Q 11-11-085JEE MainJEE Main 2023 (25 Jan, Shift 1)Easy
A Carnot engine with efficiency $50\%$ takes heat from a source at $600\ \text{K}$. In order to increase the efficiency to $70\%$, keeping the temperature of sink same, the new temperature of the source will be
Answer: (B) $1000\ \text{K}$
$0.5=1-\dfrac{T_2}{600}\Rightarrow T_2=300\ \text{K}$. For $70\%$: $1-\dfrac{300}{T_1}=0.7\Rightarrow T_1=1000\ \text{K}$.
Solution by Sreeraj P, M.Sc Physics