A Carnot engine operates between a source and a sink. The efficiency of the engine is 40 % and the…

Initial efficiency of the engine is $40\%$. Therefore,

$$\begin{gathered} \frac{40}{100}=1-\frac{T_{\sin k}}{T_{source}} \\ \Rightarrow 0.4=1-\frac{{\displaystyle \left(27+273\right)}}{{\displaystyle T_{source}}} \\ \Rightarrow T_{source}=500 \mathrm{K} \end{gathered}$$

Efficiency in second case of the engine is $50\%$. Therefore,

$$\begin{gathered} \frac{50}{100}=1-\frac{T_{\sin k}}{T_{source}} \\ \Rightarrow 0.5=1-\frac{{\displaystyle \left(27+273\right)}}{{\displaystyle T_{source}}} \\ \Rightarrow T_{source}=600 \mathrm{K} \end{gathered}$$

So clearly, the increase in temperature will be of $100 \mathrm{K}$