Question

an electric heater suppliers heat to a systematic at of hundred what if the system performs work at a rate of 75 June at what rate is the internal energy increasing ​

Answer 1

heat energy supplied per second by the heater

Q=100W=100J/s

work done by the system = 75J/s

rate of change in internal energy = ?

according to the first law of thermodynamics,

U=Q-W

U=25W

Answer 2

∆U = 100W = 100J/s

∆W = 75J/s

increase in internal energy, ∆U = ?

∆Q = ∆U + ∆W

•°• ∆U = ∆Q - ∆W = 100 - 75

= 25J/s