Answer to Question #204097 in Physical Chemistry for Arooba

Question

3000 J of heat is added to a system and 2500 J of work is done by the system. What is the change in internal energy of the system.

Answer 1

$\Delta{U}=Q-W$ ,

where Q is the heat added to the system, and W is the work done by the system. Therefore,

$\Delta{U}=3000J-2500J=500J$

Answer: 500 J