Q. A system is provided with 200 cal of heat and the work done by the system on the surrounding is 40 J. Then its internal energy
A) Increases by 600 J
B) Decreases by 800 J
C) Increases by 800 J
D) Decreases by 50 J
Explanation
(c) Given, $dQ = + 200\ cal = 200 \times 4.2 = 840\ J$
$dW = + 40J$
From first law of thermodynamics
$dQ = dU + dW$
$dU = dQ - dW$
=840-40=800 J
So, the internal energy of the system increase by 800 J.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.