I thought that the function will follow the Square-cube Law since we heat water at the surface (which is x^2) while we heat the volume (which is x^3) as it happens in animals according to Bergmann's rule. I did some home experiments and got a linear dependency (the kettle is heated at the bottom by stove):

volume (ml): 150  300  450  600  900 1200 1800
time (s): 90 130 190 260 370 540 675

My explanation: while we heat the surface of the kettle, the water moves inside it - cold water descends while hot water goes up. Therefore we don't heat at the surface, we heat the whole volume of water at every moment of time. If water molecules were not moving up/down, it would've followed the Square-cube Law.

In animals the molecules are not freely moving. Of course there is a blood stream, but when the blood plasma is getting out of the vessel it's still moving by diffusion. And there is also a cell membrane that slows down water flow even more. Also blood has a lot of proteins which are not conducting heat as good as water. So overall animal is very different from the kettle with respect to heat exchange through the surface.

  • $\begingroup$ Does the heating element get larger with larger pots? (the answer could be yes or no depending on your stove) $\endgroup$
    – user10851
    Dec 13, 2015 at 0:51
  • $\begingroup$ It's not electric - it's heated by stove. Added this info into the question, sorry for confusion. $\endgroup$ Dec 13, 2015 at 7:09

2 Answers 2


Usually, for a "regular" container such as a pot or kettle, most of the heat transfer occurs through the bottom of the container, and the heat transfer rate is pretty much independent of the amount of water in the kettle. This means that, with constant heat input the time to boil will also be roughly proportional to the volume of the water.

  • $\begingroup$ There are so called Bergmann's rule (added a link to the question) and Allen's rule according to which bigger animals live in colder climate. That's where I got the idea about Square-cube law. But as it appeared it doesn't work equally for liquids in kettles :) $\endgroup$ Dec 12, 2015 at 22:02
  • $\begingroup$ @StanislavBashkyrtsev - Well, to be fair, you don't find many animals who generate internal heat by sitting on a stove burner. $\endgroup$ Dec 12, 2015 at 22:23
  • $\begingroup$ Yes, but they still radiate heat through the surface (skin) while the heat is generated by the volume (all cells inside). Same as kettle's content (water) is heated by stove at the surface. $\endgroup$ Dec 13, 2015 at 7:12
  • $\begingroup$ No, that is not the same. You are comparing static with transient situations. The volume production term balances heat outflux for the animal. In the stove heat influx balances rate of temperature increase. This is a complete different situation $\endgroup$
    – Bernhard
    Dec 13, 2015 at 7:44

Assuming that the the rate of energy delivered to the water in the kettle is constant (i.e. varies linearly with time), and that the temperature of the water varies linearly with its internal energy - then you'd expect the temperature of the water to vary linearly with time.

I suppose this linear relationship depends on good mixing of the water, but I don't see that the 'square-cube' law would apply. If the heating element in the kettle has a specific wattage then it's going to deliver that power/energy to the water in the kettle pretty well regardless of contact area between water and element.

  • $\begingroup$ I should've mentioned from the beginning - the kettle is heated by stove, not by electric element. So the amount of energy that is getting through depends on the conduction of the surface. $\endgroup$ Dec 13, 2015 at 7:24
  • $\begingroup$ Whether the kettle is heated by a stove or by an electric element, if the power delivered to the water is constant the water will heat at a constant rate. Of course, as the kettle gets hotter it will give out heat at an increased rate - but that will happen whether the kettle is heated from below or from within. $\endgroup$ Dec 14, 2015 at 9:31

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