Q. In the certain process, 400 cal of heat are supplied to a system and at the same time 105 J of mechanical work was done on the system. The increase in its internal energy is
A) 20 cal
B) 303 cal
C) 404 cal
D) 425 cal
Explanation
(d) $dQ = 400\ cal,\ dW = - 105\ J$
$= 105/4.2\ cal = - 25\ cal\ ;dU = ?$
$dU = dQ - dW$
$dU = 400 - ( - 25) = 425$ cal
Note $dW$ is negative because work is done on the system
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.