Topic: Thermodynamics
A heat engine operates between a cold reservoir at temperature T2 = 300K and a hot reservoir at Temperature T1. It takes 200J of heat from the hot reservoir and deliver 120J of heat to the cold reservoir in a cycle. What will be the efficiency of the engine?
- A. 30%
- B. 40%
- C. 60%
- D. 20%
Answer: B. 40%
Explanation: The efficiency of a heat engine is given by the formula eta = W / Q_S = (Q_S - Q_R) / Q_S = 1 - (Q_R / Q_S). Given the heat supplied Q_S = 200 J and the heat rejected Q_R = 120 J, the efficiency is 1 - (120 / 200) = 1 - 0.6 = 0.4, which corresponds to 40%.