A Carnot cycle has a thermal efficiency of 32%. If the work developed is 1000 watt-hour, determine the heat supplied.
Given quantities:
η=0.32 A1=1000 A2−?\eta = 0.32 \space \space \space A_1=1000 \space \space \space A_2-?η=0.32 A1=1000 A2−?
η=A1−A2A1→\eta = \large\frac{A_1-A_2}{A_1} \toη=A1A1−A2→ A2=A1(1−η)=1000(1−0.32)=680A_2 = A_1(1-\eta)=1000(1-0.32)=680A2=A1(1−η)=1000(1−0.32)=680
Answer: 680 watt-hour
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