A heat engine operates between 400 K and 800K. In order to increase the efficiency to 60%, the temperature of source has to be

efficiency of heat engine =η=1-T2T1; here T1,T2 are temperatures of source and sink;givenT1=800K ,T2=400K; efficiency required is 60%⇒η=60100then new source temperature T11=?   substitute in above equation⇒60100=1−400T11⇒400T11=1−35⇒400T11=25⇒T11=1000K∴ΔT1=T11−T1=1000−800=200Ktemperature of source has to be increased by 200K