A Carnot’s engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased to increase the efficiency to 50%?

efficiency of carnot engine is n1=1−T2T1⇒40100=1−300T1⇒300T1=1−40100=100-40100=60100=35⇒300T1=35⇒T1=500Knew effficiency is 50%  50100=1−T2T11 T2T11=1−50100=50100=12⇒300T11=12⇒T11=600KΔT1=T11-T1 =600−500=100K= the temperature of source to be increased