In a certain process, 675 J of heat is absorbed by a system while 290 J of work is done on the system. What is the change in internal energy for the process?

As we know that if the quantity of heat transferred from the surrounding to the system is q and work done in the process is w, then the change in internal energy,

ΔU = q + w
where heat absorbed q = 675 J and
work done on the system, w = 290 J

Therefore change in internal energy ∆U will be equal to:
ΔU = 675 J + 290 J = 965 J.

Category: First Law of Thermodynamics

More Questions

Leave a Reply

Copyright © All rights reserved. TheBigger.com | Privacy Policy | Contact Us | Copyright Policy | Useful Resources