Question
Physics Question on Thermodynamics
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?
A
275 K
B
175 K
C
250 K
D
225 K
Answer
250 K
Explanation
Solution
Temperature of sink TL = 300 K
Original efficiency η = 40% = 0.4
Let the Initial temperature of the source be TH
Using η = 1−THTL
∴ 0.4 = 1−TH x 300 ⟹ TH = 500 K
Now the efficiency of the engine is increased by 50% of original efficiency,
∴ New efficiency η′=40% +20% =60%
∴ η′ = 1−TH'TL
OR 0.6 = 1−TH′ x 300
OR TH′ x 300 = 0.4 ⟹TH′ = 750 K
Increase in source temperature ΔTH = TH′ − TH = 750 − 500 = 250 K
so, the correct option is (C): 250 K