Question

A body at a temperature \(T\) is brought into contact with a reservoir at temperature \(2T\). Thermal equilibrium is established at constant pressure. The heat capacity of the body at constant pressure is \(C_p\). The total change in entropy of the body and the reservoir in units of \(C_p\) is ............ (Round off to 2 decimal places)

Hint:

When a body exchanges heat with a reservoir, the total entropy change is always positive, indicating irreversibility of the process.

Answer 1

Correct Answer: 0.19 - 0.2

Step 1: Entropy change of the body. 
The body is heated from temperature \(T\) to \(2T\) at constant pressure. \[ \Delta S_{\text{body}} = C_p \ln{\frac{2T}{T}} = C_p \ln{2} \]

Step 2: Entropy change of the reservoir. 
The heat lost by the reservoir is equal to the heat gained by the body. \[ Q = C_p (2T - T) = C_p T \] The reservoir temperature remains constant at \(2T\), so its entropy change is: \[ \Delta S_{\text{res}} = -\frac{Q}{T_{\text{res}}} = -\frac{C_p T}{2T} = -\frac{C_p}{2} \]

Step 3: Total entropy change. 
\[ \Delta S_{\text{total}} = \Delta S_{\text{body}} + \Delta S_{\text{res}} \] \[ \Rightarrow \frac{\Delta S_{\text{total}}}{C_p} = \ln{2} - \frac{1}{2} = 0.693 - 0.5 = 0.193 \]

Step 4: Conclusion. 
Hence, the total change in entropy of the system and reservoir in units of \(C_p\) is \(0.19\).