An engine operating between `150^(@)C` and `25^(@)C` takes 500 J heat from a higher temperature reservoir if there are no frictional losses, then work done by engine is

Options

A. 147.7 J
B. 157.75 J
C. 165.85 J
D. 169.95 J

Explanation

To solve the problem of calculating the work done by an engine operating between two temperature reservoirs, we will follow these steps:

### Step 1: Convert temperatures from Celsius to Kelvin
The temperatures given are:
- Higher temperature (T_H) = 150°C
- Lower temperature (T_L) = 25°C

To convert these to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]

Calculating:
- \( T_H = 150 + 273 = 423 \, K \)
- \( T_L = 25 + 273 = 298 \, K \)

### Step 2: Calculate the efficiency of the engine
The efficiency (η) of a Carnot engine is given by the formula:
\[ \eta = \frac{T_H - T_L}{T_H} \]

Substituting the values we found:
\[ \eta = \frac{423 - 298}{423} \]
\[ \eta = \frac{125}{423} \]

### Step 3: Calculate the work done by the engine
The work done (W) by the engine can be calculated using the formula:
\[ W = \eta \times Q_H \]
where \( Q_H \) is the heat absorbed from the higher temperature reservoir.

Given that \( Q_H = 500 \, J \):
\[ W = \frac{125}{423} \times 500 \]

Calculating:
\[ W = \frac{125 \times 500}{423} \]
\[ W \approx 147.7 \, J \]

### Final Answer
The work done by the engine is approximately **147.7 Joules**.

---