.........(i) Where, T2= Temperature of sink and T1= temperature of source According to first condition, T2=27+273=300K η=40%=0.4 From Eq. (i), 0.4=1−
300
T1
⇒
300
T1
=0.6⇒T1=
300
0.6
=500K ⇒T1=500K Again, when, η=50%=0.5, then 0.5=1−
300
T1′
⇒
300
T1′
=0.5⇒T1′=600K ∴ Increase of source temperature =T′−T1 =600−500=100K