Question

Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures, $T_1\text{and}{T}_2$. The temperature of the hot reservoir of the first engine is $T_1$ and the temperature of the cold reservoir of the second engine is $T_2$. $T$ is the temperature of the sink of the first engine, which is also the source for the second engine. How is $T$ related to $T_1\text{and}{T}_2$, if both the engines perform equal amount of work ?

• A. $T=\frac{2T_1{T}_2}{T_1+T_2}$
• B. $T=\frac{T_1+T_2}{2}$
• C. $T=\sqrt{T_1{T}_2}$
• D. $T=0$

Answer 1

The correct option is B $T=\frac{T_1+T_2}{2}$

Let
$Q_H=\text{Heat absorbed by first engine}$
$Q_L=\text{Heat rejected by first engine}$
$Q_2=\text{Heat rejected by second engine}$

Given that,
Work done by $1^{st}$ engine = work done by $2^{nd}$ engine

$W_1=W_2\Rightarrow Q_H-Q_L=Q_L-Q_2$

$\Rightarrow 2Q_L=Q_H+Q_2$

$\Rightarrow 2=\frac{Q_H}{Q_L}+\frac{Q_2}{Q_L}---\left(1\right)$

For $1^{st}$ Carnot Engine,
$\frac{Q_H}{Q_L}=\frac{T_1}{T}$

For $2^{nd}$ Carnot Engine,
$\frac{Q_L}{Q_2}=\frac{T}{T_2}$

Substituting above expressions in $\left(1\right)$

$\Rightarrow 2=\frac{T_1}{T}+\frac{T_2}{T}$

$\Rightarrow T=\frac{T_1+T_2}{2}$

Hence, option $\left(B\right)$ is correct.