Black body radiation entropy question

[madorangepand]

1. The problem statement, all variables and given/known data
An evacuated container with volume V and at a temperature T contains black body radiation with an energy density equal to 4\sigmaT4/c
I Determine the heat capacity at constant volume of the radiation
II Hence show that the entropy of the radiation is given by

S(T,V)=16\sigmaVT3/3c (can't seem to get the sigma to display correctly here)

III The container is placed in thermal contact with a heat bath at temperature Tr. If the heat capacity of the cavity material itself is negligible, show that the overall change in entropy of the universe after the system and heat bath have reached thermal equilibrium is

\DeltaStot=(4\sigmaVTr3/3c)(1-t3(4-3t))

t= T/Tr

IV comment on the sign of \DeltaStot as a function of t


2. Relevant equations



3. The attempt at a solution
Have done I and II correctly. IV looks fine.
Just need a little nudge as to how I should start III, I just can't think of the appropriate formulae at the moment. So any links to references etc are greatly appreciated.

[diazona]

Well, think about it this way: the total change in entropy of the universe comes from only two systems, the radiation and the heat bath. So you can calculate ΔStot as
\Delta S_{\text{tot}} = \Delta S_{\text{rad}} + \Delta S_{\text{bath}}

To calculate the entropy change of the heat bath, recall the (well, a) definition of entropy:
\Delta S = \int_{T_i}^{T_f} \frac{\mathrm{d}Q}{T}
Remember that the heat bath is at a constant temperature - how does that fact allow you to simplify the equation?

Calculating the entropy change of the radiation can be done in the patently obvious manner :wink: Recall the definition of the Δ symbol. Other than that, you already have the only formula you need.

[Andrew Mason]

An evacuated container with volume V and at a temperature T contains black body radiation with an energy density equal to 4\sigma T^4/c
I Determine the heat capacity at constant volume of the radiation
II Hence show that the entropy of the radiation is given by

S(T,V)=\frac{16\sigma VT^3}{3c}
(can't seem to get the sigma to display correctly here)
the latex "tex" tokens are intended be used on a separate line. For inline latex use "itex".

III The container is placed in thermal contact with a heat bath at temperature Tr. If the heat capacity of the cavity material itself is negligible, show that the overall change in entropy of the universe after the system and heat bath have reached thermal equilibrium is

\Delta S_{tot}=\left{(}\frac{4\sigma VT_r^3(1-t^3(4-3t))}{3c}\right{)}

t= T/T_r

IV comment on the sign of \Delta S_{tot} as a function of t


3. The attempt at a solution
Have done I and II correctly. IV looks fine.
Just need a little nudge as to how I should start III, I just can't think of the appropriate formulae at the moment. So any links to references etc are greatly appreciated.Keep in mind that entropy change is defined as the integral \int dQ/T over the reversible path between two states. The reversible path would be one in which the cavity is cooled by means of a carnot heat engine operating between the cavity and the heat bath. In that case, the carnot engine absorbs heat at the cavity temperature (which decreases with heat flow) and the heat bath absorbs heat from the carnot engine at constant temperature.

AM

[madorangepand]

Thanks guys, I knew it was just staring me in the face.

[blueyellow]

i also have III as a homework question.
but it looks confusing cos it goes 1-t^3(4-3t)
please help