Q112 / 180 · NEET 2013 (Karnataka) · Physics

Two Carnot engines $\mathrm{A}$ and $\mathrm{B}$ are operated in series. The engine $A$ receives heat from the source at temperature $T_1$ and rejects the heat to the sink at temperature $T$. The second engine $\mathrm{B}$ receives the heat at temperature $\mathrm{T}$ and rejects to its sink at temperature $T_2$. For what value of $T$ the efficiencies of the two engines are equal:

  • A
    $\frac{T_2-T_2}{2}$
  • B
    $T_1 T_2$
  • Correct answer
    C
    $\sqrt{T_1 T_2}$
  • D
    $\frac{T_1+T_2}{2}$

Solution

We know that efficiency of carnot engine $\eta=1-\frac{T_2}{T_1}$ Where $T_1=$ temperature of source $T_2=$ temp of $\operatorname{sink}$ For engine $A, n_{\mathrm{A}}=1-\frac{T}{T_1}$ an engine $B, n_{\mathrm{B}}=1-\frac{T_2}{T}$ From question $\begin{array}{l} n_A =n_B \\ 1-\frac{T}{T_1} =1-\frac{T_2}{T} \\ \Rightarrow \frac{T}{T_1}=\frac{T_2}{T} =\mathrm{T}^2 \\ \Rightarrow \mathrm{T}^2 =\mathrm{T}_1 \mathrm{~T}_2 \\ \Rightarrow \mathrm{T} =\sqrt{T_1 T_2} \end{array}$

Hence, the correct answer is option: 'C'.

Relevant Questions