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:
Text Solution
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The efficiency of Carnot engine is defined as the ratio of work done to the heat supplied i.e.,
η = 
= 
= 1 – 
Here, T1 is the temperature of source and T 2 is the temperature of sink
As given, η = 40% = 
and T 2 = 300K
So, 0.4 = 1 – 
⇒ T 1 = 
Let temperature of the source be increased by xK, then efficiency becomes
η ' = 40% + 50% of η
= 
= 0.4 + 0.5 × 0.4 = 0.6
Hence, 0.6 = 1 – 
⇒ 
⇒ 500 + x = 
∴ x = 750 – 500 = 250 K
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