Question

What is the change in internal energy if 70J of heat is added to a system and the system does 30J of work on the surroundings

Answer 1

Answer: Δ 100 J

Step-by-step explanation:

Since, Q = 70 J (heat added to the system) and W = -30 J (work done by the system on the surroundings), we;

Substitute these values into the equation, we have:

ΔU = 70 J - (-30 J)

ΔU = 70 J + 30 J

ΔU = 100 J