Two Carnot engines operated in series with equal work… | Thermodynamics

JEE MAIN 2019

09-01-19 S2

Question

Two Carnot engines A and B are operated in series. The first one, A, receives heat at $T_1(=600 \mathrm{~K})$ and rejects to a reservoir at temperature $T_2$. The second engine B receives heat rejected by the first engine and, in turn, rejects to a heat reservoir at $T_3(=400 \mathrm{~K})$. Calculate the temperature $T_2$ if the work outputs of the two engines are equal (MFST-1)

Select the correct option:

300 K

400K

600K

500K

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✗ Incorrect. Try again or view the solution.

Solution

$\begin{aligned} & \left.\frac{\mathrm{Q}_1}{\mathrm{~T}_1}=\frac{\mathrm{Q}_2}{\mathrm{~T}_2}=\frac{\mathrm{Q}_3}{\mathrm{~T}_3} \right\rvert\, \\ & \mathrm{W}_1=\mathrm{Q}_1-\mathrm{Q}_2 \quad \Rightarrow \quad \mathrm{~W}_2=\mathrm{Q}_2-\mathrm{Q}_3 \\ & \text { Given } \mathrm{W}_1=\mathrm{W}_2 \\ & \Rightarrow 2 \mathrm{~T}_2=\mathrm{T}_1+\mathrm{T}_3 \\ & \mathrm{~T}_2=\frac{\mathrm{T}_1+\mathrm{T}_3}{2} \\ & \mathrm{~T}_2=500 \mathrm{~K}\end{aligned}$

Question Tags

JEE Main

Physics

Easy

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