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A reversible power cycle whose thermal efficiency is 40% receives 50 kJ by heat transfer from a...

A reversible power cycle whose thermal efficiency is 40% receives 50 kJ by heat transfer from a hot reservoir at 600 K and rejects energy by heat transfer to a coldreservoir at temperature Tc. Determine the energy rejected in KJ
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Answer #1
Effiiency=1-(Qc/Qh)=1-(Tc/Th)
0.4=1-(Qc/50)
Qc=30 kJ

1-Tc/600=0.4
Tc=360 K
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Answer #2
nth=40 percent

1-(tc/th)=0.4

tc=360k


for reversible engine

1-qc/qh=0.4

qc=30kj
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Answer #3

nth=40 percent

1-(tc/th)=0.4

tc=360k

by first law of thermodynamics

for reversible engine

1-qc/qh=0.4

qc=30kj

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