# Optimizing water boiling

I'm about to prepare some spaghettis and to do so I need 2 liter of boiling water.

For getting the water to certain temperature (lets say 100 Celsius degrees) from a starting temperature of 15 Celsius degrees, I might opt by pouring the 2 liters of water in one single action or might choose by adding in 4 times (1/2 liter each time) so I would start with 1/2 liter, when it starts boiling add another 1/2 liters and so on until I have the 2 liters boiling.

What requires less energy for getting the 2 liters?

Assumptions:

1. There is only one pot size.
2. The time required for pouring additional water is 0s.
3. Pouring water despite it takes 0 seconds, does not spit any water and then there is no additional loose of temperature other than the one coming from mixing the water.
4. The surface of the pot distributes the temperature homogeneously.

Assuming the total heating power entering the system will be constant, the only factor to minimise is the wasted heat leaving by air convection and radiation. To minimise that, you want to keep the pot as cold as long as possible, because all such heat transfer mechanisms are vastly more efficient when the temperature difference (as well as absolute temperature) is high.

If we treat the wasted heat entirely as a second-order correction, the temperature curves for both your proposed procedures look something like this:

You see the pot's temperature is most of the time substantially higher if you heat in portions, and that is bad because it increases convection etc. losses and also reduces heat transfer from the stove to the pot, as the other answers remarked (though that's probably less of an issue, certainly not for gas- or induction stoves).

• But the whole point is that the power entering is not constant; it depends on the temperature difference, and it is greater when heating everything together. – Davidmh Apr 1 '14 at 11:54
• @Davidmh: that depends strongly on the particular design of the stove. For an induction stove there isn't even any temperature difference; only the change in resistivity plays a part. Even for heat-transfer based stoves, the temperature difference between heating element and pot will at all times be vastly larger than the difference between pot and environment, so that's still the more crucial parameter. — Anyway both effects just amplify each other: less power means longer time for energy to get lost through convection, the rate of which depends on temperature. – leftaroundabout Apr 1 '14 at 12:09
• Also, I think the premise is wrong. Actually the bottom of the pot will in either scenario reach close to $100^\circ\mathrm{C}$ much before the water is boiling: the heat transfer through the water is driven by liquid convection but also to a good part through vapour bubbles. – leftaroundabout Apr 1 '14 at 12:13

The energy it takes to boil the water is independent of the process: it is equal to the difference in enthalpy between liquid water at 15c and boiling water (assumed to be saturated liquid). To go further you need to make assumptions regarding the heat losses.

If you have forced heat flow by lets say an electric stove the temperature increases linearly with time and both ways should take the same time and also same energy (constant power, no heat losses on stove and pot assumed). If heat transfer is driven by temperature difference as with a gas stove ($T(t)= T_{15}+(T_{100}-T_{15})(1-e^{-\frac{k A }{m c_p}t})$, no heat losses with pot only), the heating rate is higher when the temperature difference between stove and water is larger for a longer time. So the first method should be quicker than the second and consequently more efficient too (less gas consumed).

Heating up some mass to a certain temperature requires a certain amount of energy no matter how you heat is up. However heat transfer between your hotplate and your pot depends on the temprerature gradient (temperature difference), as well as the efficiency of the heating process. So if you boil your water in smaller protions you work with smaller efficiency, so assuming your hotplate dissipates a certain amount of energy, with less efficent methods you need more time.