Find the efficiency of carnot engine whose source and sink are at `927^(@)C` and `27^(@)C`.

Explanation

To find the efficiency of a Carnot engine with a hot reservoir (source) at \(927^\circ C\) and a cold reservoir (sink) at \(27^\circ C\), we can follow these steps:

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin**:
- The temperature of the hot reservoir \(T_H\) in Kelvin:
\[ T_H = 927 + 273 = 1200 \, K \]
- The temperature of the cold reservoir \(T_L\) in Kelvin:
\[ T_L = 27 + 273 = 300 \, K \]

2. **Use the Carnot Efficiency Formula**:
- The efficiency \(\eta\) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_L}{T_H} \]

3. **Substitute the Values**:
- Substitute \(T_L\) and \(T_H\) into the efficiency formula:
\[ \eta = 1 - \frac{300}{1200} \]

4. **Calculate the Efficiency**:
- Simplify the fraction:
\[ \frac{300}{1200} = 0.25 \]
- Therefore, the efficiency becomes:
\[ \eta = 1 - 0.25 = 0.75 \]

5. **Convert Efficiency to Percentage**:
- To express the efficiency as a percentage:
\[ \eta = 0.75 \times 100 = 75\% \]

### Final Answer:
The efficiency of the Carnot engine is \(75\%\).

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