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A perfect carnot engine utilizes an ideal gas the source temperature is500K and sink temperature is 375K. If the engine takes 600k cal per cyclefrom the source, calculate(i) The efficiency of engine(ii) Work done per cycle(iii) Heat rejected to sink per cycle.

Vaibhav shukla , 6 Years ago
Grade 11
anser 1 Answers
Arun

Last Activity: 6 Years ago

Dear Vaibhav
 
T1 = 500 K
T2 = 375 K
Q1 = heat absorbed = 600 k cal
i) efficiency (n) = 1 -(T2/T1) = 125/500 = 0.25
 = 25%
ii) n = W/Q1
W = n* Q1 = 0.25 * 600 = 150 k cal
iii)- 
W = Q1 - Q2
Q2 = Q1 - W
Q2 = 600 - 150 = 450 k cal
 
Regards
Arun (askIITians forum expert)

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