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?
- 380 K
- 275 K
- 325 K
- 250 K
Answer
(4) 250 K
Solution
As per NCERT, efficiency of a Carnot engine: = 1 - T2T1
Initially: 0.40 = 1 - 300T1 T1 = 3000.6 = 500 K
New efficiency = 40\% + 50100 40\% = 60\%
0.60 = 1 - 300T1' T1' = 3000.4 = 750 K
Increase in source temperature = 750 - 500 = 250 K
Answer: (4)