Determine the efficiency of an electric kettle with a power of 0.6 kW, in which 1.5 liters

Determine the efficiency of an electric kettle with a power of 0.6 kW, in which 1.5 liters of water heats up from 20 to 100C in 17.5 minutes.

Given:

W = 0.6 kW = 600 watts – the power of the electric kettle;

T1 = 20 ° Celsius – initial water temperature;

T2 = 100 ° Celsius – boiling point of water;

V = 1.5 liters = 1.5 * 10-3 m3 – water volume;

ro = 1000 kg / m3 – water density;

t = 17.5 minutes = 1050 seconds – the time it takes for the water to boil;

c = 4200 J / (kg * C) – specific heat of water.

It is required to determine n (%) – the efficiency of the electric kettle.

Let’s find the mass of water:

m = ro * V = 1000 * 1.5 * 10-3 = 1.5 kilograms.

Then the amount of heat required to heat water (useful work) will be equal to:

Q = c * m * (T2 – T1) = 4200 * 1.5 * (100 – 20) = 6300 * 80 = 504000 Joules.

All the work spent by the kettle on heating the water will be equal to:

A = W * t = 600 * 1050 = 630,000 Joules.

The efficiency will be equal to:

n = Q / A = 504,000 / 630,000 = 0.8 or 80%.

Answer: The efficiency of an electric kettle is 80%.