AP EAPCET 25-Aug-2021 Shift 2 Paper

Given, temperature, T1=500K,T2=300K
Efficiency of the engine, η=1−

T2

T1

.
⇒ η=1−

300

500

⇒ η=

2

5

=0.4
For design A, work done =150J
Heat input =1000J
Efficiency of design A=

Work‌done

Heat‌input

=

150

1000

=0.15
For design B, work done =450J
Heat input =1000J
Efficiency of design B=

Work‌done

Heat‌input

=

450

1000

=0.45
For design C, work done =300J
Heat input =1000J
Efficiency of design C=

Work‌done

Heat‌input

Hence, design B is most suitable as its efficiency is almost equal to efficiency of given engine.