KEAM 2021 Physics and Chemistry Solved Paper

Given, temperature of source,T1=400 K
Initial efficiency, η=25%
Final efficiency, η′=50%
As, efficiency of the Carnot cycle,
η=1−‌

T2

T1

where, T2 is the temperature of the sink.
Let T2=T
∴‌‌ Initial efficiency, η=1−‌

T

T1

=25%
Substituting the given values, we get
‌

25

100

=1−‌

T

400

4×25=400−T
⇒ T=400−100=300K
Final efficiency, η′=1−‌

T

T1′

=50%
⇒ 1−‌

300

T1′

=‌

50

100

⇒‌‌100(T1′−300)=T1′×50
or 100T1′−30000=50T1′
or 50T1′‌=30000
⇒‌T1′‌=600K