The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is ........ $\%$
NEET 2018,JEE MAIN 2022, Medium
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Efficiency of an ideal heat engine,
$\eta = \left( {1 - \frac{{{T_2}}}{{{T_1}}}} \right)$
Freezing point of water $ = {0^ \circ }C = 273\,K$
Boiling point of water$ = {100^ \circ }C = \left( {100 + 273} \right)K$
$ = 373\,K$
$T_2$ Sink temperature$=273 K$
$T_1$ Source temperature $=373 K$
$\% \eta = \left( {1 - \frac{{{T_2}}}{{{T_1}}}} \right) \times 100 = \left( {1 - \frac{{273}}{{373}}} \right) \times 100$
$ = \left( {\frac{{100}}{{373}}} \right) \times 100 = 26.8\% $
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