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Heat and Work

Two Carnot en...

Question

Two Carnot engines $A$ and $B$ operate in series such that engine $A$ absorbs heat at $T_{1}$ and rejects heat to a sink at temperature $T$ . Engine $B$ absors half of the heat rejected by engine $A$ and rejects heat to the sink at $T_{3}$ . When workdone in both the cases is equal, the value of $T$ is :

\frac{2}{3} T_{1} + \frac{3}{2} T_{3}

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\frac{1}{3} T_{1} + \frac{2}{3} T_{3}

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\frac{3}{2} T_{1} + \frac{1}{3} T_{3}

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\frac{2}{3} T_{1} + \frac{1}{3} T_{3}

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Solution

The correct option is D $\frac{2}{3} T_{1} + \frac{1}{3} T_{3}$
The given situation can be represented as shown

Work done by engine $A$ is given as
$W_{A} = 1 - \frac{Q_{2}}{Q_{1}} = 1 - \frac{T}{T_{1}} \Rightarrow \frac{Q_{2}}{Q_{1}} = \frac{T}{T_{1}} . . (1)$
Also, work done by engine $B$ is given as
$W_{B} = 1 - \frac{Q_{3}}{(\frac{Q_{2}}{2})} = 1 - \frac{T_{3}}{T} \Rightarrow \frac{2 Q_{3}}{Q_{2}} = \frac{T_{3}}{T} . . (2)$
Now, it is given as
$W_{A} = W_{B}$
$\Rightarrow Q_{1} - Q_{2} = \frac{Q_{2}}{2} - Q_{3}$
$\Rightarrow \frac{2 Q_{1}}{Q_{2}} + \frac{2 Q_{3}}{Q_{2}} = 3$
From eq $(1)$ and $(2)$ we have
$\Rightarrow \frac{2 T_{1}}{T} + \frac{T_{3}}{T} = 3$
$\Rightarrow \frac{2 T_{1}}{3} + \frac{T_{3}}{3} = T$
Hence, option (d) is correct.

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