Question

An electric heater supplies heat to a system at a rate of 100 W. If system performs work at a rate of 75 Joules per second. At what rate is the internal energy increasing?

Answer 1

Given ∆Q = 100w = 100J/*

∆W = 75 J/*

∆Q = ∆U + ∆W

∆U = ∆Q – ∆W = 100 – 75 = 25J/*