Why a kettle takes its time

Put one litre of water, which weighs a kilogram, in a kettle at twenty degrees. To warm it by one degree you need about four point two kilojoules of energy, and this number is the same for every degree.

One kilogram of water, $m=1$ kg, at $20{}^{\circ }\text{C}$. One degree warmer needs about $4.2$ kJ, the same for every degree: $Q=mc\Delta T$ with $c\approx 4.2$ kJ per kg per degree.

Water $m=1$, start $T_0=20{}^{\circ }\text{C}$, heat $Q=mc\Delta T$, $c\approx 4.2$.

To reach one hundred degrees, the water must warm by eighty degrees. That is eighty times four point two: about three hundred and thirty-six kilojoules. A kettle of two kilowatts gives two kilojoules every second, so it needs one hundred and sixty-eight seconds, less than three minutes.

To reach $100{}^{\circ }\text{C}$ the water must warm by $\Delta T=100-20=80$, so $Q=1\times 4.2\times 80=336$ kJ. A $2$ kW kettle gives $2$ kJ per second: $336/2=168$ s, less than $3$ min.

$\Delta T=100-20=80\Rightarrow Q=1\times 4.2\times 80=336$; time $t=336/2=168\text{s}$.

Now the surprise. Once the water boils, its temperature stops at one hundred degrees. More heat changes nothing there; the energy now turns water into steam. Turning one kilogram of water into steam takes two thousand two hundred and sixty kilojoules, nearly seven times what the heating took.

Surprise: once boiling, the temperature stays at $100{}^{\circ }\text{C}$. More heat does not make it hotter, it makes steam. One kg of steam takes $2260$ kJ, which is $2260/336\approx 6.7$ times more.

Boiling: $T=100{}^{\circ }\text{C}$ fixed; steam $Q_s=2260$, so $2260/336\approx 6.7$.

So the same kettle, left on, needs about nineteen minutes to boil all the water away: almost seven times as long as it took to start boiling.

Same kettle, left on: $2260/2=1130$ s, about $19$ min, and $1130/168\approx 6.7$ times the time to start boiling.

$t=2260/2=1130\text{s}\approx 19\text{min}$; $1130/168\approx 6.7$.