Answer:
2.67 hours
Step-by-step explanation:
If three quarters of the mass are in liquid phase, I have a mass of water:
m1 = 3/4 * 5 = 3.75 kg
The latent heat of vaporization of water is of
ΔHvap = 2257 kJ/kg
The heat needed to vaporize it is:
Q = m * ΔHvap
Q = 3.75 * 2257 = 8464 kJ
If I have a resistor connected to 110 V, with a current of 8 A, the heat it dissipates through Joule effect will be:
P = I * V
P = 8 * 110 = 880 W = 0.88 kW
With that power it will take it
t = Q / P
t = 8464/0.88 = 9618 s = 160 minutes = 2.67 hours2.