Question

Efficiency of a Carnot engine is $30 \%$ whose heat $\sin k$ is maintained at $17^{\circ} C$. By how much amount (in $K$ ) should the temperature of source be changed in order to increase the efficiency of the engine by $20 \%$ of its original efficiency?

Answer 1

For $T _{2}=17^{\circ} C =(17+273) K =290 K$,
$\eta=30 \%$
But, $\eta=1-\frac{ T _{2}}{ T _{1}}$
$\therefore \frac{ T _{2}}{ T _{1}}=1-\frac{30}{100}=\frac{70}{100}$
$\therefore T _{1}=\frac{ T _{2} \times 100}{70}$
$=\frac{290 \times 100}{70}=414.29\, K$
When efficiency is increased by
$20 \% $ of $ \eta=\frac{20 \times 30}{100}=6 \% $
$\eta^{\prime}=30+6=36 \% $
Also,$ \eta^{\prime}=1-\frac{ T _{2}}{ T _{1}^{\prime}}$
$ \therefore \frac{ T _{2}}{ T _{1}^{\prime}} =1-\eta^{\prime}$
$=1-\frac{36}{100}=\frac{64}{100}=\frac{16}{25}$
$ \therefore T _{1}^{\prime} =\frac{290 \times 25}{16} $
$=\frac{36.25 \times 25}{2} $
$=18.125 \times 25$
$=453.13 \,K$
$ \therefore \Delta T = T _{1}^{\prime}- T _{1}$
$=453.13-414.29=38.84\, K $