A Carnot engine has efficiency of 50 % . If the temperature of sink is reduced by 40 ° C , its…

Initial efficiency ${\eta }_0=50\%$, the increase in efficiency is $30\%$. Therefore, new efficiency $\eta =50\left(1+\frac{30}{100}\right)=65\%$.

Efficiency of the Carnot engin is given by $\eta =1-\frac{T_L}{T_H}$

$\Rightarrow {\eta }_0=\frac{1}{2}=1-\frac{T_L}{T_H}$

When the temperature of the sink is decreased by $40°\mathrm{C}=40 \mathrm{K}$.

$\eta =\frac{1}{2}\left(1·3\right)=1-\left(\frac{T_L-40}{ T_H}\right)$

$\Rightarrow \frac{1}{2}\left(1·3\right)=\frac{1}{2}+\frac{40}{T_H}$

$\Rightarrow T_{\mathrm{H}}=266.7 \mathrm{K}$