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.

Question

Arun · Answer

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)

Thank you for registering.

Register Now and Win Upto 25% Scholorship for a Full Academic Year !

Enter Your Details

Registration done!

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.

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.

1 Answers

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Thermal Physics

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship