KEAM 2014 Physics and Chemistry Paper

By Carnot's ideal heat engine
η=1−‌

T2

T1

where η1=0.2,‌‌η2=0.4
For the first condition
η1=1−‌

T2

T1

0.2=1−‌

T2

T1

or ‌‌0.2=‌

T1−T2

T1

.....(i)
For the second condition
η2=1−‌

T2−50

T1

0.4=‌

T1−(T2−50)

T1

0.4=‌

T1−T2+50

T1

0.4=‌

T1−T2

T1

+‌

50

T1

0.4=0.2+‌

50

T1

[From Eq. (i)]
0.4−0.2=‌

50

T1

T1=‌

50

0.2

T1=250K
Putting the value of T1 in Eq. (i), we get
0.2=‌

T1−T2

T1

T2=T1−0.2T1
T2=250−0.2×250
T2=250−50
T2=200K
So, ‌‌T1=250K,T2=200K