Q 11-11-099JEE MainJEE Main 2022 (24 Jun, Shift 1)Medium
A Carnot engine whose heat sink is at $27^\circ\text{C}$ has an efficiency of $25\%$. By how many degrees should the temperature of the source be changed to increase the efficiency by $100\%$ of the original efficiency?
Answer: (B) Increases by $200^\circ\text{C}$
$\eta = 1 - \dfrac{T_2}{T_1}$ with $T_2 = 300\ \text{K}$.
Initially $0.25 = 1 - \dfrac{300}{T_1}\Rightarrow T_1 = 400\ \text{K}$.
The efficiency doubles to $50\%$: $0.5 = 1 - \dfrac{300}{T_1'}\Rightarrow T_1' = 600\ \text{K}$.
The source temperature must increase by $200\ \text{K} = 200^\circ\text{C}$.
Solution by Sreeraj P, M.Sc Physics