JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Heat & Thermodynamics

Two Carnot engines A and B are operated in series. The first one, A, receives heat at

T_{1} (= 600 K)

and rejects to a reservoir at temperature

T_{2}

. The second engine B receives heat rejected by the first engine and, in turns, rejects to a heat reservoir at

T_{3} (= 400 K)

. Calculate the temperature

T_{2}

if the work outputs of the two engines are equal:

(A)600 K
(B)400 K
(C)300 K
(D)500 K✓ Correct

Answer: (D)

Solution

Two Carnot engines share the intermediate reservoir at

T_{2}

. Engine A does

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

and rejects

Q_{2}

T_{2}

; engine B takes that

Q_{2}

and does

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

. For Carnot engines the heat drop is proportional to the temperature drop, and setting the two work outputs equal reduces to equal temperature steps,

T_{1} - T_{2} = T_{2} - T_{3}

. Hence

T_{2} = \frac{T _{1} + T _{3}}{2} = \frac{600 + 400}{2} = 500 K

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Heat & Thermodynamics

Solution

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