An electric boiler supplies heat to . . .

Question : An electric boiler supplies heat to the system at the rate of 100 joule per second. If the system performs work at the rate of 75 joule per second, at what rate is the internal energy increasing?

Doubt by Saif

Solution : 

dQ/dt = 100 J/s
dW/dt = 75 J/s

dU/dt = ?

We know, According to the first law of thermodynamics

Q=U+W
differentiation both sides w.r.t. time
dQ/dt = dU/dt+dW/dt

dU/dt = dQ/dt - dw/dt
dU/dt = 100-75
dU/dt = 25 J/s

Hence, the rate of increase in internal energy is 25 J/s.