Question

The current in an electromagnet connected to a240-V line is 22.5 A . At what rate must cooling water pass over the coils if the water temperature is to rise by no more than7.50 ∘C ? The specific heat of water is 4186 J/(kg⋅∘C).

Answer 1

The apparent power of an electrical circuit is
given by:

P = VI = 240*22.5 W = 0.24*22.5 kJ/s

This power will cause heating in the circuit.

The heat generated by the circuit will be carried
away by the flowing water.

Since water has a specific heat capacity of:
C = 4.186 kJ/kg -C; then the heat absorbed by
water will be: Q = mCΔT

And, since the water is flowing at a rate: (m/t)
P = dQ/dt = (m/t)CΔT = (m/t)*(4.186 kJ/kg-C)*7.5

(m/t) = 0.24*22.5/(4.186*7.5) = 0.172 kg/s