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.
0.4 kcal, 0.6 kcal, 60%
0.4 kcal, 0.6 kcal, 40%
0.7 kcal, 0.4 kcal, 40%
0.6 kcal, 0.4 kcal, 40%
A
0.4 kcal, 0.6 kcal, 40%
B
0.6 kcal, 0.4 kcal, 40%
C
0.4 kcal, 0.6 kcal, 60%
D
0.7 kcal, 0.4 kcal, 40%
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Solution
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Calculation of work done: T′=227+273=500 K T=27+273=300 K q′=1 1 kcal wq′=T′−TT′ w=q(T′−TT′) Putting the various values in the above reaction w=1(500−300500)=1×200500=0.4 kcal.
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