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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?

1. 325K
2. 250K
3. 380K
4. 275K

Correct answer: 380K

Solution

The efficiency of a Carnot engine is given by η = 1 - (T_c/T_h). Initially, η = 0.4, so T_h = T_c / (1 - η) = 300 / (1 - 0.4) = 500 K. To increase efficiency by 50%, the new efficiency becomes η' = 0.4 + 0.2 = 0.6. Using η' = 1 - (T_c/T_h'), we get T_h' = T_c / (1 - η') = 300 / (1 - 0.6) = 750 K. The increase in source temperature is ΔT = T_h' - T_h = 750 - 500 = 250 K.

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