 # 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, `
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, $T_{1}$ 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 temperature of the sink of first engine which is also the source for the second engine. How is $T$ related to $T_{1}$ and $T_{2}$, if both the engines perform equal amount of work ?

1. $T=\frac{2 T_{1} T_{2}}{T_{1}+T_{2}}$

2. $T=\frac{T_{1}+T_{2}}{2}$

3. $T=\sqrt{T_{1} T_{2}}$

4. $\mathrm{T}=0$

Correct Option: , 2

Solution:

(2) Let $Q_{H}=$ Heat taken by first engine

$Q_{L}=$ Heat rejected by first engine

$Q_{2}=$ Heat rejected by second engine

Work done by $1^{\text {st }}$ engine $=$ work done by $2^{\text {nd }}$ engine

$W=Q_{H}-Q_{L}=Q_{L}-Q_{2} \Rightarrow 2 Q_{L}=Q_{H}+Q_{2}$

$2=\frac{\theta_{H}}{\theta_{L}}+\frac{\theta_{2}}{\theta_{L}}$

Let $\mathrm{T}$ be the temperature of cold reservoir of first engine. Then in carnot engine.

$\frac{Q_{H}}{Q_{L}}=\frac{T_{1}}{T}$ and $\frac{Q_{L}}{Q_{2}}=\frac{T}{T_{2}}$

$\Rightarrow 2=\frac{T_{1}}{T}+\frac{T_{2}}{T}$                       using (i)

$\Rightarrow 2 T=T_{1}+T_{2} \Rightarrow T=\frac{T_{1}+T_{2}}{2}$