Q: Efficiency of a heat engine whose sink is at temperate of 300 K is 40%. To increase the efficiency to 60%, keeping the sink temperature constant, the source temperature must be increased by

Sol: $\large \eta = 1 – \frac{T_2}{T_1} $

$\large \frac{T_2}{T_1} = 1 – \eta $

$\large \frac{T_2}{T_1} = 1 – \frac{40}{100} = \frac{3}{5} $

$\large T_1 = \frac{5}{3} \times T_2 = \frac{5}{3} \times 300 $

T_{1} = 500 K

New efficiency η’ = 60 %

$\large \frac{T_2}{T_1′} = 1 – \eta’ $

$\large \frac{T_2}{T_1′} = 1 – \frac{60}{100} = \frac{2}{5} $

$\large T_1′ = \frac{5}{2} \times T_2 = \frac{5}{2} \times 300 $

T_{1}‘ = 750 K

∆T = 750-500 = 250 K