A Carnot engine absorbs $1000\,J$ of heat energy from a reservoir at $127\,^oC$ and rejects $600\,J$ of heat energy during each cycle. The efficiency of engine and temperature of sink will be
JEE MAIN 2014, Diffcult
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Given: $Q_1=1000 J$
$Q_2=600 J$
${T_1} = {127^ \circ }C = 400\,K$
${T_2} = ?$
$\eta = ?$
Efficiency of carnot engine,
$\eta = \frac{W}{Q_1} \times 100\% $
$or,\,\,\eta = \frac{{{Q_2} - {Q_1}}}{{{Q_1}}} \times 100\% $
$or,\,\,\,\eta = \frac{{1000 - 600}}{{1000}} \times 100\% $
$\eta = 40\% $
$Now,for\,carnot\,cycle\frac{{{Q_2}}}{{{Q_1}}} = \frac{{{T_2}}}{{{T_1}}}$
$\frac{{600}}{{1000}} = \frac{{{T_2}}}{{400}}$
${T_2} = \frac{{600 \times 400}}{{1000}} = 240\,K = 240 - 273$
$\therefore {T_2} = - {33^ \circ }C$
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