Q. What is $\mathrm{\Delta}E$ for system that does 500 cal of work on surrounding and 300 cal of heat is absorbed by the system?
A) -200 cal
B) -300 cal
C) +200 cal
D) +300 cal
Explanation
(a)
From first law of thermodynamic.
$\mathrm{\Delta}E = q + W\ Given,\ q = + 300$ cal ($\because$Heat is absorbed)
$W = - 500$ cal
($\because$Work is done on surroundings)
$\therefore$ $\mathrm{\Delta}E = q + W = 300 + ( - 500)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - 200$ cal
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.