Q. If heat given to a system is $6\ kcal$ and work done is $6\ kJ$. Then change in internal energy is
A) $19.1\ kJ$
B) $12.5\ kJ$
C) $25\ kJ$
D) Zero
Explanation
(a) $\mathrm{\Delta}Q = \mathrm{\Delta}U + \mathrm{\Delta}W \Rightarrow \mathrm{\Delta}U = \mathrm{\Delta}Q - \mathrm{\Delta}W$
$= 6 \times 4.18 - 6 = 19.08kJ \approx 19.1kJ$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.