A reversible engine takes heat from a reservoir at `527^(@)C` and gives out to the sink at `127^(@)C.` The engine is required to perform useful mechanical work at the rate of 750 watt. The efficiency of the engine is

Explanation

To find the efficiency of the reversible engine, we can follow these steps:

### Step 1: Convert the temperatures from Celsius to Kelvin
The temperatures of the hot and cold reservoirs need to be converted to Kelvin using the formula:
\[ T(K) = T(°C) + 273.15 \]

- For the hot reservoir at \( 527°C \):
\[ T_H = 527 + 273.15 = 800.15 \, K \]

- For the cold reservoir at \( 127°C \):
\[ T_C = 127 + 273.15 = 400.15 \, K \]

### Step 2: Use the formula for efficiency of a reversible engine
The efficiency (\( \eta \)) of a reversible engine is given by:
\[ \eta = 1 - \frac{T_C}{T_H} \]

### Step 3: Substitute the values into the efficiency formula
Now, we can substitute the values of \( T_C \) and \( T_H \) into the efficiency formula:
\[ \eta = 1 - \frac{400.15}{800.15} \]

### Step 4: Calculate the efficiency
Now, we can perform the calculation:
1. Calculate \( \frac{400.15}{800.15} \):
\[ \frac{400.15}{800.15} \approx 0.500 \]

2. Substitute this value back into the efficiency formula:
\[ \eta = 1 - 0.500 \]
\[ \eta = 0.500 \]

### Step 5: Convert efficiency to percentage
To express efficiency as a percentage, multiply by 100:
\[ \eta = 0.500 \times 100 = 50\% \]

### Final Answer
The efficiency of the engine is approximately **50%**.

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