the filament is when as much energy as it is capable of generating in response to the electrical current is truly accounted for not simply allowed to escape as it designed to do – it is a light bulb after all.
All that is demonstrated is that the glass is being heated by energy originating from a high energy source and trapping all THAT energy – that previously escaped because that is what light bulbs are designed for – will obviously cause a heating effect but I do not believe it can be claimed that the bulb was at a TRUE equilibrium originally because the glass of the bulb is not an active source of energy – it is completely passive.
Both of these experiments DO NOT demonstrate what is claimed.
It is converting the visible light that previously escaped into “heat” that causes the increases in temperature.
There is no possible way to determine what response is caused by the “reflected” thermal energy !!!

Rosco
May 28, 2013 2:49 pm

I believe the REAL physics has been misunderstood in this orgy of insult and revenge for name calling by all concerned.
Insuly leads to anger.
Neither is conducive to informed reasoning.

Myrrh
May 28, 2013 2:52 pm

An incandescent lightbulb radiates around 5% visible light and 95% thermal infrared, aka longwave infrared. Visible light cannot heat matter, it is not a thermal energy. What were you measuring?

dorsai123
May 28, 2013 2:52 pm

@BillMArsh,
I notice you didn’t actually answer my one question …
but you did try and make hay of my other statements …
1) There is no requirement that the objects be ‘similar’ for the experiment to work
there should be some sort of control … it is an “scientific” experiment …
2) The Aluminum foil is an ‘object’ just as the light bulb is an ‘object’.
my point is that your definition of object is pretty nebulous … which you verify here … the light bulb is hardly similar to the foil … several different materials for one …
3) Please elucidate on the several mechanisms you believe would slow the dissipation of the energy
the lack of air movement … I’ll think about some others … I’m sure you could “eludicate” some too … did you really need to ask that ? eludicate … cool word …
4) I think you’re mixing energy absorption and temperature, they aren’t the same thing.
no I’m not, I’m asking how the absorption of the radiation from the cooler object into the warmer object was measured … just because radiation is directed at an object doesn’t prove it was absorbed by said object … you are assuming that higher temp = energy absorption not me …
In general I find the discussions here interesting …. this thread appears to have broken that pattern …

Rosco
May 28, 2013 2:53 pm

Obviously I meant “Insult” – Insuly isn’t even a word.

JeffC
May 28, 2013 2:53 pm

@BillMArsh,
I notice you didn’t actually answer my one question …
but you did try and make hay of my other statements …
1) There is no requirement that the objects be ‘similar’ for the experiment to work
there should be some sort of control … it is an “scientific” experiment …
2) The Aluminum foil is an ‘object’ just as the light bulb is an ‘object’.
my point is that your definition of object is pretty nebulous … which you verify here … the light bulb is hardly similar to the foil … several different materials for one …
3) Please elucidate on the several mechanisms you believe would slow the dissipation of the energy
the lack of air movement … I’ll think about some others … I’m sure you could “eludicate” some too … did you really need to ask that ? eludicate … cool word …
4) I think you’re mixing energy absorption and temperature, they aren’t the same thing.
no I’m not, I’m asking how the absorption of the radiation from the cooler object into the warmer object was measured … just because radiation is directed at an object doesn’t prove it was absorbed by said object … you are assuming that higher temp = energy absorption not me …
In general I find the discussions here interesting …. this thread appears to have broken that pattern …

Curt
May 28, 2013 2:55 pm

Bryan, you say, “Curt’s summary implies that the presence of a colder object will ALWAYS increase the temperature of a warmer one. It contradicts common sense. Try putting a large block of ice in your living room.”
You misinterpret my argument. The key follow-on question is, “Compared to what?” If I were worried about keeping warm, would I want a block of ice near me compared to objects at ambient temperatures of 20+C? No! But would I want a block of ice near me rather than having a direct path in that direction to a clear winter night sky? Absolutely! It’s why the Inuit build igloos. More to the point, it’s why I always carry an aluminized mylar “space blanket” when I am hiking in cold weather or skiing. If I get stuck in the cold and have to use it, that blanket will be the temperature of a block of ice, but my body temperature would be higher with it than without it.
I’ll turn the argument around on you. Take a thermostatically controlled room held at an air temperature, say, of 23C. I’m brought into the room suffering from hypothermia, which means that my body is generating as much heat as it can and my body temperature is still low. The Slayers would claim that it would make no difference to me whether the room was filled with large blocks of stone at 23C or large blocks of ice at 0C, because both are colder than my body temperature and could not possibly affect my higher body temperature. I say that is wrong, and I think that my experiment demonstrates it.
The foil shells, even though colder than the bulb, do increase the bulb’s temperature by radiating back to it compared to the control case where there is less radiation coming back to the bulb (from the room walls). Slayers have often claimed this was not possible.

Anne Ominous
May 28, 2013 2:56 pm

This experiment, as described it, is completely laughable.
Anthony, if you are going to pretend to do a scientific experiment, you should know that you have to isolate your variables, which you utterly failed to do. For just ONE of the many flaws, nothing you are attempting to show here means anything in the presence of the conductive and convective conditions you describe.
Hell, even your initial assumptions are fatally flawed. While glass may be transparent to the near-infrared initially emitted by the bulb, anything warmed by your bulb by conduction or convection will be radiating strongly in the MID-infrared, which is almost completely reflected by normal glass, invalidating what you are trying to measure. THE GLASS ITSELF will be radiating strongly in the mid-infrared. While some of your observations about plain reflection may have some validity, they are going to be completely drowned out by the other effects.
If this is seriously intended to demonstrate your point, you fell flat on your face.
All other arguments aside, don’t even bother if you don’t have a vacuum chamber. You won’t prove anything.

Rosco
May 28, 2013 2:57 pm

I think people need to think about the real physics – the glass of the bulb is not the source of the energy which is designed to escape in the formof visible light.
The glass is never hot enough to emit visible light on its own and that simple fact plus the Inverse square law invalidate much of what is claimed.

Curt
May 28, 2013 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.
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.

May 28, 2013 3:06 pm

How much wasted time, experimental tools, and brain-electricity etc. for determining the obvious.
Just a question for all of you.
Why do you think that the inside of a thermos flask is painted mirror-silver?
Let see who knows the answer.

Jim Clarke
May 28, 2013 3:11 pm

I have not read all the posts, but it seems to me that in the above experiment and in the GHE we are not seeing colder objects give off radiation that heats up hotter objects. We are simply seeing colder objects slow the dissipation of the energy from the hotter objects to the outside environment, causing the hotter objects to get hotter still. There is no violation in the laws of thermodynamics in such a system.
The cooler objects are not heating up the hotter objects. That would be impossible. If we turn off the energy supply to the hotter objects, the whole system begins to cool immediately. The cooler objects do not keep the hotter objects hot (or make them hotter) by radiating heat back to them.
This isn’t about one object heating another with E/M radiation. This is about one object altering the energy dissipation of another object by interfering with the radiation’s journey to the surrounding environment.

May 28, 2013 3:16 pm

thermal equilibrium is reached when the system is dissipating 35 watts to the room as well

Soooo, the bulb does not become brighter after all?

Rosco
May 28, 2013 3:17 pm

It really is simple.
the energy escaping originally NEVER came from the glass of the bulb – it came from the filament.
A portion causes a thermal response in the glass both by radiant energy and diffusion/convection the rest escapes.
Stop that escape and cause the visible light to heat some object and of course the temperature will rise.
This easily explains a significant proportion of what is happening here.
There is no way to determine what contribution “reflected” thermal energy originating in the glass is contributing.
1. It is watts per square metre that is related to temperature.
2. All of the energy is originating in a point source – the filament.
3. The Inverse square law insists that intensity in watts per square metre DECREASES as you move away from that source.
4. The bulb is designed to be a source of light and allow it to escape – of course trapping the light by absorption will cause an increase it temperature !
5. Light is not the same as “heat” – light can cause “heat” by being absorbed.
5. As there is no measure of the power flux before and after there is no valid basis for any conclusions as drawn.
I don’t believe these experiments actually demonstrate what they claim when a true appraisal is applied.
A better experiment is to heat an object with external energy sources.
1. Use one at a certain power flux and record the temperature
2. Double its power and record the temperature.
3. At the original power flux add another equal and record the temperature both combine to generate.
That way you have real data to work with !

thefordprefect
May 28, 2013 3:21 pm

Jim Clarke says: May 28, 2013 at 3:11 pm
The cooler objects are not heating up the hotter objects. That would be impossible. If we turn off the energy supply to the hotter objects, the whole system begins to cool immediately. The cooler objects do not keep the hotter objects hot (or make them hotter) by radiating heat back to them.
————————-
I have just completed an experiment here The results show that an object at 65°C will cause a rise in temperature of an object at 75°C. (0.3°C). I cannot explain this other than radiation from cool plate increases the temperature of the hot plate.
Feel free to criticise the experiment or make suggestions to improve the test.
Results here:
http://climateandstuff.blogspot.co.uk/2013/05/back-radiation-early-results-no-fan.html
setup here:
http://climateandstuff.blogspot.co.uk/2013/05/proposed-back-radiation-test-setup.html

Gary Hladik
May 28, 2013 3:27 pm

chris y says (May 28, 2013 at 11:29 am): “GE R&D center developed an incandescent bulb back in the late 1980′s (and GE lighting commercialized it in the 1990′s) that placed a spherical clear glass shell around the filament. The shell was coated with a multilayer (anywhere from 15 – 30 separate layers) optical filter that reflects mid infrared back onto the filament, while allowing visible light to pass through.”
I remember chris y gave this example back in 2009, in the “Steel Greenhouse” thread.
http://wattsupwiththat.com/2009/11/17/the-steel-greenhouse/#comment-227006
It seems the Slayers/trolls at WUWT have been “in the dark” about this light for over three years. 🙂

Richard Dupuis
May 28, 2013 3:30 pm

Do the same experiment in a vacuum.

Kasuha
May 28, 2013 3:34 pm

As far as I understand the matter, the major problem with thinking of people around the Principia web is that they think that when you have an object (e.g. a volume of gas) at thermal equilibrium at certain temperature, all its particles will have exactly the same energy. You can easily come to all their wrong conclusions if you take that as a fact. The only problem is that it is not true.

F. Ross
May 28, 2013 3:35 pm

P Gosselin says:
May 28, 2013 at 11:28 am
“…
“Radiation from a cooler object”? That statement ought to dismiss anything else written in the essay. A cooler object doesn’t radiate energy to a warmer one.

[+emphasis]
Does the “cooler” object have some sort of, say, ESP or precognition, to “know” that a nearby body is warmer than it, and therefore “withhold” its radiation in that direction. If you suddenly remove the “warmer” body, does the “cooler” body breathe a sigh of relief and start radiating again — assuming that it is above absolute zero?

AndyG55
May 28, 2013 3:35 pm

“Bulb covered by inner reflective foil shell inside glass container 177C 68C (foil)
Bulb covered by inner reflective foil shell alone 176C 50C”
It seems that the bulb is being heated not by energy “emmitted” by the foil, but “reflected”.
You have effectively proven that a light bulb dissipates its heat mostly by radiative means. Well done.

Gary Hladik
May 28, 2013 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. 🙂

May 28, 2013 3:44 pm

rgb,
you slightly overstate your case here:
So LWIR heads up, encounters gas molecules, is absorbed (or not, depending on wavelength and chance) and if absorbed, is radiated back down at the surface just under 50% of the time.”
I don’t know the numbers, but quite often the “extra” energy that a photon gives to a CO2 molecule is transferred by collision to another nearby molecule. The mean free path in the atmosphere is of the order of 10^-7 m, and the particles are move with speeds on the order of 10^2 m/s, so they collide ~ every 10^-9 second. Depending on how long the CO2 molecule can stay in its excited state, there is a very good chance that a collision will occur before an emission.
Of course, *other* collisions will be giving the CO2 molecules extra energy from time to time, so eventually they should emit ~ 1/2 of the photons up and ~ 1/2 of the photons down.

Myrrh
May 28, 2013 3:48 pm

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.

May 28, 2013 3:51 pm

to: Curt says: May 28, 2013 at 3:05 pm
Then I wasted all the money I spent on that high dollar 3M Sun Control Film that only blocks 15% of the visible light and 97% of the IR? (ratings from 3M) The Sample they gave me seems near invisible. If So, why is my “sun” room (windows on 3 sides) 20-30 degrees cooler in the summer? Also wouldn’t this make all Solar Film a scam?

May 28, 2013 3:58 pm

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! 😀

mikerossander
May 28, 2013 4:04 pm

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.

May 28, 2013 4:14 pm

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.

May 28, 2013 4:16 pm

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?

Myrrh
May 28, 2013 4:19 pm

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.

vivendi
May 28, 2013 4:22 pm

“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.

Carrick
May 28, 2013 4:27 pm

usurbrain:

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.

No, that’s a quite small effect, but possibly economically significant still.
The main effect is because it prevents free-air convective heat loss.

mikerossander
May 28, 2013 4:36 pm

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.

graphicconception
May 28, 2013 4:37 pm

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?

Carrick
May 28, 2013 4:37 pm

Max:

Soooo, the bulb does not become brighter after all?

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:

The bulb was heated by the power supplied to it, not back radiation, obviously.

Gee last I looked back-radiation referred to radiation emitted in response to an external stimulus. So no *** Sherlock.

It started out dissipating 35 W to the room, it wound up dissipating 35 W to the room.

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.

Curt
May 28, 2013 4:39 pm

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?

May 28, 2013 4:41 pm

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.

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.

Carrick
May 28, 2013 4:48 pm

Max, you are mistaken.
There is an effect from blocking IR, though it’s small under most nocturnal conditions.
Google “IR blocking greenhouse plastic”.

May 28, 2013 4:48 pm

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.

joeldshore
May 28, 2013 4:51 pm

Bryan says:

See what you and RJB think of this situation
A hollow sphere at a higher temperature and black body emissivity has a colder object placed at its centre.
Lets say that there is a vacuum between the cold and hotter sphere to eliminate conduction and convection.
Now the cold object is a new source of radiation.
Its radiation falls on the spheres inner surface and is absorbed.
Yet the sphere will not increase in temperature.
Quite the opposite in fact.

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:

“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.”

Joel the above quote is from the write up of the experiment. This does not square with your explanation.

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.

May 28, 2013 4:52 pm

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.

joeldshore
May 28, 2013 4:54 pm

Max says:

The bulb was heated by the power supplied to it, not back radiation, obviously.

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.

Carrick
May 28, 2013 5:01 pm

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:

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.

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.

Carrick
May 28, 2013 5:03 pm

ururbrain:

BINGO – that is why the box whether it be glass, tinfoil, black, etc. makes the bulb hotter.

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.

OldWeirdHarold
May 28, 2013 5:20 pm

Moon? Mann? I think somebody’s trying to do a funny.

Olaf Koenders
May 28, 2013 5:24 pm

“Joseph E Postma says: “It actually makes worse mistakes. Keep an eye at PSI”

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.

May 28, 2013 5:33 pm

Max, you are mistaken.
There is an effect from blocking IR, though it’s small under most nocturnal conditions.
Google “IR blocking greenhouse plastic”.

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!

Gary Hladik
May 28, 2013 5:36 pm

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. 🙂

JeffC
May 28, 2013 5:38 pm

bring a 6 inch diameter pot of water to 100 degrees in a room with ambient temp of 60 … turn off the heat and suspend a very thin (12″x12″) sheet of glass 12 inches above the pot and measure how fast the water returns to 60 degrees …
now repeat heating the same pot to 100 and suspend a 12″x12″x12″ block of aluminum 12 inches over the pot …
so now we should have the same convective cooling going on with the addition of the block radiating energy back into the pot of water …
based on the experiment described in the article I would say it would now cool slower since the block must be radiating energy back into and thus “re-heating” the water …
the experiment in the article has too many uncontrolled variables for my taste … convection being the primary one …

Greg House
May 28, 2013 5:42 pm

[snip]

commieBob
May 28, 2013 5:53 pm

Myrrh says:
May 28, 2013 at 3:48 pm
… Since visible light cannot be heating either t-shirt, just what do you think a first grader can understand here?

Go away Myrrh and come back when you can explain why Maxwell’s equations do not apply to visible light just as they do for infra-red and radio waves. (hint: you won’t be able to)
https://en.wikipedia.org/wiki/Maxwell%27s_equations

May 28, 2013 5:59 pm

I think what he was saying is that there is a difference between vibrational, rotational, and other modes of absorption and emission.
Light of different wavelengths produces different effects when it interacts with matter.
This is why visible light is transmitted through air with little to no distortion while other wavelengths are partially or completely blocked.
Similarly this is why a microwave oven is able to warm your food, while the same microwaves in a different use are able to transfer information through your walls from your wifi router to your computer.
Photons at visible wavelengths do not produce the same sort of rotational or vibrational heating as microwave and infrared do, respectively, visible light involves electrons hopping between energy levels.
Physics, yo, it works.

Olaf Koenders
May 28, 2013 6:09 pm

“Baa Humbug says:
May 28, 2013 at 12:55 pm
I have some questions that better minds than mine can hopefully answer.
In which case, whether we place the foil on the outside of the glass or on the inside of the glass should make no difference to the result (radiative transfer restricted in both cases) but there is a substantial difference i.e. bulb T = 138DegC with foil on the outside and 176DegC with foil on the inside. How do you explain the difference?”

Hello Baa. The difference in this case is that the foil on the inside is a smaller box,giving less scattering to clear space and increasing the radiative power to the bulb surface. It’s like looking at the mirror from the bulb’s perspective and moving the mirror closer, the bulb in the mirror gets bigger = more power directly reflected to the bulb surface.

“Furthermore, the instance of foil cover alone should return the same result as foil on the inside or outside of the glass, yet it returns 148DegC. Again, how do you explain the difference.”

The difference would be through conductionof the heat in the foil to the glass, therefore the glass acting as a large heat sink to the outside world.
But once again, there’s the difference of the foil on the inside being the smaller box and greater reflector, whereas on the outside it’s larger and the IR can be scattered more easily within that space. Note also that the glass on the inside will absorb some of the IR as well, transferring that heat to the foil on the outside through conduction.
The different permutations of this experiment affect all 3 things differently – reflectivity, convection and conduction – which is why they were all taken into account.

clt
May 28, 2013 6:14 pm

Max:

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!

This is a straw man argument since it a addresses and refutes a point that was never made.
While the predominant effect is blocking convective heat loss, the point was the other effect exists too.

I think what he was saying is that there is a difference between vibrational, rotational, and other modes of absorption and emission.

Perhaps, but it doesn’t matter, since the claim above was visible light doesn’t heat objections. You still get absorption of photons, regardless of the mechanism, and a heating up of the matter that did the absorption.
As you said, physics works.

commieBob
May 28, 2013 6:15 pm

OK you folks who think that visible light doesn’t heat things. All light that strikes an object, and doesn’t reflect, is absorbed and results in heat. Energy is conserved.

Curt
May 28, 2013 6:32 pm

JeffC, you say, “the experiment in the article has too many uncontrolled variables for my taste … convection being the primary one …”
One of the key aspects of the experimental design was to control for convection. That is why the cases with the metal shells were compared to that of the glass shell, and not to the bulb in open air. In all cases, direct convection to the room was suppressed.
I stated that I could not come up with any reason the conductive/convective losses would be less with the metal shells than with the glass shells, and several why they should be greater. (So the higher temperatures with the metal shells must be from radiative effects.) No one in the over 200 comments so far has even suggested a reason why the metal shells might have lower conductive/convective losses.

Gary Hladik
May 28, 2013 6:41 pm

JeffC says (May 28, 2013 at 5:38 pm): “the experiment in the article has too many uncontrolled variables for my taste … convection being the primary one …”
What experiment did you read? The one I read explicitly controlled for convection. Any “leftover” convection would only have diminished the quite dramatic “back radiation” effect.

May 28, 2013 6:49 pm

This is a straw man argument since it a addresses and refutes a point that was never made.

Uh, no, I’m pretty sure the point I was responding to was about IR blocking being important in greenhouses when it is at best a secondary or tertiary effect.
You can build a greenhouse just fine with glass that doesn’t block IR, you can not build a greenhouse without stopping convection, the point I responded to was implying that the IR blocking of the glass was worth considering, it would be with the variation I suggested, where there would be a sheet of glass between the bulb and the mirror, without otherwise stopping convection.

Gary Hladik
May 28, 2013 7:08 pm

graphicconception says (May 28, 2013 at 4:37 pm): “My concerns include:
o) the use of glass in two places while still expecting IR to pass;”
Let your heart not be troubled. If IR passes through, fine. If it’s reflected, fine. If it’s absorbed, the glass temp increases and it loses more heat via convection and, yes, IR.
“o) the lack of consideration for any of the surfaces to absorb radiation and become warmer as a result resulting in conduction/convection;”
All of the surfaces become warmer when added to the experiment, and all of them lose heat via convection and radiation. The “inside” foil/glass/plate surfaces radiate/reflect radiation toward the bulb, toward the bench, and toward each other. Can you explain your concern in more detail?
“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?”
Since the infrared emissivities of, for example, gaseous nitrogen and oxygen are orders of magnitude less than, for example, water vapor and carbon dioxide gas, yes, infrared radiation from these gases can be ignored, despite their greater abundance.

Curt
May 28, 2013 7:16 pm

usurbrain, you say, “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?” What did I say that would even imply that? Do you think that the absorption in the visible spectrum does not affect the heating of the object?
Actually, you have a good test in your sun room since you have blocked out the IR part of the sun’s spectrum. Under similar solar conditions, cover everything in white fabric, then in black fabric. See if you can tell the difference.

wayne
May 28, 2013 7:21 pm

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.

Stated as it is that is blatantly incorrect in science! I’m ashamed of your lose terms with word Anthony and this is what keeps this misunderstanding to continue ad infinitum. If you would have included the words that this only occurs if and only if the warmer mass is an energy source, creating new energy into the system you would have been correct. Surely you are not doing this to generate traffic and comments for your site? Taken point blank you are being as bad as the ‘slayers’!
[Reply: You might want to note that Anthony is not the author of this test or article – mod]

Curt
May 28, 2013 7:31 pm

Max, you say, “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! :D”
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. In all cases with a 35W input from the electrical system and 35W losses to the room (with these numbers actually being slightly smaller when the temperatures are higher). That is the key point of the results.
And this is what the Slayers say is impossible. Postma says the idea that putting a highly absorptive/emissive shell around a body with an internal power source could increase the temperature of that body “is so intellectually offensive, so mentally incompetent, that it beggars the imagination.” (That’s a direct quote.) And yet that is what the black-anodized aluminum shell in the experiment does.

Greg Mansion
May 28, 2013 7:39 pm

Curt says May 28, 2013 at 11:49 am “wikeroy, you say: “I did the basic course in thermodynamics 30 years ago. I went to the [attic] and looked through the books. Couldnt find the word [backradiation]. Probably some new definition.”
In engineering heat transfer texts, it’s usually referred to with a term like “radiative exchange” between bodies. Right near the beginning of the chapter on radiative heat transfer in any of these texts, you will see an equation for the radiative heat transfer between two bodies with some constants multiplying the term (Th^4 – Tc^4), where Th is the absolute temperature of the hotter body and Tc^4 is the temperature of the colder body. The Tc^4 term is the “back radiation”, whatever you choose to call it.”

=========================================================
This is not true. The Tc^4 term refers to the temperature Tc of the colder body, it is not the “back radiation” and has nothing to do with back radiation. “Th” is the temperature of the hotter body. The term (Th^4 – Tc^4) refers to the heat transfer from the hot to the cold body and does not imply any “back radiation”. There is no room for any “exchange” in this term, only transfer.
Wikeroy has hit the nail on the head.

markx
May 28, 2013 8:07 pm

Nice bit of work. Clearly shows cooler object adding energy to a warmer one. (and, I’m pretty sure the warmer object does not know or care whether that energy is reflected or radiated).
Michael Moon says: May 28, 2013 at 8:52 am
The Temperature of the Filament is the only temperature that matters here. Did the Filament warm from back-radiation? Did heat actually “flow uphill?”
Geez Michael, I for one will bet you a dollar the filament will be at least a bit warmer if the bulb is so much hotter.

chris y
May 28, 2013 8:09 pm

Gary Hladik-
“I remember chris y gave this example back in 2009, in the “Steel Greenhouse” thread.”
How did you remember that? I didn’t even remember that! I do however remember Willis’ excellent post on the steel greenhouse.
“It seems the Slayers/trolls at WUWT have been “in the dark” about this light for over three years. :-)”
🙂

May 28, 2013 8:15 pm

Can I get class credits for digesting the education offered by rgbatduke?

itronix
May 28, 2013 8:24 pm

Notes for critics of this experiment.
1. Nobody disputes that convection is significant in moving heat away from the bulb.
2. With the glass cube around the bulb, temperature increases until energy loss through convection, conduction and radiation is equal to the electrical energy input. This temperature will remain steady until something else changes.
3. Foil is wrapped around the glass cube and the temperature increases until energy loss through convection, conduction and radiation is equal to the electrical energy input. This temperature will remain steady until something else changes.
4. The glass cube is removed and the foil cube put back around the bulb. Temperature increases until energy loss through convection, conduction and radiation is equal to the electrical energy input. This temperature will remain steady until something else changes.
Everyone understands that the glass or foil cube slows convection so the bulb heats. Can any critic of the experiment explain the difference between glass only /glass and foil /foil only? Do you think the foil was a better insulator (of conduction or convection) than the glass? The difference is radiation effects.

Rob
May 28, 2013 8:44 pm

This actually proves very little , the filament of a light bulb can reach temperatures in accesses 2500c which is over 4500 in Fahrenheit .

May 28, 2013 8:44 pm

Radiation is one of the three methods of heat transmission. Radioactivity is different phenomenon. Your expt. does not prove what you claim. Heat is always transmitted from body with higher temperature to the body of lower temperature, like water from higher to lower height to level up the height.

Darren Potter
May 28, 2013 8:52 pm

rgbatduke says: “The conductive insulation capability of aluminum foil is zilch — try using a piece as a potholder if you doubt that.”
Now that is funny, I don’t care who you are!
Well-done (pardon pun) there rgbatduke.

May 28, 2013 9:08 pm

From the article:

“Second, the foil, which is a small fraction of a millimeter thick,”

It looks as though there is more than one layer of Al foil especially at the top where the side panels are folded on one another creating many layers. With all of those alternating layers of foil and air, it creates a better blanket than the single layer of glass or blackboard does.
Please can you reveal to us Curt Wilson how many layers of foil there are? And where are they placed — it looks like there are more layers on top where you say the heat mostly accumulates. It’s a critical issue for this experiment.
It’s just automatically assumed by the author that the difference between glass/board and foil (a difference which is rather small) is due to the reflection issue rather than the foil being a more effective blanket to convection and conduction.

Ray Boorman
May 28, 2013 10:24 pm

Kurt, might not a better experiment be to have 2 light bulbs, say 100W & 25W. Turn on the 100W & find its equilibrium. Then position the 25W nearby, turn it on & measure the changes in the 100W bulb. I would also suggest that you should position the bulbs at least a few feet away from any other surface. This method would eliminate the greenhouse effect you cause by enclosing the bulb as you did in your experiment, & also minimise any other sources of reflected energy.
[Reply: actually, Anthony tells me he has an experiment already in the works like that – mod]

Editor
May 28, 2013 10:27 pm

Myrrh says:
May 28, 2013 at 4:19 pm

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

Sigh. Back in the debate about {snip – I don’t want to drag that up again}
someone pointed out a NASA (I think) page that supported his (wrong) position.
I wrote the author of the page, and he agreed with me and got the page fixed.
In this case, NASA has already fixed it – the new page at http://missionscience.nasa.gov/ems/07_infraredwaves.html says things like “The region from 8 to 15 microns (µm) is referred to by Earth scientists as thermal infrared since these wavelengths are best for studying the longwave thermal energy radiating from our planet.” And “A typical television remote control uses infrared energy at a wavelength around 940 nanometers. While you cannot “see” the light emitting from a remote, some digital and cell phone cameras are sensitive to that wavelength of radiation.” (Gone is the misleading “Shorter, near infrared waves are not hot at all – in fact you cannot even feel them.”)
BTW, the reason an IR remote control doesn’t feel hot is because it emits something like 30 or 40 milliwatts of power. Not enough to feel on your skin. A flashlight with an incandescent bulb doesn’t get hot because it puts out a fraction of a watt.
On that T-shirt description – what happens when a photon is absorbed by a black T-shirt? Photons have energy, when they are absorbed they wind up leaving some heat which can show up as a warmer T-shirt. It’s really that simple. Just follow the photons. It doesn’t matter if they’re from a visible light laser, a LED, or hot tungsten – if there are photons a T-shirt can absorb, it will, and it will get warmer. Unless, of course, it’s radiating or conducting heat away faster than it’s coming in….

May 28, 2013 10:29 pm

In radiative terms, surely the smaller inner foil on its own, and the larger outer foil on its own, are very similar. Yet in the table there is a difference of 28C.
Bulb covered by inner reflective foil shell alone: 176C
Bulb covered by outer reflective foil shell alone: 148C
If these temperature differences are due to radiative effects, not convective/conductive ones, how is this difference to be explained? Obviously the smaller size of the inner shell has a significant impact on the temperature, probably because it restricts convection more due to the smaller volume.

May 28, 2013 11:14 pm
Curt
May 28, 2013 11:27 pm

Paul Clark, you say, “If these temperature differences are due to radiative effects, not convective/conductive ones, how is this difference to be explained? Obviously the smaller size of the inner shell has a significant impact on the temperature, probably because it restricts convection more due to the smaller volume.”
I agree with you on this one. In my analysis, I stated, “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. ”
You ask a very good question about how many layers of foil there were. Over most of the surface, just one, but there were a few areas where there were two or even three – I did not spend much time on the construction. If I get the chance, I will build one more carefully with minimal overlap. Thanks for pointing that out.

Somebody
May 28, 2013 11:34 pm

[snip – this comment is too stupid to publish, about what you’d expect from “somebody” – mod]

May 28, 2013 11:55 pm

Convection is not free in this experiment. Because of this the air between the bulb and the glass jar acts as an insulator.

May 29, 2013 12:33 am

Charles Gerard Nelson says:
May 28, 2013 at 11:55 pm
Convection is not free in this experiment. Because of this the air between the bulb and the glass jar acts as an insulator
Indeed, but in all cases the convection was blocked and the insulation was practically identical. So the different outcomes of the experiments are not caused by differences in convection…

Konrad
May 29, 2013 12:37 am

A far simpler experiment showing essentially the same thing simply requires a 100mm square of aluminium foil. Pick a cold still night and go outside. Hold the foil in a vertical orientation close to one cheek (not so close as to restrict convection) with the shiny side toward your face. One cheek will feel warmer. IR emitted from your skin is being reflected back to your skin, slowing its cooling rate. Simple.
AGW however remains a physical impossibility. There is no net radiative GHE on earth.
The reason the radiative GHE hypothesis fails is not due to major problems with radiative physics, but rather with fluid dynamics and gas conduction.
There is one minor problem with the radiative physics. The interface between surface of water and the atmosphere is a special case. Incident LWIR has no significant effect on the cooling rate of liquid water that is free to evaporatively cool. Given this only effects 71% of the earth’s surface, not what you might call a huge problem. It just means that surface Tav under a non radiative atmosphere would be cooler but not 33C cooler.
The critical error in the radiative GHE hypothesis is in fluid dynamics and the physics of gas conduction. Radiative gases are critical to convective circulation below the tropopause. Without these gases, gas conductively heated at the surface could still convect to altitude, however it would no longer be able to lose energy and descend. In such an atmosphere the lapse rate would disappear and the temperature profile would trend to isothermal. The temperature of such a non radiative atmosphere would be set conductively by surface Tmax, not surface Tav. A non radiative atmosphere would be far hotter than an atmosphere with radiative gases and convective circulation.
The net effect of radiative gases in our atmosphere is cooling. The cooling effect of changing CO2 concentrations from 300 to 400ppm would be immeasurably small and indistinguishable from 0C.

Reply to  Konrad
May 29, 2013 1:35 am

Joel Shore and Curt
Climate ‘Science’ publications are unique in the way that the word ‘warm’ or ‘to warm’ or ‘warmer’ is used.
It is routinely used to describe a situation where an object is actually cooling.
Can you find anywhere else in the rich resource of English literature where ‘warm’ is used in that way?
That’s a rhetorical question because you cannot!
Three objects or entities are at different temperatures
A is at temperature 350K
B is at temperature 300K
C is at temperature 250K
All agree that A will lose heat to B and C
However Climate ‘Science’ will explain that if C is replaced with B then B has ‘warmed’ A.
Is there a reason for torturing language in this way?
Roy Spencer admits in his blog that he does this to ‘wind up’ the ‘slayers’.
Now John O’ Sullivan is a real pain in the ###, but I suggest that reduced heat loss is not best described by ‘warming’ or ‘heating’ the true heat source.

May 29, 2013 12:47 am

Anthony –
Coming into this a bit late, so you may not read this, but I hope so.
“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."
There is fourth factor. That low emissivity of the foil that you mentioned earlier is important. That heat that isn't emitted is being contained! Your experiment is very similar to my own real world experience when working in R&D with heated molds for hot runner co-injection blow molding. We had 425°F hot cores (male part of the molds) that were highly polished, and attempts to measure their temperatures with IR sensors were unsuccessful, just as you experienced. Our cores, however, were much thicker, made of solid bars of a stainless steel high-strength alloy.
The really weird thing, for one’s brain was this:
We could not feel any of the 425°F heat coming off the surface of the cores, even when we got well within a millimeter. It was even something we would show off to people. The temps were essentially shop-floor ambient, even that close to the surface. It was amazing to play with, to see how close we could get and not touch it – and no matter HOW close we got, we could feel no heat. But when we actually TOUCHED them, of course, we got burned and burned pretty badly. *** There was essentially no conductivity or convectivity.
Why? We were not sure, but we concluded that the highly polished surface of the cores were containing the heat, as if it were being reflected back inside. hy did we conclude that?Because other surfaces nearby with less polish emitted/radiated and convected just fine.
But the emissivity was not constrained to only radiating the heat. The heat transfer through radiation, through convection, and through conduction to the air literally was stopped, for all intents and purposes. Whatever heat was getting out was very, very little, below our ability to feel it no matter how close we got, as long as we didn’t touch it.
So, your statement, “The foil shell also inhibits direct conductive/convective losses to the room atmosphere, but it will not inhibit them to any greater extent.,” is not correct, based on my own extensive experience with polished/shiny metal surfaces. In my experience that shiny metallic surface DOES inhibit conduction and convection, as well as radiation/emissivity.
Therefore the foil
does contain more heat within the enclosure than the glass, because convection and conduction don’t happen (or is vastly lower). I don’t agree with your conclusion at all.
Try the experiment again and move a thermometer within close proximity to the foil and compare to doing it with the glass. You should see much less temperature variations (and cooler) with the foil than with the glass.
Steve Garcia
*** This was in the late 1980s, just before CFCs were banned, including freon. When we burned ourselves on the cores – because dammit we couldn’t even tell when we were getting too close, we couldn’t feel the heat of the cores – we had spray cans of freon that we would use. It was absolutely the best thing ever for treating burns. One spray and the pain would go away, plus the burns never blistered, though the outer skin separated from the lower dermis. No other treatment was ever necessary, other than a 1-2 second spray of freon. When they banned freon I nearly went ballistic. No one even considered this application of freon when they banned it. And to boot, I completely disagreed with the conclusion that CFCs were destroying the ozone hole above the South Pole. The logic was flawed right from the start. There was a variety of reasons that I thought that, but I won’t go into now.

richard verney
May 29, 2013 2:06 am

I have not read all the comments so apologies if I say something that has already been said, but I am tempted to agree with Michael Moon when he says ( May 28, 2013 at 8:52 am) ” The Temperature of the Filament is the only temperature that matters here. Did the Filament warm from back-radiation?”
As I see matters, the flaw in this experiment is that it is not really addressing the crux of Joseph’s claim.
What we have here is the behavoir of glass bulb which is intially at room temperature but which then heats up when the electric bulb is switched on. The glass bulb is being heated by photons which come from a source at about 4000K. So it is no surprise that the glass bulb heats up tp 368K. When convection is hindered the heat loss from the glass bulb is restricted so it then obtains a new equalibrium temperature of 378K
Now instead of bringing a second object which has its own internal energy source and which is radiating photons at say 300K close to the proximity of the glass bulb to see whether photons radiating from an object which has a temperature of about 300K can warm an object which is at 368K, alternatively at 378K, instead the experimentor reflects back photons at about 4000K (ie, the same temperature as the filament source). I would suggest that one should get an object which only radiates and have that object at say 1 degC above the ambient room temperature and then pladce taht object near to the elctric bulb and see what effect it has on the temperature of the glass bulb. This object can be placed say 9 inches away, then 8 inches away, then 7 inches away etc until it is only 1 inch away.
Surely we all know that one can collect and focuse high energy photons from the sun. One can do this with an appropriately curved mirror. It can heat an object placed at its focal point. If we add another similar mirror, we get more power. We see this in some of the solar energy arrays which may have hundreds of mirrors. Each one is collecting some of the power and directing the power collected to a focal point such that the energy at the focal point is increased by the more mirrors that are collecting and focusing the power.
This experiment appears to be doing little more than that.
What we need to see is the effect of photons eminating from a completely different source which source is cooler if one is to refute the claim by Joseph. Of course, Joseph needs to explain what is wrong with the experiment, and it would be useful if he would conduct an experiment which establishes what he claims to be the case.

richard
May 29, 2013 2:11 am

Postma says the idea that putting a highly absorptive/emissive shell around a body with an internal power source could increase the temperature of that body.
Went to bed last night, freezing , wrapped the duvet around me the internal heat source, didn’t warm me at all so had to put my undersheet electric blanket on.

Paul Aubrin
May 29, 2013 2:20 am

“Michael Tremblay says: May 28, 2013 at 10:44 am
Your conclusion is fundamentally flawed. The statement: “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.” is false. What you have is solid experimental evidence that reducing the amount of energy (heat) leaving a system while maintaining the amount of energy (heat) entering the system will result in a temperature rise in the system. The temperature rise is due to the increasing amount of heat in the system and this heat is coming from the electric current being supplied to the bulb, not from the back radiation.”
This seems one of the very few rational comments here: “The temperature of a radiative body is modified by its environment. This is exactly what is violently rejected by any warmist who will claim that the increase of temperature can be nothing but the result of an increase of the power received by the bulb. In the case of the experiment above, warmists pretend that the light bulb in the blackbody enclosure must receive an extra 20W of power from the black enclosure. And from where did this power come from? From the bulb itself, they say.
In the real world, those 20W of extra “back-radiation” power are nowhere to be found.

Steve C
May 29, 2013 2:22 am

As usual, this topic has generated far more heat than light (pun fully intended). As far as I can see, it will continue to do so forever until Slayer and Antislayer sit down together and design an experiment satisfactory to both sides – with complete agreement on what to measure, the correct interpretation of those measurements, and so on. Until then, my position remains that neither side has actually shown anything – certainly not in terms recognisable / acceptable to the other side – and that meanwhile perhaps it’s time I got round to re-reading my aging copy of “Engineering Thermodynamics, S.I. Units” by Rogers and Mayhew (Longmans, 2nd ed, 1969 reprint).

richard verney
May 29, 2013 2:29 am

Ray Boorman says:
May 28, 2013 at 10:24 pm
Kurt, might not a better experiment be to have 2 light bulbs, say 100W & 25W. Turn on the 100W & find its equilibrium. Then position the 25W nearby, turn it on & measure the changes in the 100W bulb. I would also suggest that you should position the bulbs at least a few feet away from any other surface. This method would eliminate the greenhouse effect you cause by enclosing the bulb as you did in your experiment, & also minimise any other sources of reflected energy.
[Reply: actually, Anthony tells me he has an experiment already in the works like that – mod]
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I don’t think that works..
I consider that you require not a second light bulb (whose filiament will be running far hotter than the glass bulb of the primary bulb), but rather you need some form of rock, perhaps made of soapstone, or the like, which has high heat retention and is a good radiator.
You keep your 40 watt light bulb as is. You put the rock (soapstone) in an oven say at 30 to 50degC and heat it up until it is about 40degC, so that it is warmer than ambient room air temperature but not by so much. This rock will then radiate energy.
You switch on your 40watt light bulb as before and let it reach its equalibrium. You then introduce the rock/soapstone close (but not too close) to the 40watt light bulb so that the glass bulb of 40 light bulb receives not only the energy of the photons from its filiament but it additionally receives the photons from the rock/soapstone and then see what effect this has on the temperature of the glass bulb.

richard
May 29, 2013 2:32 am

Up at Castle Howard they have a great heating system. Pipes run through the lake in the grounds back to the house. Even though the lake is much colder than the house there is heat to be extracted, to do this you need a Geothermal or ground source heat pump.

richard
May 29, 2013 2:33 am

if I ran just cold pipes from the lake through the house I imagine it would take away the heat.

richard verney
May 29, 2013 2:55 am

MODERATORS
I note that you are moderating my latest comment timed at 2:29hrs. I have slightly reviesed my comment as set out below. Please post the revised version once approved. Thanks.
Ray Boorman says:
May 28, 2013 at 10:24 pm
Kurt, might not a better experiment be to have 2 light bulbs, say 100W & 25W. Turn on the 100W & find its equilibrium. Then position the 25W nearby, turn it on & measure the changes in the 100W bulb. I would also suggest that you should position the bulbs at least a few feet away from any other surface. This method would eliminate the greenhouse effect you cause by enclosing the bulb as you did in your experiment, & also minimise any other sources of reflected energy.
[Reply: actually, Anthony tells me he has an experiment already in the works like that – mod]
///////////////////////////////////////////////
I don’t think that works. The key issue here is not the effect of back-radiation but rather the effect of radiation. Back means nothing more than the direction.
I consider that you require not a second light bulb (whose filament will be running far hotter than the glass bulb of the primary bulb and will therefore be supplying high energy photons to the glass of the primary bulb), but rather you need some form of rock, perhaps made of soapstone, or the like, which has high heat retention and is a good radiator. What we need to see is the effect of low energy radiation.
You keep your 40 watt light bulb as is. You put the rock (soapstone) in an oven say at 30 to 50degC and heat it up until it is about 40degC, so that it is warmer than ambient room air temperature but not by so much. This rock will then radiate energy (low energy photons).
You switch on your 40watt light bulb as before and let it reach its equilibrium temperature (presumably around 95degc as before). You then introduce the rock/soapstone close (but not too close) to the 40watt light bulb so that the glass bulb of the 40watt light bulb receives not only the high energy of the photons from its filament but it additionally receives the low energy photons from the rock/soapstone and then see what effect this has on the temperature of the glass bulb. One should also measure the surface temperature of the rock/soapstone.
The rock/soapstone will not lose its heat quickly due to its latent heat characteristics and because it is not that much hotter than ambient room temperature.

richard
May 29, 2013 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.

Edim
May 29, 2013 4:29 am

(Th^4 – Tc^4)
I find it unbelievable that someone can claim that increasing the temperature of the cooler body cannot warm the warmer body, by reducing the radiative cooling rate of the warmer body. It’s in the plain sight. Those who claim that should go find a heat transfer textbook and solve all the examples in it. Then, they can discuss the matter – before that they are just a distraction.
Of course, this doesn’t mean that the so-called GHGs warm the Earth’s surface. On the face of it, they cool the atmosphere and therefore the surface.

May 29, 2013 4:36 am

I would REALLY like to see what the profile from the side–perpendicular to the mirror placement–would be.
One side of the bulb should be warmer than the other side, shouldn’t it?

Alex
May 29, 2013 4:37 am

GHGs aside, all that has been shown with the experiment is that the gas in the chamber has increased in temperature. This hotter gas has heated the globe externally, as would be expected There is no measurement of the heat source (filament). I admire your efforts, unfortunately they were misdirected. You measured the wrong thing.

richard
May 29, 2013 4:39 am

“I find it unbelievable that someone can claim that increasing the temperature of the cooler body cannot warm the warmer body, by reducing the radiative cooling rate of the warmer body”
is this used in industry to extract more heat out of a cooler body, sounds wonderfully efficient.

richard
May 29, 2013 4:40 am

or is it called insulation.

johnmarshall
May 29, 2013 4:50 am

Still the wrong conclusions. Yes the bulb envelope increased its temperature but did the element, ie., the thing that produces the heat and light, increase in temperature? this would still be producing the same temperature, a current change of 1mA would not make a measurable difference and might be a system variation, clip on ammeters are not that accurate, since they work by induction and are normally used for industrial purposes not to measure micro currents in a laboratory. (sensitive but not highly accurate).

thefordprefect
May 29, 2013 4:50 am

richard verney says: May 29, 2013 at 2:55 am
see experiment here
http://www.climateandstuff.blogspot.co.uk/2013/05/back-radiation-early-results-no-fan.html

steveta_uk
May 29, 2013 4:52 am

“Covering the bulb with foil has no effect on the filament temperature. The concept of heating by back-radiation is not supported”
This is one that can be tested at home so simply that it is almost unbelievable that the writer hasn’t tried it.
Get two similar bulbs, same power, same supplier, preferably a highish rating, which will make the test so much faster. If you can get 100W incandescants they’d be great.
Cover one tightly in a single layer of foil (small overlaps won’t matter). Make sure it’s a nice tight fit, so that the foil will be at almost the same temperature as the glass of the bulb.
Then turn on both lights, and wait.
See if you can guess which filament overheats and burns out in no time at all?
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.

thefordprefect
May 29, 2013 4:59 am

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.

richard