A Carnot engine has efficiency of 50 % . If the temperature of sink...

Initial efficiency ${η}_{0} = 50$, the increase in efficiency is $30$. Therefore, new efficiency $η = 50\left(1 + \dfrac{30}{100}\right) = 65$.

Efficiency of the Carnot engin is given by $η = 1 - \dfrac{{T}_{L}}{{T}_{H}}$

$⇒ {η}_{0} = \dfrac{1}{2} = 1 - \dfrac{{T}_{L}}{{T}_{H}}$

When the temperature of the sink is decreased by $40 ° C = 40K$.

$η = \dfrac{1}{2}\left(1 \cdot 3\right) = 1 - \left(\dfrac{{T}_{L} - 40}{{T}_{H}}\right)$

$⇒ \dfrac{1}{2}\left(1 \cdot 3\right) = \dfrac{1}{2} + \dfrac{40}{{T}_{H}}$

$⇒ {T}_{H} = 266 . 7K$