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A Carnot engine has an efficiency of $1/6$. When the temperature of the sink is reduced by $62\,^oC$, its efficiency is doubled. The temperatures of the source and the sink are, respectively
$62\,^oC,\, 124\,^oC$
$99\,^oC,\, 37\,^oC$
$37\,^oC,\, 99\,^oC$
$124\,^oC, \,62\,^oC$
Solution
Efficiency of Carnot engine $=1-\frac{T_{\sin k}}{T_{\text {source }}}$
Given,
$\frac{1}{6}=1-\frac{T_{\text {sink }}}{T_{\text {soure }}} \Rightarrow \frac{T_{\text {sink }}}{T_{\text {source }}}=\frac{5}{6}\,,\,\,\,\,\,\,\,\,\,\,\,\,\,…(i)$
Also,
$\frac{2}{6}=1-\frac{T_{\sin k}-62}{T_{\text {rource }}} \Rightarrow \frac{62}{T_{\text {source }}}=\frac{1}{6}\,\,\,\,\,\,\,\,\,\,\,…(ii)$
$\therefore \quad \mathrm{T}$ source $=372 \mathrm{K}=99^{\circ} \mathrm{C}$
Also, $T_{\sin k}=\frac{5}{6} \times 372=310 \mathrm{K}=37^{\circ} \mathrm{C}$
(Note: Temperature of source is more than temperature of sink)