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JEE Main 20207 JanuaryEvening Shift

Exam: JEE Main
Year: 2020
Date: 7 January
Shift: Evening Shift
Subject: Physics
Chapter: Thermodynamics

JEE Main 2020 · 7 January · Evening Shift · Physics · Thermodynamics

Two ideal Carnot engines operate in cascade (all heat given up by...

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?

Options

A $\text{T} = \dfrac{2{\text{T}}_{1}{\text{T}}_{2}}{{\text{T}}_{1} + {\text{T}}_{2}}$
B $\text{T} = \dfrac{{\text{T}}_{1} + {\text{T}}_{2}}{2}$
C $\text{T} = \sqrt{{\text{T}}_{1}{\text{T}}_{2}}$
D $T = 0$

Correct answer

B. $\text{T} = \dfrac{{\text{T}}_{1} + {\text{T}}_{2}}{2}$

Step-by-step solution

{Q}_{1}:

Heat input to

{I}^{st}

engine

Q:

Heat rejected from

{I}^{st}

engine

{Q}_{2}:

Heat rejected from

I{I}^{nd}

engine
Work done by

{I}^{st}

engine = work done by

I{I}^{nd}

engine.

Q - {Q}_{2} = {Q}_{1} - Q

2Q = {Q}_{1} + {Q}_{2}

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

\text{T} = \dfrac{{\text{T}}_{1} + {\text{T}}_{2}}{2}

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