Topic: Thermodynamics
A heat engine is supplied with 250 KJ/s of heat at a constant fixed temperature of $227^{\circ}C$. The heat is rejected at $27^{\circ}C$ and the cycle is reversible. What is the amount of heat rejected?
- A. 273 KJ/s
- B. 200 KJ/s
- C. 180 KJ/s
- D. 150 KJ/s
Answer: D. 150 KJ/s
Explanation: Convert temperatures to Kelvin: $T_H = 227 + 273 = 500~K$, $T_L = 27 + 273 = 300~K$.
For a reversible cycle (Carnot engine), $\frac{Q_H}{T_H} = \frac{Q_L}{T_L}$.
$\frac{250}{500} = \frac{Q_L}{300} \Rightarrow Q_L = 300 \times 0.5 = 150~KJ/s$.