Discussion
A capacitor of capacitance C farads is charged from a battery of emf V volts. Out of the work done by the battery an amount \(\frac{1}{2}CV^{2}\) is stored in the capacitor and the rest is released to the surroundings. If the process is carried at temperature T the change in the entropy of the universe would be :
Answer: C
The entropy of the universe increases by \(\frac{CV^{2}}{2T}\).
The battery moves charge \(Q = CV\) through emf V, so it does work \(W = QV = CV^{2}\). Half of it, \(\frac{1}{2}CV^{2}\), is stored in the capacitor's field, which is ordered energy carrying no entropy. The other half is dissipated as heat into the surroundings at temperature T:
\[ \begin{aligned} Q_{heat} &= CV^{2} - \frac{1}{2}CV^{2} = \frac{1}{2}CV^{2} \\ \Delta S &= \frac{Q_{heat}}{T} = \frac{CV^{2}}{2T} \end{aligned} \]
\(CV^{2}/T\) counts the whole work as heat, and \(CV^{2}/4T\) and \(CV^{2}/8T\) halve the heat again.
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