The kettle heats up 2 liters of water at an initial temperature of 20 in 10 minutes, the current consumed is 6A

The kettle heats up 2 liters of water at an initial temperature of 20 in 10 minutes, the current consumed is 6A, the voltage in the 220V network. The specific heat capacity of water is considered equal to 4.2 * 10 ^ 3.

V = 2 l = 0.002 m3.

ρ = 1000 kg / m3.

t1 = 20 ° C.

t2 = 100 ° C.

T = 10 min = 600 s.

I = 6 A.

U = 220 V.

C = 4200 J / kg * ° С.

Efficiency -?

Let us write down the definition for efficiency: efficiency = Qw * 100% / Qt, where Qw is the amount of heat that goes to heat the water, Qt is the amount of heat that is released during the passage of an electric current.

Qw = C * mw * (t2 – t1).

mv = V * ρ.

mw = 0.002 m3 * 1000 kg / m3 = 2 kg.

Qw = 4200 J / kg * ° C * 2 kg * (100 ° C – 20 ° C) = 672000 J.

Qv = I * U * T.

Qv = 6 A * 220 V * 600 s = 792000 J.

Efficiency = 672000 J * 100% / 792000 J = 85%.

Answer: an electric kettle has an efficiency of 85%.