A Carnot engine whose sink is at $$300\,K$$  has an efficiency of $$40\% .$$  By how much should the temperature of source be increased so as to increase its efficiency by $$50\% $$  of original efficiency ?

Solution :
The efficiency of Carnot engine is defined as the ratio of work done to the heat supplied i.e.
$$\eqalign{ & \eta = \frac{{{\text{Work done}}}}{{{\text{Heat supplied}}}} = \frac{W}{{{Q_1}}} = \frac{{{Q_1} - {Q_2}}}{{{Q_1}}} \cr & = 1 - \frac{{{Q_2}}}{{{Q_1}}} = 1 - \frac{{{T_2}}}{{{T_1}}} \cr} $$
Here, $${{T_1}}$$ is the temperature of source, $${{T_2}}$$ is the temperature of sink, $${{Q_1}}$$ is heat absorbed and $${{Q_2}}$$ heat rejected
As given, $$\eta = 40\% = \frac{{40}}{{100}} = 0.4\,\,{\text{and}}\,\,{T_2} = 300\;K$$
So $$0.4 = 1 - \frac{{300}}{{{T_1}}} \Rightarrow {T_1} = \frac{{300}}{{1 - 0.4}} = \frac{{300}}{{0.6}} = 500\,K$$
Let temperature of the source be increased by $$x K,$$  then efficiency becomes
$$\eqalign{ & \eta ' = 40\% + 50\% {\text{ of }}\eta \cr & = \frac{{40}}{{100}} + \frac{{50}}{{100}} \times 0.4 \cr & = 0.4 + 0.5 \times 0.4 = 0.6 \cr} $$
Hence, $$0.6 = 1 - \frac{{300}}{{500 + x}}$$
$$\eqalign{ & \Rightarrow \frac{{300}}{{500 + x}} = 0.4 \cr & \Rightarrow 500 + x = \frac{{300}}{{0.4}} = 750 \cr & \therefore x = 750 - 500 = 250\;K \cr} $$
NOTE
All reversible heat engines working between same temperatures are equally efficient and no heat engine can be more efficient than Carnot engine (as it is ideal).