A heat engine absorbs 760 kJ heat from a source at 380kK . It rejects (1) 650 kJ, (ii) 560 kJ, (iii) 504 kJ of heat to sink at 280 K . State which of these represent a reversible , an irreversible and an impossible cycle .

Explanation

To determine which of the given heat rejection scenarios represents a reversible, irreversible, or impossible cycle, we will use the efficiency formula for a Carnot engine and compare the efficiencies of the given cycles.

### Step 1: Understand the Efficiency of a Carnot Engine
The efficiency (\( \eta \)) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_C}{T_H} \]
where:
- \( T_H \) = temperature of the heat source (in Kelvin)
- \( T_C \) = temperature of the heat sink (in Kelvin)

### Step 2: Calculate the Efficiency of the Carnot Engine
Given:
- \( T_H = 380 \, K \)
- \( T_C = 280 \, K \)

Using the formula:
\[ \eta_{Carnot} = 1 - \frac{280}{380} \]
Calculating this gives:
\[ \eta_{Carnot} = 1 - 0.7368 = 0.2632 \]

### Step 3: Calculate the Efficiency for Each Case
Now we will calculate the efficiency for each case where heat is rejected.

#### Case (i): Heat rejected = 650 kJ
\[ \eta_1 = \frac{Q_H - Q_C}{Q_H} = \frac{760 - 650}{760} = \frac{110}{760} \approx 0.1447 \]

#### Case (ii): Heat rejected = 560 kJ
\[ \eta_2 = \frac{Q_H - Q_C}{Q_H} = \frac{760 - 560}{760} = \frac{200}{760} \approx 0.2632 \]

#### Case (iii): Heat rejected = 504 kJ
\[ \eta_3 = \frac{Q_H - Q_C}{Q_H} = \frac{760 - 504}{760} = \frac{256}{760} \approx 0.3368 \]

### Step 4: Compare the Efficiencies
Now we compare the calculated efficiencies with the Carnot efficiency (\( \eta_{Carnot} \approx 0.2632 \)):

1. **Case (i)**: \( \eta_1 \approx 0.1447 < \eta_{Carnot} \) (Irreversible)
2. **Case (ii)**: \( \eta_2 \approx 0.2632 = \eta_{Carnot} \) (Reversible)
3. **Case (iii)**: \( \eta_3 \approx 0.3368 > \eta_{Carnot} \) (Impossible)

### Conclusion
- **Case (i)**: Irreversible cycle
- **Case (ii)**: Reversible cycle
- **Case (iii)**: Impossible cycle