Topic: Thermodynamics
A Carnot engine receiving heat at 400 k has an efficiency of 25%. What is the COP of a Carnot refrigerator working between same temperature limits?
- A. 1
- B. 2
- C. 3
- D. 4
Answer: C. 3
Explanation: Efficiency of Carnot engine, $\eta = 1 - \frac{T_L}{T_H} = 0.25 \Rightarrow \frac{T_L}{T_H} = 0.75$.
COP of a Carnot refrigerator working between the same temperatures is $COP_R = \frac{T_L}{T_H - T_L} = \frac{\frac{T_L}{T_H}}{1 - \frac{T_L}{T_H}} = \frac{0.75}{1 - 0.75} = \frac{0.75}{0.25} = 3$.