• # question_answer The efficiency of a Carnot engine operating between temperatures of $100{}^\circ C$ and $-23{}^\circ C$ will be A) $\frac{100-23}{273}$                      B) $\frac{100+23}{273}$C) $\frac{100+23}{100}$                    D) $\frac{100-23}{100}$

Efficiency of Carnot engine is given by $\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}}$ Given, ${{T}_{1}}=$temperature of reservoir $=100+273=373\,K$ ${{T}_{2}}=$temperature of sink $=-\,23\,+\,273$ = 250 K Substituting in Eq.(i), we get $\therefore$$\eta =\frac{373-250}{373}=\frac{123}{373}$ $=\frac{100+23}{373}$ Hence, the correction option is [b].