## Solution to 1996 Problem 74

 We use the formula $dS = \delta Q/T$. We assume the system has reached thermal equilibrium. This means that the two bodies must have the same temperature. Conservation of energy implies that this temperature must be $300 \mbox{ K}$. If the equilibrium temperature were greater than $300 \mbox{ K}$, the amount of heat enters the body originally at $100 \mbox{ K}$ would be greater than the amount of heat that leaves the body originally at $500 \mbox{ K}$. If the equilibrium temperature were less than 300 K, the amount of heat that leaves the body originally at $500 \mbox{ K}$ would be greater than the amount of heat that enters the body originally at $100 \mbox{ K}$. Both situations violate conservation of energy (this assumes, of course, that the heat capacity $C$ is a constant). The entropy change of the body originally at $500 \mbox{ K}$ is\begin{align*}\int \frac{\delta Q}{T} = \int_{500} ^{300} \frac{m C dT}{T} = m C \ln(3/5)\end{align*}The entropy change of the body originally at $100 \mbox{ K}$ is\begin{align*}\int \frac{\delta Q}{T} = \int_{100} ^{300} \frac{m C dT}{T} = m C \ln(3/1)\end{align*}Therefore, the total entropy change is \begin{align*}m C \ln(3/5) + m C \ln(3/1) = m C \ln (9/5)\end{align*}Hence answer (B) is correct.