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Standard XII

Physics

Carnot Cycle

Two Carnot en...

Question

Two Carnot engines A and B are operated in succession. The first one, A receives heat from a source at $T_{1}$ = 800 K and rejects to sink at $T_{2}$ K. The second engine B receives heat rejected by the first engine and rejects to another sink at $T_{3}$ = 300 K. If the work outputs of two engines are equal, then the value of $T_{2}$ is

100K

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300K

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550K

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700K

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Solution

The correct option is C. 550 K.
The efficiency of a Carnot engine is given by,
$η = \frac{T_{S o u r c e} - T_{S i n k}}{T_{S o u r c e}}$
$\Rightarrow η_{A} = \frac{T_{1} - T_{2}}{T_{1}} = \frac{W_{A}}{Q_{1}}$ and
$η_{B} = \frac{T_{2} - T_{3}}{T_{2}} = \frac{W_{B}}{Q_{2}}$
$∴ \frac{Q_{1}}{Q_{2}} = \frac{T_{1}}{T_{2}} \times \frac{T_{2} - T_{3}}{T_{1} - T_{2}}$
$∵ W_{A} = W_{B} and \frac{Q_{1}}{Q_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{T_{1}}{T_{2}} = \frac{T_{1}}{T_{2}} \times \frac{T_{2} - T_{3}}{T_{1} - T_{2}}$
$∴ T_{2} = \frac{T_{1} + T_{3}}{2} = \frac{800 + 300}{2} = 550 K$ .

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The correct option is C. 550 K.The efficiency of a Carnot engine is given by,η=TSource−TSinkTSource⇒ηA=T1−T2T1=WAQ1 andηB=T2−T3T2=WBQ2∴Q1Q2=T1T2×T2−T3T1−T2∵ WA=WB and Q1Q2=T1T2T1T2=T1T2×T2−T3T1−T2 ∴T2=T1+T32=800+3002=550 K.