Solution:
Initial source temperature $T_1 = \frac{300}{1-0.4} = 500\text{ K}$. New efficiency $\eta' = 1.5 \times 0.4 = 0.6$. New source temperature $T_1' = \frac{300}{1-0.6} = 750\text{ K}$. Increase $\Delta T = 750 - 500 = 250\text{ K}$.
A Carnot engine whose sink is at $300\text{ K}$ 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? (2006)
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