PHYSICS Homework....380 J of energy extracted from a hot reservoir and 120 J expelled to the cold reservoir at temperature 850 K...?
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(a) How many joules are converted to work? (b) What is the temperature of the hot reservoir in kelvins? Visual aid: http://tinypic.com/r/9r6m93/8
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Answer:
∆U=Qh-|Qc|-|w| And because the heat engine works in a cycle so ∆U=0 (a) : So Qh-|Qc|=|w| ---------> 380-120=260 (b): Efficiency= |w|/Qh And also the other formula for the max efficiency is η=1- (Tc/Th) Which comes from : η = |W|/Qh=(Qh-Qc)/Qh=(Qh/Qh)-(Qc /Qh)=1-(Qc-Qh)=1-(Tc/Th) so we have : |W|/Qh= 260/380=(1-(850/Th)) So Th=2691.67 (which is a little weird)
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