Q 11-11-028NEETJEE MainMedium
A Carnot engine has an efficiency of $40\%$ with its sink at $300$ K. By how much must the source temperature be raised to make the efficiency $50\%$?
Answer: (C) $100$ K
Concept. Carnot efficiency $\eta = 1 - \dfrac{T_2}{T_1}$.
Step 1. Initial source. $1 - \dfrac{300}{T_1} = 0.4 \Rightarrow T_1 = 500$ K.
Step 2. New source. $1 - \dfrac{300}{T_1'} = 0.5 \Rightarrow T_1' = 600$ K.
Step 3. Increase. $100$ K.
Answer. $100$ K (Option C).
Solution by Sreeraj P, M.Sc Physics