Solution:
Source temperature $T_1 = \frac{T_2}{1-\eta_1} = \frac{500}{0.5} = 1000\text{ K}$. New sink temperature $T_2' = T_1(1-\eta_2) = 1000(1 - 0.6) = 400\text{ K}$.
The efficiency of Carnot engine is $50\%$ and temperature of sink is $500\text{ K}$. If temperature of source is kept constant and its efficiency raised to $60\%$, then the required temperature of the sink will be: (2007)
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