Welcome to The Nonlinear Library, where we use Text-to-Speech software to convert the best writing from the Rationalist and EA communities into audio. This is: Does descaling a kettle help? Theory and practice, published by philh on May 2, 2023 on LessWrong.

I've heard that descaling a kettle makes it more efficient, and can save me time and money.

Okay, but how much? For some reason my my intuition says it'll be basically unnoticable. Let's try to figure it out, and then actually try it.

The Models

There's a first-order approximation which says this should be impossible: heating things up is 100% efficient. What's the limescale going to do, convert some electricity into forms of energy other than heat?

No, but it might cause the heat to go to the wrong places. I think there are two ways that could happen. The first is if the limescale itself has a high heat capacity, meaning it takes a lot of energy to heat up. That doesn't seem likely to me; I think there's likely less than 10 g of it, and I think it probably has less heat capacity than the same mass of water (because I think water's is higher than most other things). I don't think adding 10 ml water (weighing 10 g) to my kettle will significantly effect the boiling time.

Spot check: water's specific heat capacity is about 4 J/K·g. So to heat 10 g water by 100 K needs about 4000 J. My kettle is 2200 W according to the email I got when I bought it, so an extra 10 ml should take about 2 s longer to boil.

Also, I normally boil about 500 ml of water, so we'd expect 10 ml extra to make about a 2% difference. (I don't have a strong intuition for how long my kettle normally takes to boil. 1-3 minutes? The above calculation suggests it should be around 100 s.)

The second way the limescale could matter is if it's a good thermal insulator. Then the metal heating element gets significantly above 100°C before the water boils. And from my googling, it seems like this is the reason people give that descaling a kettle is helpful. How much effect would this have?

This page says the thermal conductivity is 2.2 W/m·K. I don't remember studying this in school, and I don't find that wikipedia page very clear. But I think maybe this means: the rate of energy transfer (W) from the heating element is 2.2 W/m·K, multiplied by the surface area of the heating element (m²), divided by the thickness of the limescale (m), multiplied by the temperature difference (K) between the heating element and the water. The units check out at least.

This implies, if there's no limescale, that the power transfer should be infinite, regardless of the size of the heating element. That's clearly not how things work. Still, it seems fine to have as a model, it just means that the water and heating element will stay the same temperature.

But doing some field research (i.e. looking at my kettle) makes me think it's a bad model in this case. Here's what it looks like:

(This kettle was last descaled around August 2020. I know because I bought vinegar for it, and thanks to online shopping the receipt is in my email.)

It seems that I have some limescale over some of the heating element (bits of the metal plate at the bottom, and the tube around where it passes through the plate), and zero limescale over some of it. The "no limescale gives ∞ W/K" model is going to give ∞ W/K no matter how small the no-limescale area is. So scratch that model.

I don't feel like trying to model the rate of power transfer from an un-limescaled heating element to water. I don't know what material it's made of (some kind of steel I suppose?), and I don't know what property of that material to look up. Plus it probably depends on how fast water convects.

Here's another way to think about it: assume that the amount of metal the manufacturer used is roughly as small as possible without compromising too much on boiling speed.

When we turn the kettle on, we start putting 2200 W into the heating element, and the heating element starts putting some of ...