In a Carnot's engine if the temperatures of the source and the sink are decreased by $100 \mathrm{~K}$ each…
In a Carnot's engine if the temperatures of the source and the sink are decreased by $100 \mathrm{~K}$ each then the efficiency of the engine
increases
decreases
remains constant
becomes one
Solution
Initial efficiency is given by
$\eta=1-\frac{T_2}{T_1}=\frac{T_1-T_2}{T_1}$
Decreased by $100 \mathrm{~K}$ each the efficiency
$\begin{aligned} & \eta=1-\frac{T_2-100}{T_1-100} \\ & =\frac{T_1-100-T_2+100}{T_1-100}=\frac{T_1-T_2}{T_1-100}\end{aligned}$