Topic: Thermodynamics
What is the maximum rate that a heat pump which uses 1 kW of electric power can supply heat to a house at 27°C when the outside temperature is 12°C
- A. 50 J/s
- B. 25 kJ/s
- C. 20 kJ/s
- D. 30 J/s
Answer: C. 20 kJ/s
Explanation: The maximum Coefficient of Performance (COP) for a heat pump is given by T_H / (T_H - T_L), where temperatures are in Kelvin. T_H = 27+273=300K, T_L = 12+273=285K. COP = 300 / (300-285) = 20. The rate of heat supply Q_H = COP * W_in = 20 * 1 kW = 20 kW = 20 kJ/s.