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Q. An ideal heat engine exhausting heat at $ 27^{\circ}C $ is to have $ 25\% $ efficiency. It must take heat at

UPSEEUPSEE 2008Thermodynamics

Solution:

$T_{2}=27^{\circ}C=\left(27+273\right)$
$K=300\,K$,
$\eta=25\%=\frac{1}{4}$
We know that,
$\eta=1-\frac{T_{2}}{T_{1}}$
$\Rightarrow \frac{1}{4}=1-\frac{300}{T_{1}}$
or $\frac{300}{T_{1}}=1-\frac{1}{4}$
$\Rightarrow \frac{300}{T_{1}}=\frac{3}{4}$
or $ T_{1}=\frac{300\times4}{3}$
$\Rightarrow T_{1}=400\,K$
or $T_{1}=\left(400-273\right)^{\circ}C$
$=127^{\circ}C$