A carnots engine whose sink is at the temperature of 300K has an efficiency of 40. By how much should the temperature of the source be increased so as to increase the efficiency to 50%?
A
250K
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B
275K
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C
300K
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D
325K
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Solution
The correct option is A250K Efficiency of heat engine n=(T2−T1)T2
where T2 is source temperature and T1 is the sink temperature ...
For n=0.4
0.4=(T2−T1)/T2
T1/T2=0.6
As T1=300K...T2=3000.6=500K
Case 2 : for efficiency n to be increased 50% of its original efficiency, i.e. n=0.4(1+0.5)=0.6=60%
Source temperature can be calculated by the same formula n=(T2−T1)T2
0.6=(T2−300)T2 ... (sink temperature is constant)
T2=3000.4=750K
Increase in source temperature
Del T=(750−500)=250K...
That's your answer..
The source temperature should be increased by 250 K so as to increase the efficiency of heat engine by 50% of its original efficiency.