The efficiency of carnot engine is $50\%$ and temperature of sink is $500\;K$. If temperature of source is kept constant and its efficiency raised to $60\%$, then the required temperature of the sink will be
AIPMT 2002, Medium
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(b) $\eta = 1 - \frac{{{T_2}}}{{{T_1}}}$==>$\frac{1}{2} = 1 - \frac{{500}}{{{T_1}}}$==> $\frac{{500}}{{{T_1}}} = \frac{1}{2}$…..$(i)$
$\frac{{60}}{{100}} = 1 - \frac{{{T_2}'}}{{{T_1}}}$==>$\frac{{{T_2}'}}{{{T_1}}} = \frac{2}{5}$…..$(ii)$
Dividing equation $(i)$ by $(ii)$, $\frac{{500}}{{{T_2}'}} = \frac{5}{4}$==>${T_2} = 400K$
$\frac{{60}}{{100}} = 1 - \frac{{{T_2}'}}{{{T_1}}}$==>$\frac{{{T_2}'}}{{{T_1}}} = \frac{2}{5}$…..$(ii)$
Dividing equation $(i)$ by $(ii)$, $\frac{{500}}{{{T_2}'}} = \frac{5}{4}$==>${T_2} = 400K$
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