Q 11-11-131JEE MainJEE Main 2021 (27 Aug, Shift 2)Easy
A heat engine operates between a cold reservoir at temperature $T_2 = 400$ K and a hot reservoir at temperature $T_1$. It takes $300$ J of heat from the hot reservoir and delivers $240$ J of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be ______ K
Numerical value type. Enter your answer.
Answer: 500
Efficiency $= \dfrac{300 - 240}{300} = 0.2$.
No engine can beat the Carnot efficiency: $0.2 \le 1 - \dfrac{400}{T_1} \Rightarrow T_1 \ge 500$ K.
Solution by Sreeraj P, M.Sc Physics