Discussion
The entropy of black-body radiation is given by \(S = \frac{4}{3}\sigma V^{1/4}U^{3/4}\) where \(\sigma\) is a constant V is the volume and U is internal energy of the system. The temperature of the radiation is :
Answer: A
The temperature is \(T = U^{1/4}/(\sigma V^{1/4})\).
Temperature follows from the thermodynamic relation \(1/T = (\partial S/\partial U)_{V}\):
\[ \begin{aligned} \frac{1}{T} &= \frac{4}{3}\sigma V^{1/4}\cdot\frac{3}{4}U^{-1/4} \\ &= \sigma V^{1/4}U^{-1/4} \end{aligned} \]
Inverting gives \(T = U^{1/4}/(\sigma V^{1/4})\), equivalent to \(U \propto VT^{4}\), the Stefan-Boltzmann law for the energy of cavity radiation. Option B drops the volume, and options C and D have the wrong powers.
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