A heat engine operating between 227 deg c and 27 deg c absorbs 1 kcal of heat from the 227 deg c reservoir per cycle. Calculate (1) the amount of heat discharged into the low temperature reservoir. (2) the amount of work done per cycle. (3) the efficiency of cycle.
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Answer:
0.4 kcal, 0.6 kcal, 40%
Explanation:
Calculation of work done:
T
′
=227+273=500 K
T=27+273=300 K
q
′
=1 1 kcal
q
′
w
=
T
′
T
′
−T
w=q(
T
′
T
′
−T
)
Putting the various values in the above reaction
w=1(
500
500−300
)=1×
500
200
=0.4 kcal.
Calculation of heat discharged q
q
′
−q=w
1−q=0.4
q=0.6 kcal
Calculation of efficiency
η=
q
′
w
=0.4 kcal or 40%
Answered by
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Calculation of work done:
Calculation of work done:T′=227+273=500 K
Calculation of work done:T′=227+273=500 KT=27+273=300 K
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