A Carnot engine with efficiency 50 % takes heat from a source at...

Efficiency of Carnot engine is given by $η = 1 - \dfrac{{T}_{sink}}{{T}_{source}} = 1 - \dfrac{{T}_{2}}{{T}_{1}}$

Given: Initial efficiency $η = \dfrac{1}{2}$

$∴ \dfrac{1}{2} = 1 - \dfrac{{T}_{2}}{600}$

$⇒ \dfrac{{T}_{2}}{600} = \dfrac{1}{2}$

$⇒ {T}_{2} = 300K$

When efficiency is increased to $70$ and ${T}_{2} = 300K$,

Let ${T}^{'}$ be new temperature of source

$⇒ \dfrac{7}{10} = 1 - \dfrac{300}{{T}^{'}}$

$⇒ \dfrac{300}{{T}^{'}} = 1 - \dfrac{7}{10}$

$∴ {T}^{'} = 1000K$.