Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: DGriffiths on 04/11/2009 23:44:03
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I bought an infra red sensor from a car boot sale and was messing about at home. From the surface of my hot coffee cup i was getting about 110 unit of IR detected at a distance of about 1cm (the cup is ceramic and white) from the surface of the kettle (nickel chrome surface/ shiny silver anyway) I was picking up about 30 units of IR. Now the surface of the kettle was hotter (to the very quick touch) than the surface of the coffee cup. I though the amount of IR emitted from a surface depended only on its temperature (was it Wien's law trying to remember my A level physics??).
What is the mechanism that makes a shiny surface (at these sort of temperatures) a poor emitter of infra red). I have tried to find the answer to this on google to no avail. [8)]
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I'll make a guess; some knowledgeable soul may need to correct me. [:)] The shiny surface is a good reflector and so is less likely to be warmed by incident radiation. When you touch the surface, you feel the conducted heat. Radiated heat depends upon how close the surface is to being a black body.
Just Google with your thread title (http://physics.indiana.edu/~brabson/p510/selectivesurfaces.html) You will be rewarded with this thread on top, then with:
Good absorbers are good emitters: An ideal absorber is often called a black body. It absorbs all the radiation that hits it. It is easy to see that an ideal absorber of a particular wavelength of radiation is also the best possible emitter at that wavelength. That is, no object at temperature T can emit more radiation than a black body.
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The reason that shiny surfaces are poor emitters comes from the conservation of energy.
Imagine your kettle next to your coffee cup, but imagine that the kettle, in addition to being a good reflector is also a good emitter.
The kettle will cool by radiation- it's a good emitter. But, since it's a good reflector it must be a poor absorber. So, any energy radiated from the cup will be reflected back towards the cup keeping it warm, but the energy radiated from the kettle will be absorbed by the cup so the kettle gets colder.
Eventually, with no external work having been done on the system, the cup ends up warmer than the kettle.
If I could get that to happen I could run a very small steam engine on the temperature difference.
That would be a perpetual motion machine so something must have been wrong with the story I told.
What went wrong was that I ignored Kirchoff's radiation law.
Anyway, I'm pleased to see that people are still doing things like this "I bought an infra red sensor from a car boot sale and was messing about at home.".
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The cup will never be warmer than the kettle at best they'll equalise and the kettle is being warmed by electricity so it's entirely losing heat and no perpetual motion to worry about.
The shiny surface also has a lower surface area than a scratched surface so it loses less heat.
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The cup will never be warmer than the kettle at best they'll equalise and the kettle is being warmed by electricity so it's entirely losing heat and no perpetual motion to worry about.
You have entirely missed the point.
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This looks like a shiny metal kettle (you can see the reflection of the person holding it) and a ceramic teapot
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Making tea, thermogram
http://www.sciencephoto.com (http://www.sciencephoto.com/images/imagePopUpDetails.html?pop=1&id=725840104&pviewid=&country=67&search=thermal+AND+%28image+AND+%28kettle+AND+cup%29%29&matchtype=EXACT)
Despite the metal kettle being as hot, or hotter, than the ceramic teapot, in this image the kettle emits less IR at the frequencies this camera is sensitive to than the ceramic teapot.
The shiny surface also has a lower surface area
Sounds a plausible explanation.
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The simplest way of thinking of this problem is that the shiny surface reflects both ways. it reflects the infra red energy in the material back into the kettle as well as reflecting light fom its surface. the process is simply the reflection of electromagnetic radiation at a mismatch and the mismatch between a smooth conducting surface and free space is just about as big as you can get.
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That explanation is correct but I don't think we have the same idea of "simple".
Ordinarily I try to avoid solving the Schrodinger equation for a many-body problem then applying time dependent perturbation theory so I can have a look at the effect of different boundary conditions.
I grant you it works.
Incidentally, that's a nice picture.
The water is the brightest thing in the picture. It will be very smooth.
Water is an excellent absorber of IR; it therefore emits well.
If you ever find anything which emits well and reflects well patent it quickly. It will be a free energy machine. Another way to look at that is that you won't find something that's a good emitter and good reflector because it would break the conservation of energy.
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You don't need that level of complexity in the theory. Maxwells equations are quite good enough for solids up to ultraviolet because the spacing between the atoms is much smaller than the wavelength of the radiation you are talking about. Atomic spacing and/or motions and quantum theory only becomes significant at X Ray frequencies.
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Fair point, but I think the appeal to the conservation of energy is still easier than Maxwell's equation.
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True [:)]