1.

A Carnot engine, having an efficiency of eta=1//10 as heat engine, is used as a refrigerator. If the work done on the system is 10J, the amount of energy absorbed from the reservoir at lower temperature is

Answer»

99 J
90 J
1 J
100 J

Solution :Here `ETA=1/10 , W=10 J theta_2 = ? `
`beta=(1-eta)/eta=(1-1//10)/(1//10)=9`
As ` beta = theta_2 / W " or " theta_2 = 90J`


Discussion

No Comment Found

Related InterviewSolutions