Volatile Yard 挥洒自如 Write anything I want to write...

Introduction

I have been wondering how long an electrical kettle [en.wikipedia.org] takes to boil water. There are thousand types of kettles in the world, but they have simple and similar circuit designs. If you have a kettle, you can check its power rating and its maximum capacity easily. With these figures, along with the initial temperature of the water (which is normally equal to the room temperature), you should be able to make a good estimate of the time it takes.

The concept behind working out a good estimate should not be too difficult, and involves only high-school physics. However I have neither seriously sat down to do the calculation nor any experiment. A few months ago, I decided to do all these to see what I can get.

I think that the calculation here is very educational to high-school science students. Wonder why I didn't see similar exercises from my physics textbooks.

Theoretical Derivation of the Estimation Formula

The amount of electrical energy $W$ supplied to a kettle with a power rating $P$ within time $t$ is given by

$!W=Pt$.

Unless you have a lousy kettle, most (probably more than 95%) of the energy here should be converted to heat for boiling the water. When you heat up the water, a small amount of the energy will be 'wasted' for heating up the kettle, for controlling other components of the kettle (e.g., LED status indicators), for the heat leaked to the air around the kettle, etc. However, the total mass and the specific heat of the kettle components are very little compared to those of the water. Moreover, a good kettle has (and should) been designed to minimize the wasted energy. Therefore it is generally safe to neglect the 'wasted' energy in our estimation. (However, when we round up some figures later to get a simpler formula later, we round up in the way that would somehow account for the 'wasted' energy.)

By neglecting the little energy not used for heating up the water, the amount of energy $Q$ for heating up the water is:

$!Q \approx W = Pt$.

Thus,

$!t \approx \frac{Q}{P}$.

To heat up some water with an increase in temperature of $\triangle T$, the amount of energy required is

$!Q=mc\triangle T$,

where $m$ and $c$ are the mass and the specific heat of the water respectively. To boil the water, we have

$!\triangle T = T_b - T_i$,

where $T_i$ and $T_b$ are the initial temperature and the boiling point of the water.

The mass of the water with volume $V$ is

$!m=\rho V$,

where $\rho$ is the density of the water.

Combining the formulas above, we obtain

$!t \approx \frac{\rho Vc(T_b-T_i)}{P}$

The specific heat of pure water, $c$, is $4.184 \times 10^{3}\mathrm{J}\,\mathrm{kg}^{-1}\mathrm{K}^{-1}$. The water we boil is hardly pure and generally contains a lot of minerals, and even bacteria and viruses. However, since we are dealing with rough estimation, we assume that we are using pure water and stick to this value.

The density of pure water, $\rho$, varies with its temperature. When we boil water, the initial temperature normally varies from a few Celcius (in winter) to 35 degree Celcius (in summer). The density of the water within this range of temperature varies from $999.8 \mathrm{kg}\,\mathrm{m}^{-3}$ to $995.6 \mathrm{kg}\,\mathrm{m}^{-3}$ (refer to, say, here [www.simetric.co.uk]). Since the density doesn't change much and we are only doing rough estimation, it's reasonable to stick to a value of, say, $997 \mathrm{kg}\,\mathrm{m}^{-3}$ (ok for 24-27 degree Celsius).

The boiling point of pure water, $T_b$, varies with the pressure applied on it. Well, most people boil water near the ground level (unless you stay in Tibet or other high terrains), so we can assume that the pressure applied is one standard atmosphere pressure [en.wikipedia.org]. With this assumption, most people know that water boils at 100 degree Celsius, or 373.15K.

By substituting these values in to the formula above, we obtain:

$!t \approx \frac{997 \times V \times 4.184 \times 10^{3} \times (373.15\mathrm{K}-T_i)}{P}$.

Let's choose the units that are convenient for our calculation and usage. Let's use litre for water volume $V$, degree Celsius for temperature $T_i$, watt for power rating $P$, and second for time $t$. After some manipulation, we obtain

$!t \approx \frac{4171 \times V \times (100-T_i)}{P} \; \mathrm{seconds}$.

A bad thing about this formula is the 'odd' figure 4171. It's good if we can round it to a multiple of a hundred. Throughout the derivation, we know that we have underestimated the time taken. (In particular, more time and thus more energy is required to heat up the whole kettle, to control other components, etc.) So we can round it up and obtain

$!t \approx \frac{4200 \times V \times (100-T_i)}{P} \; \mathrm{seconds}$

This formula looks handy enough for the estimation. Let's try it with a real-world kettle to check how well it works.

Experimental Verification of the Formula

I use my kettle to test the formula. First, check the power rating.

I stay in Malaysia, and the AC voltage here is 240V. It corresponds to the power rating of 2200W. If you use 220V, then the power rating should be 1850W. You can actually find out the relationship between the power rating and the AC voltage easily. Let see.

The main component of a kettle is a heating rod. The resistance $R$ of the heating rod varies only a little under different operating voltages and temperatures. Since the heating rod should consume most electrical power, we have

$!P \approx \frac{V^2}{R}$,

where $P$ is the power rating and $V$ is the operating voltage. If we fix $R$, and manipulate this formula, we get

$!\frac{P_1}{P_2} \approx \frac{{V_1}^2}{{V_2}^2}$.

Let's check the specification with this formula, with $V_1=220V$, $V_2=240V$, $P_1=1850W$, and $P_1=2200W$. The LHS is $\frac{1850W}{2200W} \approx 0.8409$, while the RHS is $\frac{(220V)^2}{(240V)^2} \approx 0.8403$. So the figures here are quite consistent. Great!

Back to our estimation. Now check the maximum capacity of the kettle. It reads 1.7L.

A room temperature of 28 degree Celsius is common here. By substituting $P=2200W$, $V=1.7L$, $T_i = 28^{\circ}\mathrm{C}$ into the formula, we get:

$!t \approx \frac{4200 \times 1.7 \times (100-28)}{2200} \mathrm{s} \approx 234 \mathrm{s} = 3\mathrm{min}\;54\mathrm{s}$

I then took a stop watch and boiled 1.7L of water. I have timed it a few times over a few days, and the time I get lies within 230-250 seconds. So the estimation of 234 seconds is good enough. (If we had used 4171 instead of 4200 in the formula, we get about 232 seconds, which doesn't differ much.) It also shows that the kettle is rather energy efficient.

How about yours?

Related Post: How expensive does a kettle boil water?