# How long does it take for an electric kettle with a current of 4 A, voltage of 220 V to heat 2 kg

**How long does it take for an electric kettle with a current of 4 A, voltage of 220 V to heat 2 kg of water from 20 degrees to a boil.**

I = 4 A.

U = 220 V.

m = 2 kg.

C = 4200 J / kg * ° C.

t1 = 20 ° C.

t2 = 100 ° C.

T -?

To heat water of mass m from temperature t1 to boiling point t2, we express the formula: Qw = C * m * (t2 – t1), where C is the specific heat capacity of water.

The amount of heat Qh is released in the electric kettle, which we express by the formula: Qh = I * U * T, where I is the current strength, U is the current voltage, T is the operating time of the kettle.

The loss of thermal energy during the transfer from the kettle to water can be neglected: Qh = Qw.

I * U * T = C * m * (t2 – t1).

T = C * m * (t2 – t1) / I * U.

T = 4200 J / kg * ° C * 2 kg * (100 ° C – 20 ° C) / 4 A * 220 V = 763.6 s = 12 min 43.6 s.

Answer: to heat water in an electric kettle, T = 12 min 43.6 s.