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NEET 2013
Exam
NEET
Year
2013
Subject
Physics
Chapter
Thermodynamics

NEET 2013 · Physics · Thermodynamics

Two Carnot engines A and B are operated in series. The engine A...

Two Carnot engines A\mathrm{A} and B\mathrm{B} are operated in series. The engine AA receives heat from the source at temperature T1T_1 and rejects the heat to the sink at temperature TT. The second engine B\mathrm{B} receives the heat at temperature T\mathrm{T} and rejects to its sink at temperature T2T_2. For what value of TT the efficiencies of the two engines are equal:

Options

  1. AT2−T22\frac{T_2-T_2}{2}
  2. BT1T2T_1 T_2
  3. CT1T2\sqrt{T_1 T_2}
  4. DT1+T22\frac{T_1+T_2}{2}

Correct answer

C. T1T2\sqrt{T_1 T_2}

Step-by-step solution

We know that efficiency of carnot engine η=1−T2T1\eta=1-\frac{T_2}{T_1} Where T1=T_1= temperature of source T2=T_2= temp of sink⁡\operatorname{sink} For engine A,nA=1−TT1A, n_{\mathrm{A}}=1-\frac{T}{T_1} an engine B,nB=1−T2TB, n_{\mathrm{B}}=1-\frac{T_2}{T} From question nA=nB1−TT1=1−T2T⇒TT1=T2T=T2⇒T2=T1 T2⇒T=T1T2\begin{array}{l} n_A =n_B \\ 1-\frac{T}{T_1} =1-\frac{T_2}{T} \\ \Rightarrow \frac{T}{T_1}=\frac{T_2}{T} =\mathrm{T}^2 \\ \Rightarrow \mathrm{T}^2 =\mathrm{T}_1 \mathrm{~T}_2 \\ \Rightarrow \mathrm{T} =\sqrt{T_1 T_2} \end{array}
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