A Cannot engine works between 600K and 300K, In each cycle of operations the engine draws 1000 joule of energy from the source at 600K. The efficiency of the engine is:

Options

A. 0.2
B. 0.5
C. 0.7
D. 0.9

Explanation

To find the efficiency of a Carnot engine operating between two temperatures, we can use the formula:

\[ \text{Efficiency} (\eta) = 1 - \frac{T_L}{T_H} \]

where:
- \(T_L\) is the absolute temperature of the cold reservoir (sink),
- \(T_H\) is the absolute temperature of the hot reservoir (source).

### Step-by-Step Solution:

1. **Identify the temperatures**:
- The hot reservoir temperature \(T_H = 600 \, K\)
- The cold reservoir temperature \(T_L = 300 \, K\)

2. **Substitute the values into the efficiency formula**:
\[ \eta = 1 - \frac{T_L}{T_H} = 1 - \frac{300 \, K}{600 \, K} \]

3. **Calculate the fraction**:
\[ \frac{300}{600} = 0.5 \]

4. **Complete the calculation for efficiency**:
\[ \eta = 1 - 0.5 = 0.5 \]

5. **Express the efficiency as a percentage**:
\[ \eta = 0.5 \times 100 = 50\% \]

### Final Answer:
The efficiency of the Carnot engine is **50%** or **0.5**.