Correct Answer - Option 2 : 200

**Concept:**

The efficiency of the heat engine

\(η_{HE} = 1-\frac{Q_L}{Q_H}\)

Where Q_{L} is heat rejected to the sink, Q_{H} is heat received from the source

Carnot heat engine efficiency

\(η = 1-\frac{T_L}{T_H}\)

Where T_{L} is sink temperature, T_{H} is Source temperature

**Calculation:**

**Given:**

Q_{H} = 600 kJ, T_{H} = 627 °C = 627 + 273 = 900 K, T_{L} = 27 °C + 273 = 300 K

Since the above engine is working on the carnot cycle, Therefore,

Carnot efficiency = Heat engine efficiency

η = η_{HE}

\(1-\frac{T_L}{T_H} = 1-\frac{Q_L}{Q_H}\)

\(\frac{300}{900} = \frac{Q_L}{600}\)

**∴ Q**_{L} = 200 kJ.