Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
A perfect carnot engine utilizes an ideal gas the source temperature is500K and sink temperature is 375K. If the engine takes 600k cal per cyclefrom the source, calculate(i) The efficiency of engine(ii) Work done per cycle(iii) Heat rejected to sink per cycle.
Dear Vaibhav T1 = 500 K T2 = 375 K Q1 = heat absorbed = 600 k cal i) efficiency (n) = 1 -(T2/T1) = 125/500 = 0.25 = 25% ii) n = W/Q1 W = n* Q1 = 0.25 * 600 = 150 k cal iii)- W = Q1 - Q2 Q2 = Q1 - W Q2 = 600 - 150 = 450 k cal Regards Arun (askIITians forum expert)
Find solution of question on doubt solving app Instasolv
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -