A Carnot engine whose low temperature reservoir is at $7\,°C$ has an efficiency of $50\%$. It is desired to increase the efficiency to $70\%$. By how many degrees should the temperature of the high temperature reservoir be increased ....... $K$
Medium
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(d) Initially $\eta = \frac{{{T_1} - {T_2}}}{{{T_1}}}$==> $0.5 = \frac{{{T_1} - (273 + 7)}}{{{T_1}}}$
==> $\frac{1}{2} = \frac{{{T_1} - 280}}{{{T_1}}}$==> ${T_1} = 560K$
Finally ${\eta _1}' = \frac{{{T_1}'\, - {T_2}}}{{{T_1}^\prime }}$==>$0.7 = \frac{{{T_1}'\, - \,(273 + 7)}}==>{{{T_1}^\prime }}$==>${T_1}' = 933K$
$\therefore$ increase in temperature $ = 933 - 560 = 373K \approx 380K$
==> $\frac{1}{2} = \frac{{{T_1} - 280}}{{{T_1}}}$==> ${T_1} = 560K$
Finally ${\eta _1}' = \frac{{{T_1}'\, - {T_2}}}{{{T_1}^\prime }}$==>$0.7 = \frac{{{T_1}'\, - \,(273 + 7)}}==>{{{T_1}^\prime }}$==>${T_1}' = 933K$
$\therefore$ increase in temperature $ = 933 - 560 = 373K \approx 380K$
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