A Carnot engine whose sink is at $300\, K$ has an efficiency of $40\%.$ By how much should the temperature of source be increased so as to increase its efficiency by $50\%$ of original efficiency ..... $K$
AIPMT 2006, Diffcult
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Efficiency of a Carnot engine,$\eta = 1\frac{{{T_2}}}{{{T_1}}}$
$or,\,\,\frac{{{T_2}}}{{{T_1}}} = 1 - \eta = 1 - \frac{{40}}{{100}} = \frac{3}{5}$
$\therefore {T_1} = \frac{5}{3} \times {T_2} = \frac{5}{3} \times 300 = 500\,K.$
$Increases\,in\,efficiency = 50\% \,of\,40\% = 20\% $
$New\,efficiency,\eta ' = 40\% + 20\% = 60\% $
$\therefore \,\,\frac{{{T_2}}}{{{T_1}'}} = 1 - \frac{{60}}{{100}} = \frac{2}{5}$
${T_1}' = \frac{5}{2} \times {T_2} = \frac{5}{2} \times 300 = 750\,K.$
$Increases\,in\,temperature\,of\,source = {T_1}' - {T_1}$
$ = 750 - 500 = 250\,K.$
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