# Step 1: Understanding the Energy Transfer When a metal bar connects a hot reservoir (temperature \( T_H \)) and a cold reservoir (\( T_C \)), energy is transferred by heat from the hot to the cold reservoir. The heat transferred from the hot reservoir is denoted \( Q_H \). **Explanation:** This process is an example of heat conduction, where energy flows spontaneously from higher to lower temperature. In the direct process, \( Q_H \) leaves the hot reservoir, and a corresponding amount is absorbed by the cold reservoir. --- # Step 2: Expression for Change in Entropy The change in entropy (\( \Delta S \)) in this process is the sum of the entropy change in both reservoirs: \[ \Delta S = \Delta S_H + \Delta S_C \] where \[ \Delta S_H = \frac{-Q_H}{T_H} \quad \text{(hot reservoir loses heat)} \] \[ \Delta S_C = \frac{Q_H}{T_C} \quad \text{(cold reservoir gains heat)} \] So, \[ \Delta S = \frac{Q_H}{T_C} - \frac{Q_H}{T_H} \] **Explanation:** Entropy decreases in the hot reservoir (heat lost), increases in the cold reservoir (heat gained). The total entropy change is positive if \( T_H > T_C \), reflecting the irreversibility of the spontaneous heat transfer. --- # Final Answer **Expression for the change in entropy:** \[ \boxed{ \Delta S = Q_H \left( \frac{1}{T_C} - \frac{1}{T_H} \right) } \] This shows the total entropy increases when heat flows from hot to cold spontaneously.