Thermal contact

The entropy of a system with constant (temperature independent) heat capacity is $S=c \ln(T) + const$ . (The constant term is not important here, but it is related to the fact the the heat capacity cannot be independent of the temperature, because according the the third law at $T \rightarrow 0$ , $S \rightarrow 0$ .) The entropy change is $\Delta S=c \ln(T_f/T_i)$ , where $T_f$ is the final temperature and $T_i$ is the initial temperature.

According to the notations in the problem, $c=C/2$ . The common temperature will be $T_f=(3/2) T$ . The entropy change of the first body is $\Delta S_1=(C/2) \ln \frac{3/2 T}{T} =(C/2) (0.4055)$ . Similarly, for the second body $\Delta S_2=(C/2) \ln \frac{3/2 T}{T} =-(C/2) (0.2877)$ . The total change is $\Delta S= \Delta S_1+ \Delta S_2=C (0.1178/2)=C (0.05589)$

The final temperature $T_m$ results form energy conservation $\begin{aligned} \Delta Q_1 = C_1 (T_m - T_1) &\stackrel{!}{=} -\Delta Q_2 = -C_2 (T_m - T_2) \\ \Rightarrow \quad T_m &= \frac{C_1 T_1 + C_2 T_2}{C_1 + C_2} = \frac{T_1 + T_2}{2} \end{aligned}$ The entropy change is caused by a heat flow $dQ_i = C_i dT_i$ , $i = 1,2$ , so that $\begin{aligned} dS_i &= \frac{dQ_i}{T_i} = \frac{C_i}{T_i} dT_i \\ \Rightarrow \quad \Delta S_i &= \int_{T_i}^{T_m} \frac{C_i}{T_i} dT_i = C_i \ln \frac{T_m}{T_i} \\ \Rightarrow \quad \Delta S &= \Delta S_1 + \Delta S_2 = C_1 \ln \frac{T_m}{T_1} + C_2 \ln \frac{T_m}{T_2} = \frac{C}{2} \ln \frac{T_m^2}{T_1 T_2} = C \ln \frac{T_m}{\sqrt{T_1 T_2}} \\ &= C \ln \frac{3}{2 \sqrt{2}} = 0.05889 \cdot C \end{aligned}$