Question

A heating coil inside an electric kettle delivers 2.2 kW of electrical power to the water in the kettle. How long will it take to raise the temperature of 0.50 kg of water from 30.0° C to 72.0° C

Answer 1

Answer:

0.04s

Explanation:

Specific heat of water c = 4.186J/g.C

The heat energy it would take to heat up 0.5 kg of water from 30C to 72 C is

$E = mc\Delta T = 0.5 * 4.186 * (72 - 30) = 87.9 J$

If the power of the electric kettle is 2.2kW (or 2200W) and suppose the work efficiency is 100%, then the time it takes to transfer that power is

$t = \frac{E}{P} = \frac{87.9}{2200} = 0.04 s$