Here’s a nice question that gauges your understanding of the Physics of boiling a kettle which a few of my students found tricky.

We have a 2.2 kW kettle containing 0.5 kg of water at 20

^{o}C. The kettle is switched on for 3 minutes. What mass of water remains in the kettle at the end of this time? The specific heat capacity of water is 4186 J kg^{-1}K^{-1}and its specific latent heat of vaporisation is 2.26 MJ kg^{-1}.

It’s probably a good idea to draw a sketch of what’s happening. Notice that until it boils energy is flowing into the thermal energy store of the water. As the kinetic energy of the water molecules increases the temperature increases. But as the temperature reaches 100^{o}C it stops rising. Now the energy flows into breaking bonds between water molecules completely to change state from liquid water to water vapour which is a gas.

We split the problem into two parts. The first part is to calculate how long the temperature takes to reach 100^{o}C. The second part is to work out how much water is vaporised in the remaining time until 180 seconds.

## Time To Boil

We have to increase the temperature of the water from 20^{o}C to 100^{o}C. This is a temperature increase of 100^{o}C-20^{o}C=80^{o}C.

The energy required to raise the temperature of 1 kg of water by 1^{o}C is 4186 joules. To raise half a kilogram of water by 80 centigrade will require \(0.5 \times 80 \times 4186=167,440\) joules. We’re using the equation

where

\(\Delta E\) | change in energy | joules (J) |

\(m\) | mass | kilograms (kg) |

\(c\) | specific heat capacity | J kg^{-1} C^{-1} |

\(\Delta \theta\) | change in temperature | centigrade (C) or kelvin (K) |

The kettle’s heating element is delivering energy at a rate of 2,200 joules per second (watts). The time taken to deliver 167,440 joules is \(167440 \times 2200 = 76.1\) seconds. That’s because

\(t=\frac{E}{P}\)where

\(t\) | time to deliver energy | seconds (s) |

\(E\) | energy | joules (J) |

\(P\) | power | watts (W) or joules per second (J s^{-1}) |

The remaining time is spent boiling, and as 3 minutes is 180 seconds that is 180-76.1 or 103.9 seconds.

## Amount of Water Boiled Away & Remaining

If we’re boiling the 0.5 kg of water for 103.9 seconds the energy delivered by the heating element in that time is \(103.9 \times 2200=228,560\) joules.

We know that it takes 2.26 megajoules to vaporise 1 kg of water. The amount of water that is vaporised by 228,560 joules is \(\frac{228560}{2.26 \times 10^6} = 0.101\) kg. As we began with 0.5 kg of water the remaining water has a mass of 0.5-0.101=0.399 kg. We’re using the equation

\(\Delta E=ml\)where

\(\Delta E\) | change in energy | joules (J) |

\(m\) | mass | kilograms (kg) |

\(l\) | latent heat of vaporisation | J kg^{-1} |