The efficiency of the heat engine working between 327oC and 27oC is to be increased by 10%. The temperature of the source should be increased by
A
52oC
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
67oC
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
37oC
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
77oC
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B67oC The efficiency of the cycle is given as η=1−T2T1. Here T1 is the temperature of the source and T2 is the temperature of the sink. . Thus we get η=1−300600=50%. The efficiency is to be increased by 10%. Thus the new efficiency is 55%. Thus we get 0.55=1−300T1 or T1=3000.45=667K Thus the increase in temperature of the source is 667−600=67oC