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?

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

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

$T = 0$

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

answer is B.

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Detailed Solution

$Q_{H} :$ Heat input to 1st engine
$Q :$ Heat rejected from 1st engine
$Q_{L} :$ Heat rejected from IInd engine

$η = 1 - \frac{T_{\sin k}}{T_{s o u r c e}} - - - (1)$

$η = \frac{W}{Q_{1}} = \frac{Q_{a b s o r b e d} - Q_{r e j e c t e d}}{Q_{a b s o r b e d}} = 1 - \frac{Q_{r e j e c t e d}}{Q_{a b s o r b e d}} - - - (2)$

$e q u a t i o n (1) = E q u a t i o n (2) \frac{T_{\sin k}}{T_{s o u r c e}} = \frac{Q_{r e j e c t e d}}{Q_{a b s o r b e d}}$

$A s p e r f i r s t e n g i n e, \frac{Q}{Q_{H}} = \frac{T}{T_{1}}$

$\Rightarrow Q_{H} = Q \frac{T_{1}}{T}$

$A s p e r s e c o n d e n g i n e, \frac{Q_{L}}{Q} = \frac{T_{2}}{T}$

$\Rightarrow Q_{L} = Q \frac{T_{2}}{T}$
Now, work done by 1st engine = work done by 2nd engine

$\Rightarrow Q_{H} - Q = Q - Q_{L} \Rightarrow 2 Q = Q_{H} + Q_{L}$
$\Rightarrow 2 = \frac{T_{1}}{T} + \frac{T_{2}}{T} \Rightarrow T = \frac{T_{1} + T_{2}}{2}$