Question

When two equal sized pieces of the same metal at different temperatures $T_h$(hot piece) and $T_c$ (cold piece) brought into contact into thermal contact and isolated from it's surrounding. The total change in entropy system is given by ? [$C_v$ (J/K) = heat capacity of metal]

• A. $C_v ln \frac{T_c + T_h}{2T_c}$
• B. $C_v ln \frac{T_2}{T_1}$
• C. $C_v ln \frac{(T_c + T_h)^2}{2T_h \cdot T_c}$
• D. $C_v ln \frac{(T_c + T_h)^2}{4T_h \cdot T_c}$

Answer 1

Answer: (D)

$C_v ln \frac{(T_c + T_h)^2}{4T_h \cdot T_c}$

Answer 2

The final equilibrium temperature $T_f$ is found by equating heat lost by the hot piece to heat gained by the cold piece: $C_v(T_h - T_f) = C_v(T_f - T_c)$, yielding $T_f = \frac{T_h + T_c}{2}$. The entropy change for each piece is $\Delta S = C_v \ln(T_{final}/T_{initial})$. Thus, $\Delta S_{hot} = C_v \ln(T_f/T_h)$ and $\Delta S_{cold} = C_v \ln(T_f/T_c)$. The total entropy change is the sum: $\Delta S_{total} = C_v \ln(T_f/T_h) + C_v \ln(T_f/T_c) = C_v \ln(\frac{T_f^2}{T_h T_c})$. Substituting $T_f$ gives the final result.