The efficiency of a Carnot engine operating between temperatures of $${100^ \circ }C$$  and $${-23^ \circ }C$$  will be

Solution :
Efficiency of Carnot engine is given by
$$\eta = 1 - \frac{{{T_2}}}{{{T_1}}} = \frac{{{T_1} - {T_2}}}{{{T_1}}}\,.......\left( {\text{i}} \right)$$
Given, $${{T_1}}$$ = temperature of reservoir
$$ = 100 + 273 = 373\,K$$
Given, $${{T_2}}$$ = temperature of sink
$$ = - 23 + 273 = 250\,K$$
Substituting in Eq. (i), we get
$$\eqalign{ & \therefore \eta = \frac{{373 - 250}}{{373}} = \frac{{123}}{{373}} \cr & = \frac{{100 + 23}}{{373}} \cr} $$