JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Heat & Thermodynamics

Two Carnot engines

A

and operate in series such that engine

A

absorbs heat at

T_{1}

and rejects heat to a sink at temperature T. Engine B absorbs half of the heat rejected by Engine A and rejects heat to the sink at

T_{3}

. When work done in both the cases is equal, the value of

T

is:

Answer: (C)

Solution

For the Carnot engine A working between

T_{1}

and

T

\frac{Q _{2}}{Q _{1}} = \frac{T}{T _{1}}

, so

W_{A} = Q_{1} - Q_{2} = Q_{1} (1 - \frac{T}{T _{1}})

Engine B absorbs

\frac{Q _{2}}{2}

T

and rejects at

T_{3}

, so

W_{B} = \frac{Q _{2}}{2} (1 - \frac{T _{3}}{T}) = \frac{Q _{1} T}{2 T _{1}} (1 - \frac{T _{3}}{T}) = \frac{Q _{1} ( T - T _{3} )}{2 T _{1}}

Equating the two works and dividing by

Q_{1}

1 - \frac{T}{T _{1}} = \frac{T - T _{3}}{2 T _{1}}

2 T_{1} - 2 T = T - T_{3} \Rightarrow 3 T = 2 T_{1} + T_{3}

The sink temperature of the first engine is

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

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Single correct option