Question

Three Carnot engines operate in series between a heat source at a temperature $T_1$ and a heat sink at temperature $T_4$ (see figure). There are two other reservoirs at temperature $T_2$ and $T_3$, as shown, with $T_1>T_2>T_3>T_4$. The three engines are equally efficient if :

• A. $T_2={\left(T_1{T_4}^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.};T_3={({T_1}^2{T}_4)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$
• B. $T_2={\left(T_1{T_4}^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.};T_3={({T_1}^2{T}_4)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$
• C. $T_2={\left({T_1}^2{T}_4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.};T_3={(T_1{T_4}^2)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$
• D. $T_2={\left({T_1}^3{T}_4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.};T_3={(T_1{T_4}^3)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$

Answer 1

The correct option is C

$T_2={\left({T_1}^2{T}_4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.};T_3={(T_1{T_4}^2)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Step 1: Given the information

The heat source at a temperature $T_1$

Heat sink at the temperature $T_4$

Two other reservoirs at temperature $T_2$and $T_3$

All three engines have the same efficiency($t_1,t_2,t_3$).

Step 2: Calculate whether the three engines are equally efficient.

The efficiency of three engines is,

$t_1=1-\left(\frac{T_2}{T_1}\right)$

$\Rightarrow t_2=1-\left(\frac{T_3}{T_2}\right)$

$\Rightarrow t_3=1-\left(\frac{T_4}{T_3}\right)$

So, $\frac{T_2}{T_1}=\frac{T_3}{T_2}=\frac{T_4}{T_3}$

$T_2=\sqrt{\left(T_1{T}_3\right)}$

$\Rightarrow T_2=\sqrt{T_1\sqrt{\left(T_2{T}_4\right)}}$ $\left[\because T_3=\sqrt{\left(T_2{T}_4\right)}\right]$

$\begin{array}{l}{T}_2=\left({T_1}^{\frac{1}{2}}{\left(T_2{T}_4\right)}^{\frac{1}{4}}\right)\\ T_2^1={T_1}^{\frac{1}{2}}{\left(T_2{T}_4\right)}^{\frac{1}{4}}\\ \frac{T_2^1}{T_2^{\frac{1}{4}}}={T_1}^{\frac{1}{2}}{T}_2^{\frac{1}{4}}\\ T_2^{1-\frac{1}{4}}={T_1}^{\frac{1}{2}}{T}_2^{\frac{1}{4}}\\ T_2^{\frac{3}{4}}={T_1}^{\frac{1}{2}}{T}_4^{\frac{1}{4}}\\ T_2^{\frac{3}{4}\times \frac{4}{3}}={T_1}^{\frac{1}{2}\times \frac{4}{3}}{T}_4^{\frac{1}{4}\times \frac{4}{3}}\\ T_2={T_1}^{\frac{2}{3}}{T}_4^{\frac{1}{3}}\\ \therefore T_2={\left({T_1}^2{T}_4\right)}^{\frac{1}{3}}\end{array}$

Put value $T_2$ in $T_3=\sqrt{\left(T_2{T}_4\right)}$

$$\begin{gathered} \begin{array}{l}{T}_3={({T_1}^{\frac{2}{3}}{T}_4^{\frac{1}{3}}\times T_4)}^{\frac{1}{2}}\\ T_3=\left({T_1}^{\frac{1}{3}}{T}_4^{\frac{2}{3}}\right)\\ T_3={\left(T_1{T}_4^2\right)}^{\frac{1}{3}}\end{array} \\ \therefore T_3={(T_1{T_4}^2)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \end{gathered}$$

Hence, the correct option is (c).