What is the current through the heating element of the electric kettle connected to the network

What is the current through the heating element of the electric kettle connected to the network in a network with a voltage of 220V if after 5 minutes. the water is brought to a boil from a temperature of 20 degrees celcius. The volume of water is 2 liters.

Let’s write down what is given and what needs to be found. Let’s translate everything into the SI system
t = 5min = 300 s
T2 = 20`C = 293K
T2 = 100`C = 373K
m = 2kg (mass of 1 liter of water = 1 kg)

The amount of heat required to heat the body:
Q = c * m * ΔT, where c is the specific heat capacity of water = 4200 J / kg * K, m is the mass of water, ΔT is the temperature difference
ΔT = T2-T2 = 80

The amount of heat that the heater emits:
Q = I * U * t – where I is the current in the circuit, U is the voltage t is the heating time.
Let us express from here I = Q / (U * t) = (c * m * ΔT) / (U * t), substituting the numerical values, we have:
I = (4200 * 2 * 80) / (220 * 300) = 10.2 A

Answer: 10.2 A