Thermodynamics Part 3

Given, initial efficiency of engine, η1=

1

4

Let, initial temperature of sin‌k=T2
Final temperature of sin‌k=T2−58°C
Final efficiency of engine, η2=2η1=

2

4

=

1

2

Temperature of source =T1
Since, η=1−

T2

T1

∴η1=

1

4

=1−

T2

T1

⇒

T2

T1

=1−

1

4

⇒

T2

T1

=

3

4

⇒T1=

4

3

T2 ...(i)
andη2=

1

2

=1−

T2−58

T1

⇒

T2−58

T1

=1−

1

2

⇒

T2−58

4 3 T2

=

1

2

[From Eq. (i)]
⇒2T2−116=

4

3

T2
⇒

2T2

3

=116
⇒T2=

116×3

2

=58×3=174°C