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
$1 - \dfrac{300}{T_1} = 0.4 \Rightarrow T_1 = 500$ K. For $50\%$: $T_1' = 600$ K. Increase $= 100$ K.
Solution by Sreeraj P, M.Sc Physics