Two heat engines are operated in series. The first one receives heat at 800 K and rejects to a reservoir at temperature TK. The second one receives heat rejected by first one and rejects heat to a sink at $- 73^{0} C$ . The temperature T, if the two outputs of two engines are equal

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200 K

300 K

400 K

500 K

answer is D.

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Detailed Solution

$\begin{array}{l} F o r f i r s t e n g i n e, e f f i c i e n c y o f h e a t e n g i n e = η = 1 - \frac{T_{2}}{T_{1}^{}} = 1 - \frac{Q_{2}}{Q_{1}} \\ \Rightarrow \frac{Q_{2}}{Q_{1}} = \frac{T_{2}}{T_{1}}; g i v e n T_{1} = 800 K = t e m p e r a t u r e o f s o u r c e; T_{2} = T = t e m p e r a t u r e o f \sin k, s u b s t i t u t e \\ \Rightarrow \frac{Q_{2}}{Q_{1}} = \frac{T}{800} - - - (1) \\ Q_{1} = h e a t a b s o r b e d; Q_{2} = h e a t r e j e c t e d \\ η_{1} = \frac{W_{1}}{Q_{1}} f o r f i r s t e n g i n e \\ \Rightarrow W_{1} = Q_{1} . η_{1} \\ \Rightarrow W_{1} = Q_{1} [1 - \frac{T_{2}}{T_{1}^{}}] \\ \Rightarrow W_{1} = Q_{1} [1 - \frac{T}{800}] - - - (2) \end{array}$

$\begin{array}{l} f o r s e c o n d e n g i n e \\ η_{2} = \frac{W_{2}}{h e a t a b s o r b e d b y s e c o n d e n g i n e Q_{2}^{}} = 1 - \frac{T_{2}^{1}}{T_{1}^{1}} \\ H e a t a b s o r b e d b y I I e n g i n e = H e a t r e j e c t e d b y I e n g i n e = Q_{2}^{} \\ t e m p e r a t u r e o f \sin k f o r I I e n g i n e = - 73 + 273 = 200 K = T_{2}^{1} \\ \Rightarrow W_{2} = Q_{2}^{} [1 - \frac{T_{2}^{1}}{T_{1}^{1}}] \\ W_{2} = Q_{2}^{} [1 - \frac{200}{T}] - - - (3) \end{array}$

$\begin{array}{l} g i v e n W_{2} = W_{1} (∵ o u t p u t s a r e s a m e); s u b s t i t u t e e q u a t i o n s (2) a n d (3) \\ Q_{2}^{} [1 - \frac{200}{T}] = Q_{1} [1 - \frac{T}{800}] \\ \frac{Q_{2}}{Q_{1}} [1 - \frac{200}{T}] = [1 - \frac{T}{800}]; s u b s t i t u t e e q u a t i o n (1) \\ \frac{T}{800} [1 - \frac{200}{T}] = 1 - \frac{T}{800} \\ \frac{T}{800} - \frac{1}{4} = 1 - \frac{T}{800} \\ \Rightarrow \frac{2 T}{800} = \frac{5}{4} \Rightarrow T = 500 K \end{array}$