How is the Entropy of Blackbody Radiation Calculated?

[roam]

Homework Statement

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.

The Attempt at a Solution

So, I can't understand when calculating the entropy using $S=\frac{4}{3} aT^3V$, 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 a_λ = ε_λ. And

$Flux \ = \ \frac{ac}{4} T^4 = \sigma T^4 \implies a = 7.56 \times 10^{-16}$

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

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

Any help is appreciated.

 

[andrien]

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

 

[roam]

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

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.

 

[andrien]

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.

 

[roam]

Thank you, I see what you mean now... they were after the entropy per volume (J m^-3). Thanks! :)