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Emission of infra red from hot bodies

  1. Nov 4, 2009 #1
    I've seen much mention in websites of matt black surfaces being better emitters of infra red than shiny silver ones so if a matt black and shiny silver surface are heated to about 100 deg C the black one will cool down more quickly as a result of this! Doesn't Wien's law state that the amount of I.R. emitted by a hot body is only dependent on the temperature??

    My question is then why is a shiny silver surface a much poorer emitter than a matt black one?
     
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  3. Nov 4, 2009 #2

    Andy Resnick

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    Because of the conservation of energy: absorptivity + reflectivity + transmissivity = 1. For an opaque material, transmissivity = 0. Also, absorptivity = emissivity (Kirchoff's law), so emissivity = 1 - reflectivity. More reflectivity = less emissivity.
     
  4. Nov 5, 2009 #3
    Yep. The key to understanding this is to understand Kirchhoff's law equating absorptivity to emissivity. Imagine two people of different size sitting by a camp fire. Who heats up the quickest, and who stays hot the longest?
     
  5. Nov 5, 2009 #4

    mgb_phys

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    No Wien's law says that the peak position of the blackbody curve only depends on temperature
     
  6. Nov 10, 2009 #5
    Thanks very much for all replies - I'd only heard of Kirchoff in respect to electrical circuits.

    I get the "emissivity = 1 - reflectivity. More reflectivity = less emissivity." if that's what the law says but I still don't get why? Is there any mental picture (even if not totally accurate) that may help. I can see that the bending of molecules (due to the fact that charged particles are being accelerated) can result in the emission of e.m radiation in the infra red region but I still have no model of what is going on at the surface of a shiny metal that results in the emission of i.r. and also why it is so poor at doing so.

    I appreciate I may be wanting a handwavy classical model that is not appropriate for a quantum mechanical concept but if anyone has any ideas I'd really appreciate it.
     
  7. Nov 10, 2009 #6
    Andy was right in pointing you to conservation of energy. Consider a silver ball and a black ball illuminated by the same heat source. They should come to the same temperature. But the silver ball reflects most of the light falling on it while the black ball absorbs the light. Therefore to shed the same amount of heat at the same temperature, the black ball must have a much higher emissivity.
     
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