Temperature of source is `327^(@)C`. Temperature of sink is changed in order to increase the efficiency of engine from `(1)/(5)` to `(1)/(4)`, by

Options

A. 30K
B. 303K
C. 603K
D. 60K

Explanation

To solve the problem, we need to find the change in temperature of the sink when the efficiency of the Carnot engine changes from \( \frac{1}{5} \) to \( \frac{1}{4} \).

### Step-by-Step Solution:

1. **Convert the Source Temperature to Kelvin**:
The temperature of the source is given as \( 327^\circ C \). To convert this to Kelvin:
\[ T_{\text{source}} = 327 + 273 = 600 \, \text{K} \]

2. **Set Up the Efficiency Equations**:
The efficiency \( \eta \) of a Carnot engine is given by:
\[ \eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} \]
For the first case where the efficiency is \( \frac{1}{5} \):
\[ \frac{1}{5} = 1 - \frac{T_1}{600} \]
Rearranging gives:
\[ \frac{T_1}{600} = 1 - \frac{1}{5} = \frac{4}{5} \]
Thus,
\[ T_1 = 600 \times \frac{4}{5} = 480 \, \text{K} \]

3. **Calculate for the Second Efficiency**:
For the second case where the efficiency is \( \frac{1}{4} \):
\[ \frac{1}{4} = 1 - \frac{T_2}{600} \]
Rearranging gives:
\[ \frac{T_2}{600} = 1 - \frac{1}{4} = \frac{3}{4} \]
Thus,
\[ T_2 = 600 \times \frac{3}{4} = 450 \, \text{K} \]

4. **Calculate the Change in Temperature**:
The change in temperature \( \Delta T \) of the sink is given by:
\[ \Delta T = T_1 - T_2 = 480 - 450 = 30 \, \text{K} \]

5. **Conclusion**:
The change in temperature of the sink is \( 30 \, \text{K} \).

### Final Answer:
The change in temperature of the sink is \( 30 \, \text{K} \).

---