Water at 300 K can be brought to 320 K using one of the following processes.
Process 1: Water is brought in equilibrium with a reservoir at 320 K directly.
Process 2: Water is first brought in equilibrium with a reservoir at 310 K and then
with the reservoir at 320 K.
Process 3: Water is first brought in equilibrium with a reservoir at 350 K and then
with the reservoir at 320 K.
The corresponding changes in the entropy of the universe for these processes are
∆S_1, ∆S_2 and ∆S_3, respectively. Then
Options:
(A) $\Delta S_2 > \Delta S_1 > \Delta S_3$
(B) $\Delta S_3 > \Delta S_1 > \Delta S_2$
(C) $\Delta S_3 > \Delta S_2 > \Delta S_1$
(D) $\Delta S_1 > \Delta S_2 > \Delta S_3$
Official Answer Key:
B
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.37)**
**1. Thermodynamics Principles & Entropy of Universe**:
- Total entropy change of the universe in a heating process:
$\Delta S_{\text{univ}} = \Delta S_{\text{water}} + \Delta S_{\text{res}}$
- Water is heated from $T_1 = 300$ K to $T_2 = 320$ K. Since the initial and final states of water are identical in all processes, $\Delta S_{\text{water}} = C \ln(320/300)$ is constant.
- Entropy change of heat reservoir at temperature $T_{\text{res}}$:
$\Delta S_{\text{res}} = -\frac{Q}{T_{\text{res}}} = -\frac{C(T_2 - T_1)}{T_{\text{res}}}$
- The irreversibility (total entropy generation $\Delta S_{\text{univ}}$) decreases as the temperature steps are made smaller and closer to equilibrium:
$\Delta S = \Delta S_{\text{water}} - \sum_i \frac{Q_i}{T_{\text{res}, i}}$
**2. Comparison of Processes**:
- **Process 1**: Single reservoir at $320$ K: $\Delta S_{\text{res}} = -\frac{C(20)}{320} = -0.0625 C$.
- **Process 2**: Two reservoirs in series ($310$ K then $320$ K):
$\Delta S_{\text{res}} = -\left[\frac{C(10)}{310} + \frac{C(10)}{320}\right] = -C(0.03226 + 0.03125) = -0.06351 C$
Hence $\Delta S_2$ generated is smaller than $\Delta S_1$.
- **Process 3**: Bringing into contact with a reservoir with a large thermal mismatch produces the highest irreversibility.
- Thus, the correct ranking of total entropy changes is $\Delta S_3 > \Delta S_1 > \Delta S_2$.
**3. Official Verification**:
The verified official IIT Master Answer is **Option (B)** (`\Delta S_3 > \Delta S_1 > \Delta S_2`).
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