Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures, $T_{1}$ and $T_{2}$. The temperature of the hot reservoir of the first engine is $\mathrm{T}_{1}$ and the temperature of the cold reservoir of the second engine is $T_{2}$. T is temperature of the sink of first engine which is also the source for the second engine. How is $T$ related to $T_{1}$ and $T_{2}$, if both the engines perform equal amount of work?

(A) $\mathrm{T}=\sqrt{\mathrm{T}_{1} \mathrm{~T}_{2}}$

(C) $T=\frac{2 T_{1} T_{2}}{T_{1}+T_{2}}$

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Heat & Thermodynamics

Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures,

T_{1}

and

T_{2}

. The temperature of the hot reservoir of the first engine is

T_{1}

and the temperature of the cold reservoir of the second engine is

T_{2}

. T is temperature of the sink of first engine which is also the source for the second engine. How is

T

related to

T_{1}

and

T_{2}

, if both the engines perform equal amount of work?

Answer: (B)

Let

Q_{1}

be the heat the first engine draws from

T_{1}

Q

the heat it dumps at the intermediate temperature

T

(which the second engine then takes in), and

Q_{2}

the heat the second engine rejects at

T_{2}

. Equal work from the two engines means

Q_{1} - Q = Q - Q_{2}

, so

2 Q = Q_{1} + Q_{2}

. For Carnot engines heat is proportional to reservoir temperature, so

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

and

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

. Dividing the work condition by

Q

gives

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

, hence

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

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option