The efficiency of Carnot engine is $$50\% $$  and temperature of sink is $$500\,K.$$  If the temperature of source is kept constant and its efficiency is to be raised to $$60\% ,$$  then the required temperature of the sink will be

Solution :
Efficiency of the Carnot engine is given by
$$\eta = 1 - \frac{{{T_2}}}{{{T_1}}}\,......\left( {\text{i}} \right)$$
where, $${{T_1}} =$$  temperature of source
$${{T_2}} =$$  temperature of sink
Given, $$\eta = 50\% = 0.5,\,{T_2} = 500\,K$$
Substituting in Eq. (i), we have
$$\eqalign{ & 0.5 = 1 - \frac{{500}}{{{T_1}}}\,\,{\text{or}}\,\,\frac{{500}}{{{T_1}}} = 0.5 \cr & \therefore {T_1} = \frac{{500}}{{0.5}} = 1000\,K \cr} $$
Now, the temperature of sink is changed to $${{T'}_2}$$ and the efficiency becomes $$60\% $$  i.e., 0.6.
Using Eq. (i), we get
$$\eqalign{ & 0.6 = 1 - \frac{{{{T'}_2}}}{{1000}} \cr & {\text{or}}\,\frac{{{{T'}_2}}}{{1000}} = 1 - 0.6 = 0.4\,\,{\text{or}}\,\,{{T'}_2} = 0.4 \times 100 = 4000\,K \cr} $$
NOTE
Carnot engine is not a practical engine because many ideal situations have been assumed while designing this engine which cannot be obtained practically.