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JEE Main 202127 JulyEvening Shift
Exam
JEE Main
Year
2021
Date
27 July
Shift
Evening Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2021 · 27 July · Evening Shift · Physics · Thermodynamics

Two Carnot engines A and B operate in series such that engine A...

Two Carnot engines AA and BB operate in series such that engine A absorbs heat at T1{T}_{1} and rejects heat to a sink at temperature TT. Engine BB absorbs half of the heat rejected by EngineAA and rejects heat to the sink at T3{T}_{3}. When workdone in both the cases is equal, to value of TT is :Two Carnot engines A and B operate in series such that engine A absorbs heat at T _ 1 and rejects heat to a sink at

Options

  1. A23T1+32T3\dfrac{2}{3}{T}_{1} + \dfrac{3}{2}{T}_{3}
  2. B13T1+23T3\dfrac{1}{3}{T}_{1} + \dfrac{2}{3}{T}_{3}
  3. C32T1+13T3\dfrac{3}{2}{T}_{1} + \dfrac{1}{3}{T}_{3}
  4. D23T1+13T3\dfrac{2}{3}{T}_{1} + \dfrac{1}{3}{T}_{3}

Correct answer

D. 23T1+13T3\dfrac{2}{3}{T}_{1} + \dfrac{1}{3}{T}_{3}

Step-by-step solution

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

WA=1−Q2Q1=1−TT1⇒Q2Q1=TT1{W}_{A} = 1 - \dfrac{{Q}_{2}}{{Q}_{1}} = 1 - \dfrac{T}{{T}_{1}} ⇒ \dfrac{{Q}_{2}}{{Q}_{1}} = \dfrac{T}{{T}_{1}}

WB=1−Q3(Q2/2)=1−T3T⇒2Q3Q2=T3T{W}_{B} = 1 - \dfrac{{Q}_{3}}{\left({Q}_{2} / 2\right)} = 1 - \dfrac{{T}_{3}}{T} ⇒ \dfrac{2{Q}_{3}}{{Q}_{2}} = \dfrac{{T}_{3}}{T}

Now, WA=WB{W}_{A} = {W}_{B}

Q1−Q2=Q22−Q3{Q}_{1} - {Q}_{2} = \dfrac{{Q}_{2}}{2} - {Q}_{3}

⇒2Q1Q2+2Q3Q2=3⇒ \dfrac{2{Q}_{1}}{{Q}_{2}} + \dfrac{2{Q}_{3}}{{Q}_{2}} = 3

⇒2T1T+T3T=3⇒ \dfrac{2{T}_{1}}{T} + \dfrac{{T}_{3}}{T} = 3

2T13+T33=T\dfrac{2{T}_{1}}{3} + \dfrac{{T}_{3}}{3} = T

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