Question

A heat engine is supplied with 450 kJ/s of heat at a constant fixed temperature of 227°C. The heat is rejected at 27°C. What will be the amount of heat rejected if the cycle is reversible?

1. 300 kW
2. 270 kW
3. 330 kW
4. 320 kW

Answer 1

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