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Entropy of Blackbody Radiation

  1. May 1, 2012 #1
    1. The problem statement, all variables and given/known data

    The following is from a book:

    "Terrestrial radiation, T=255 K. Emitted flux ≈ 240 W m-2. Energy density for cavity radiation ≈ 3x10-6 J m-3. Entropy for cavity radiation ≈ 1.7x10-8 J K-1 m-3."

    I can't understand how they have calculated the Entropy.


    3. The attempt at a solution

    So, I can't understand when calculating the entropy using [itex]S=\frac{4}{3} aT^3V[/itex], what value they were using for "V"? V has to be the volume of the blackbody and we are treating Earth as one. So I used the volume of Earth, however my calculations do not match the book.

    From Kirchhoff’s Law aλ = ελ. And

    [itex]Flux \ = \ \frac{ac}{4} T^4 = \sigma T^4 \implies a = 7.56 \times 10^{-16}[/itex]

    [itex]S= (4/3)(7.56 \times 10^{-16})(255)^3(1.083206 \times 10^{21}) = 1.8105 \times 10^{13}[/itex]

    The results are very different. I was wondering if I'm missing something here or the book is wrong? :confused:

    Any help is appreciated.
     
    Last edited: May 1, 2012
  2. jcsd
  3. May 1, 2012 #2
    I think you should calculate it for unit volume means calculate s/v.I think that's all and moreover have you seen the unit of entropy which is given to you
     
  4. May 1, 2012 #3
    What do you mean? The formula clearly say that I must multiply things by "V", not divide by it. If I divide I get 1.5430e-29 which is wrong.
     
  5. May 2, 2012 #4
    I mean divide the answer you are getting by 1.083206*10^21.(1.8105*10^13)/(1.083206*10^21)=1.7*10^-8j k-1m-3.this is because the answer of book is per unit volume.
     
  6. May 4, 2012 #5
    Thank you, I see what you mean now... they were after the entropy per volume (J m-3). Thanks! :)
     
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