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Q. The efficiency of a Carnot engine working between $127^{\circ}C$ and $77^{\circ}C$ is

Thermodynamics

Solution:

Efficiency of the carnot engine $ \eta=1-\frac{T_{2}}{T_{1}} $
where, $T_1 =$ temperature of source
$T_2 =$ temperature of sink
Here, $T_1 = 127^{\circ}C = 400\, K$,
$ T_2 = 77^{\circ}C = 350\, K$
$\therefore \eta = 1 -\frac{350}{400}$
$ = \frac{50}{400} \times 100$
$ = 12.5 \%$