Two Carnot engines A and B operate in series such that engine A absorbs heat at T 1 and rejects heat to a…

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 half of the heat rejected by Engine A and rejects heat to the sink at T3. When workdone in both the cases is equal, to value of T is :
  1. 23 T1+32 T3
  2. 13 T1+23 T3
  3. 32 T1+13 T3
  4. 23 T1+13 T3

Solution

WA=1-Q2Q1=1-TT1⇒Q2Q1=TT1

WB=1-Q3Q2/2=1-T3 T⇒2Q3Q2=T3 T

Now, WA=WB

Q1-Q2=Q22-Q3

⇒2Q1Q2+2Q3Q2=3

⇒2 T1 T+T3 T=3

2 T13+T33=T

Asked in: JEE Main 2021 (27 Jul Shift 2)

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