Question
$A$ reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by $ 62^oC$, the efficiency of the engine is doubled. The temperatures of the source and sink are
✓
Answer
(d) Initially $\eta = \left( {1 - \frac{{{T_2}}}{{{T_1}}}} \right) = \frac{W}{Q} = \frac{1}{6}$ ...$(i)$
Finally $\eta ' = \left( {1 - \frac{{{T_2}'}}{{{T_1}}}} \right) = \left( {1 - \frac{{({T_2} - 62)}}{{{T_1}}}} \right) = 1 - \frac{{{T_2}}}{{{T_1}}} + \frac{{62}}{{{T_1}}}$
$ = \eta + \frac{{62}}{{{T_1}}}$ ....$(ii)$
It is given that $\eta ' = 2\eta .$ Hence solving equation $ (i)$ and$(ii)$
==> ${T_1} = 372\,K = 99^\circ C$ and ${T_2} = 310K = 37^\circ C$
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