Question

A system absorb 6 KJ of heat and does 1.5 kJof work on its surrounding the change in internal energy is

Answer 1

ΔQ + ΔW = ΔU

since the heat is absorbed by the system ΔQ is positive.

And work is done on the surroundings ΔW is negative.

6 - 1.5 = ΔU

ΔU = 4.5 kJ