Two carnot engines A and B operate in series such that engine A absorbs heat at T1 and rejects heat to a sink at temperature T. Engine B absorbs 3 th 4 of the heat rejected by Engine A and rejects heat to the sink at T3 . When workdone in both the cases are equal, the value of T is:
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Work done by engine
A is given as
WA=1−(Q2/Q1)=1−(T/T1)
⇒(Q2/Q1)=(T/T1)
(1)
Also, work done by engine
B is given as
WB=1−(Q3/(Q2/2))=1−T3/T
⇒(2Q3/Q2)=(T3/T)
(2)
Now, it is given as
WA=WB
⇒ Q1−Q2=(Q2/2)−Q3
⇒ 2Q1/Q2+2Q3/Q2=3
From eq (1) and (2) we have
⇒(2T1/T)+(T3/T)=3
⇒ (2T1/3)+(T3/3)=T
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