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I Black bodies, perfect reflection and heat transfer

  1. Aug 11, 2016 #1
    Hi Guys
    While studying physics for my masters (doing it in material science engineering), huge amount of questions appeared.
    First, lets imagine two experiments
    1) One enclosure with perfect reflective walls (R=1 over the entire thermal radiation spectra) at a temperature T1 and a body inside with Temperature T2 (T2 higher than T1). Lets say there is vacuum inside so energy can only be exchanged through radiation
    After some time, since the reflected photons have the same energy of the incoming ones, there is no energy left for increasing temperature of the wall? If that statement is true, this system will never eventually reach thermal equilibrium
    2) The same enclosure but now the walls are perfect black bodies. The walls will absorb thermal radiation until it reaches the same temperature of the body inside. Is that true so the system will reach thermal equilibrium?
     
  2. jcsd
  3. Aug 11, 2016 #2

    mfb

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    If the walls are perfectly reflective, the wall does not absorb or emit any energy, it does not change its temperature, not even initially (assuming there are no effects from the outside). The system never reaches equilibrium because you removed heat transfer completely.
    Right.
     
  4. Aug 11, 2016 #3
    Well, since reflection and absorption are opposite events (i think), returning to the real world
    Let´s study a metal. Since transmittance is pretty low, we can say R + A = 1 (R stands for reflection and A for absorption)
    The extinction coefficient (K) is usually high in metals. So we think about the Absorptivity= A= 4πK/λ + scattering and R= ((n+12+K2)/((n+1)2+k2) assuming a electromagnetic wave travelling from air to a conductor. Increasing K raises the values of both A and R
    So, in the end, metal absorbs radiation, reflects, they are similar things or none of above?
     
  5. Aug 11, 2016 #4

    mfb

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    R=1 directly leads to T=A=0, as their sum is 1 and they cannot be negative.

    Such a material does not exist in reality, but metals are good reflectors for everything below their plasma frequency (that makes them shiny), and superconductors are even better reflectors, at least a lower frequencies.
     
  6. Aug 11, 2016 #5
    But how can the same coefficient (K) increases both R and A at the same time and R= 1 -A?
     
  7. Aug 11, 2016 #6

    mfb

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    Your formula for the absorption is the absorption length in the material, it is not the fraction of light that gets absorbed by the surface.
     
  8. Aug 11, 2016 #7
    And the amount of absorbed light would be evaluated through Beer Lambert law, correct?
     
  9. Aug 11, 2016 #8

    mfb

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    After you take surface reflection into account, you can use that law to calculate absorption while radiation passes through the material. Then you can consider reflection at the end of the material to find the amount transmitted to whatever comes behind.
     
  10. Aug 12, 2016 #9
    Ahhh, now it makes sense. Vielen dank Herr mfb, sie sind sehr freundlich
     
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