In an electric kettle with a power of 2 kW, water weighing 1.8 kg is heated from a temperature of 20 C

In an electric kettle with a power of 2 kW, water weighing 1.8 kg is heated from a temperature of 20 C to boiling. Determine the time for which the water boils if the efficiency of the kettle is 0.7 and the specific heat capacity of water is 4200 J / kg * K

Let’s find the amount of heat required to heat a given amount of water to the boiling point:

Q = C * m * (T2 – T1), where C is the specific heat, m is the mass, T1 and T2 are the initial and final temperatures.

Q = 4200 * 1.8 * (100 – 20) = 604.5 kJ.

Taking into account the efficiency, the energy released by the spiral will be equal to:

A = 604.5 / 0.7 = 864 kJ.

By definition of power:

P = A / t, A – work (energy), t time.

Let’s express the time:

t = A / P.

t = 864/2 = 432 s ≈ 7 min.

Answer: the required time is 7 minutes.