A carnot.engine whose sink is at 300 K has an efficiency of 40 %. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency ?

A carnot.engine whose sink is at 300 K has an efficiency of 40 %. By h

1.	A carnot.engine whose sink is at 300 K has an efficiency of 40 %. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency ?
Answer» Solution :EFFICIENCY of a carnot engine, `eta=(T_1-T_2)/T_1` , where `T_1`=TEMPERATUREOF the SOURCE `T_2`=temperature of the sink `THEREFORE 0.40=(T_1-300)/T_1` `therefore T_1-300=0.40T_1` `therefore 0.6T_1=300 rArr T_1=300/0.6=3000/6`=500 K Now, 50% efficiency increases , `therefore 0.60=(T_1-300)/T_1 rArr T_1-300=0.6T_1` `therefore 0.4T_1=300 rArr T_1 =300/0.4=(300xx10)/4` =750 Increase in temperature of source =750-500=250 K