Correct Answer - Option 2 : 270 kW

__Concept:__

For reversible heat engine efficiency

\(\begin{array}{l} \eta = 1 - \frac{{{T_R}}}{{{T_S}}}\end{array}\)

\(\begin{array}{l} \eta = \frac{W}{{{Q_S}}} = 1 - \frac{{{Q_R}}}{{{Q_S}}}\\ \end{array}\)

__Calculation:__

__Given: __

Q_{S }= 450 kJ/s, T_{R} = 27°C = 300 K, T_{S }= 227°C = 500 K

\(\begin{array}{l} \eta = 1 - \frac{{{T_R}}}{{{T_S}}} = 1 - \frac{{300}}{{500}} = 0.4\\ \eta = \frac{W}{{{Q_S}}} = 1 - \frac{{{Q_R}}}{{{Q_S}}}\\ {Q_R} = 270\;kW \end{array}\)

∴ Q = 270 kW or 270 kJ/s