Q. An engine has an efficiency of 1/3. the amount of work this engine can perform per kilocalorie of heat input is
A) 1400 cal
B) 700 cal
C) 700 J
D) 1400 J
Explanation
(a) $\eta = \frac{W}{Q_{1}}\ \Rightarrow W = \eta\ Q_{1} = \frac{1}{3} \times 1000\ cal = \frac{1000}{3} \times 4.2 = 1400\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.