Question

A system performs 61 kJ of work on its surroundings and releases 107 kJ of heat. What is the change in internal energy of the system?

Answer 1

ΔU = - 168 KJ

Step-by-step explanation:

• ΔU = Q + W.....................first law

∴ W = - 61 KJ........A system performs on its surroundings

∴ Q = - 107 KJ,,,,,,A system release heat

⇒ ΔU = - 107 KJ - 61 KJ = - 168 KJ