Question

the sink of carnot heat engine is at 300 k and has efficiency 0.4 if the efficiency increased engine is 0.5.find its klevin that temperature of the source should be increased

Answer 1

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

here's the answer...

Efficiency of heat engine n = (T2 - T1)/T2

where T2 is source temperature and T1 is the sink temperature....

For n = 0.4

0.4 = (T2 - T1)/T2

T1/T2 = 0.6

As T1 = 300 K...... T2 = 300/0.6 = 500K

Case 2: for efficiency n to be increased 50% of its original efficiency , i.e. n = 0.4( 1+ 0.5) = 0.6 = 60%

Source temperature can be calculated by the same formula n = (T2 - T1)/T2

0.6 = (T2 - 300)/T2 ………( sink temperature is constant)

T2 = 300/0.4 = 750K

Increase in source temperature

Del T = (750 - 500) = 250 K....

That's your answer....

The source temperature should be increased by 250 K so as to increase the efficiency of heat engine by 50 % of its original efficiency