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

B

Explanation

Efficiency of a carnot engine,
`eta=(T_1-T_2)/T_1` , where `T_1`=temperature of 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