There are two boilers (220V), which boil 1 liter (kg) of water for 2 minutes each. Find power, current, resistance.

Decision
First, let’s determine the amount of heat that is released during boiling.
Q = m * c * (T2-T1)
where, T2 is the boiling point, T1 is the room temperature, s is the specific heat, m is the mass
Q = m * c * (T2-T1) = 1 * 4.187 * ((273 + 100) – (273 + 20)) = 334.96 kJ
20 degrees is room temperature
Determine the resistance
Q = U ^ 2 / R
R = U ^ 2 / Q = 220 ^ 2 / 334.96 = 144.45 (Ohm)
Determine the current
I = U / R = 220 / 144.45 = 1.52 (A)
P = I ^ 2 * R = 1.52 ^ 2 * 144.45 = 2.3104 * 144.45 = 333.74 W