Topic: Thermodynamics
A engine works on Carnot cycle between $727^{\circ}C$ and $227^{\circ}C$. The efficiency of the engine is
- A. 50%
- B. 75.4%
- C. 31.2%
- D. 68.8%
Answer: A. 50%
Explanation: Convert the given temperatures to Kelvin: the lower temperature $T_L = 227 + 273 = 500 K$, and the higher temperature $T_H = 727 + 273 = 1000 K$. The efficiency of a Carnot engine is calculated by $\eta = 1 - \frac{T_L}{T_H} = 1 - \frac{500}{1000} = 0.5$, which corresponds to 50%.