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

Answer: 38.84

For $T_2=17^{\circ }C=(17+273)K=290K$,
$\eta =30\%$
But, $\eta =1-\frac{T_2}{T_1}$
$\therefore \frac{T_2}{T_1}=1-\frac{30}{100}=\frac{70}{100}$
$\therefore T_1=\frac{T_2\times 100}{70}$
$=\frac{290\times 100}{70}=414.29K$
When efficiency is increased by
$20\%$ of $\eta =\frac{20\times 30}{100}=6\%$
${\eta }^{\prime }=30+6=36\%$
Also,${\eta }^{\prime }=1-\frac{T_2}{T_1^{\prime }}$
$\therefore \frac{T_2}{T_1^{\prime }}=1-{\eta }^{\prime }$
$=1-\frac{36}{100}=\frac{64}{100}=\frac{16}{25}$
$\therefore T_1^{\prime }=\frac{290\times 25}{16}$
$=\frac{36.25\times 25}{2}$
$=18.125\times 25$
$=453.13K$
$\therefore \Delta T=T_1^{\prime }-T_1$
$=453.13-414.29=38.84K$