AP EAPCET 23-Aug-2021 Shift 1 Paper

Given, efficiencies η1=η2=η3=η
and temperature, T1>T2>T3>T4
As we know that, in case of Carnot engine,
Efficiency (η)=1−

Tsink

Tsource

∴η1=η=1−

T2

T1

⇒T2=(1−η)T1 ...(i)
Similarly,
T3=(1−η)T2 ...(ii)
T4=(1−η)T3 ...(iii)
Now, by using Eqs. (i), (ii) and (iii), we get
(T12.T4)

1

3

=[(

T2

1−η

)2.(1−η)T3]

1

3

=[

T22

(1−η)

.T3]

1

3

=[(

T22

1−η

)(1−η)T2]

1

3

=[T23]

1

3

=T2
Again, by using Eqs. (i), (ii) and (iii), we get
(T1.T32)

1

3

=[(

T2

1−η

)[(1−η)T3]2]

1

3

=[T2(1−η)T32]

1

3

=[

T3

1−η

(1−η)T32]

1

3

=[T33]

1

3

=T3