Three Carnot engines operate in series between a heat source at a temperature $\mathrm{T}_{1}$ and a heat sink at temperature $\mathrm{T}_{4}$ (see figure). There are two other reservoirs at temperature $T_{2}$ and $T_{3}$, as shown, with $T_{1}> T_{2}>T_{3}>T_{4}$. The three engines are equally efficient if:

(B) $T_{2}=\left(T_{1}^{2} T_{4}\right)^{1 / 3} ; T_{3}=\left(T_{1} T_{4}^{2}\right)^{1 / 3}$

(A) $\mathrm{T}_{2}=\left(\mathrm{T}_{1} \mathrm{~T}_{4}\right)^{1 / 2} ; \mathrm{T}_{3}=\left(\mathrm{T}_{1}^{2} \mathrm{~T}_{4}\right)^{1 / 3}$

(C) $\mathrm{T}_{2}=\left(\mathrm{T}_{1} \mathrm{~T}_{4}^{2}\right)^{1 / 3} ; \mathrm{T}_{3}=\left(\mathrm{T}_{1}^{2} \mathrm{~T}_{4}\right)^{1 / 3}$

(D) $\mathrm{T}_{2}=\left(\mathrm{T}_{1}^{3} \mathrm{~T}_{4}\right)^{1 / 4} ; \mathrm{T}_{3}=\left(\mathrm{T}_{1} \mathrm{~T}_{4}^{3}\right)^{1 / 4}$

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Heat & Thermodynamics

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Three Carnot engines operate in series between a heat source at a temperature

T_{1}

and a heat sink at temperature

T_{4}

(see figure). There are two other reservoirs at temperature

T_{2}

and

T_{3}

, as shown, with

T_{1} > T_{2} > T_{3} > T_{4}

. The three engines are equally efficient if:

(A) $T_{2} = (T_{1} T_{4})^{1/2}; T_{3} = (T_{1}^{2} T_{4})^{1/3}$
(B) $T_{2} = (T_{1}^{2} T_{4})^{1/3}; T_{3} = (T_{1} T_{4}^{2})^{1/3}$ ✓ Correct
(C) $T_{2} = (T_{1} T_{4}^{2})^{1/3}; T_{3} = (T_{1}^{2} T_{4})^{1/3}$
(D) $T_{2} = (T_{1}^{3} T_{4})^{1/4}; T_{3} = (T_{1} T_{4}^{3})^{1/4}$

Answer: (B)

Solution

Three engines in series are equally efficient only when each sheds the same fractional temperature, so

1 - \frac{T _{2}}{T _{1}} = 1 - \frac{T _{3}}{T _{2}} = 1 - \frac{T _{4}}{T _{3}}

, which forces

\frac{T _{2}}{T _{1}} = \frac{T _{3}}{T _{2}} = \frac{T _{4}}{T _{3}}

. The four temperatures thus form a geometric progression with common ratio

(T_{4} / T_{1})^{1/3}

, giving

T_{2} = T_{1}^{2/3} T_{4}^{1/3} = (T_{1}^{2} T_{4})^{1/3}

and

T_{3} = T_{1}^{1/3} T_{4}^{2/3} = (T_{1} T_{4}^{2})^{1/3}

. Equal efficiency is really the statement that the intermediate reservoirs are the geometric means splitting the overall temperature ratio into three equal steps.

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Heat & Thermodynamics

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