A carnot engine has efficiency 40 % and it absorbs heat from source at 500^@ K. How much should the temperature of source be increased so as to increase its efficiency by 60 % of original efficiency ?

A carnot engine has efficiency 40 % and it absorbs heat from source at

1.	A carnot engine has efficiency 40 % and it absorbs heat from source at 500^@ K. How much should the temperature of source be increased so as to increase its efficiency by 60 % of original efficiency ?
Answer» 1200 K 750 K 600 K Cannot increases 50% EFFICIENCY of carnot ENGINE. Solution :`eta_1`=40% =0.4, `T_1`=500 K , `eta_2`=60% = 0.6 , `T._1`=? `eta_1=1-T_2/T_1` `T_2/T_1=1-eta_1` `T_2/500=1-0.4` `THEREFORE T_2=0.6xx500` `therefore T_2`=300 K Now `eta_2=1-T_2/(T._1)` `T_2/(T._1)=1-eta_2` `300/(T._1)=1-0.6` `300/(T._1)=0.4` `therefore T._1=300/0.4` `therefore T._1`=750 K