What should be the current strength of the spiral tile in order to heat 2 liters of water from 20 to 100 degrees

What should be the current strength of the spiral tile in order to heat 2 liters of water from 20 to 100 degrees Celsius in 5 minutes. The voltage from the network is 200 V. Neglect losses due to heating of the ambient air.

T = 5 min = 300 s.

V = 2 l = 0.002 m3.

ρ = 1000 kg / m3.

t1 = 20 ° C.

t2 = 100 ° C.

U = 220 V.

С = 4200 J / kg * ° C.

I -?

For heating water, the required amount of heat Qw is expressed by the formula: Qw = C * m * (t2 – t1), where C is the specific heat capacity of water, m is the mass of heated water, t1, t2 are the initial and final water temperatures.

We express the mass of water m by the formula: m = ρ * V.

Qw = C * ρ * V * (t2 – t1).

The amount of heat Qp is released in the spiral to the tile, which is determined by the Joule-Lenz law: Qp = I * U * T, where I is the current in the spiral, U is the voltage at the ends of the spiral, T is the time the current passes along the spiral.

Since all the heat of the spiral goes to heating the water, then Qw = Qp.

C * ρ * V * (t2 – t1) = I * U * T.

I = C * ρ * V * (t2 – t1) / U * T.

I = 4200 J / kg * ° C * 1000 kg / m3 * 0.002 m3 * (100 ° C – 20 ° C) / 220 V * 300 s = 10 A.

Answer: the current strength in the spiral tile is I = 10 A.