Correct Answer - Option 2 : 0.4 kW

__Concept:__

Reversible heat engine cycle will reject least rate of heat (QL) per kW net output (W) of engine.

For reversible heat engine,

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

__Calculation:__

Given T_{H} = 800°C = 1073 K, T_{L} = 30°C = 303 K, W = 1 kW,

\(\frac{W}{{{Q_H}}} = 1 - \frac{{{T_L}}}{{{T_H}}} \Rightarrow \frac{1}{{{Q_H}}} = 1 - \frac{{303}}{{1073}}\)

⇒ Q_{H} = 1.39 kW

⇒ Q

_{L} = Q

_{H} – W = 1.39 – 1 = 0.39 kW ≈ 0.4 kW