# Energy efficiency and boiling water

I’ve noticed that my last few posts have been veering towards the metaphysical so I thought today I would talk about some kitchen science, literally. The question is what is the most efficient way to boil water.  Should one turn the heat on the stove to the maximum or is there some mid-level that should be used?  I didn’t know what the answer was so I tried to calculate it.  The answer turned out to be more subtle than I anticipated.

So consider a pot of water heated by a stove from below.  The stove can be gas or electric.  Water boils because heat is transferred from the stove to the water.  I will assume we are using a pot with high heat conductivity (e.g. stainless steel or copper) so the temperature at the bottom of the pot is the same as the burner.  I will also assume that the main source of heat transfer from the bottom of the pot to the bulk of the water is through convection.  The rate of convective heat transfer is proportional to the difference in the temperature between the bottom surface $T_s$ and the interior water temperature $T_w$.  The goal is to raise the water temperature to the boiling temperature $T_b \sim 100$ degrees Celsius.  At the same time the water is losing heat to the air and we can use Newton’s law of cooling, which is that the rate of heat transfer is proportional to the difference between the temperature of the air $T_a$ and $T_w$.  Now if we assume that the temperature of the burner and the air remain constant then we can model the change in the thermal energy of the pot $Q$ as $\displaystyle \frac{dQ}{dt} = aT_s+bT_a - (a+b)T_w$

where $a$ and $b$ are constants that are known or can be measured.  Now in equilibrium the thermal energy is proportional to the temperature.  So we will assume that $T_w = d Q$ for another measurable constant $d$.  My final assumption is that the temperature of the burner is proportional to the energy rate delivered to the burner (e.g. electrical power or gas flow).   The time integral of the energy rate over the time it takes to boil the water is the total amount of energy used. I will assume that the energy rate is constant.  By rescaling and lumping together parameters we arrive at the simple first order differential equation for the rescaled water temperature $T$ $\frac{dT}{dt} = P + C - T$    (*)

where $P$ is the input power, $C$ is the rescaled room temperature, and the boiling temperature is a rescaled $T_b$ and $C.  The energy used to boil water is $P t_b$,  where $t_b$ is the time to boil.

Assume that at time $t=0$ the temperature is $T_0$ and we want to compute the time it takes $T$ to reach $T_b>T_0$.  The solution of (*) is $T= T_0 e^{-t} + (P+C)(1-e^{-t})$

which can be easily obtained by solving (*) with the integrating factor $e^{-t}$ or inferred intuitively because at time zero the temperature is $T_0$ and at time infinity it is $P+C$ and the approach to the asymptotic temperature is exponential.  Setting $T=T_b$ gives the condition $\displaystyle t_b = \ln\left[\frac{T_0-C-P}{T_b-C-P}\right]$

which is a monotonically decreasing function of $P$ that is infinity at $P=T_b-C$ and zero as $P$ goes to infinity.  So if the temperature is below boiling you will never boil water and if it infinite then it will take no time to boil.

The total energy used is thus $\displaystyle P \ln\left[\frac{P+C-T_0}{P+C-T_b}\right]$        (**)

The goal is to find the minimum of (**).  Given that this is a product of a monotonically increasing function with a monotonically decreasing function then it is possible that there could be a minimum between $T_b-C$ and infinity.  We thus need to do a bit more of analysis.

The variables in equation (**) are expressed in scaled temperature units.  If we assume that the initial water temperature is approximately the room air temperature then we can set $C=T_0$, which simplifies the expression.  We can then show that the  energy is monotonically decreasing (since the derivative is always negative) so the most efficient way to boil water is to set the burner to the maximum.  This would also hold true if the initial water temperature is below the room temperature.  However, if the initial water temperature is higher than room temperature then there could be a setting less than maximum that is more efficient.  I generally like to use cold water for cooking (since hot water has been sitting in the water heater) so I crank the dial to the maximum when I boil.

Since $C$ and $T_0$ are not scaled identically, $C=T_0$ doesn’t imply that the room temperature is the same as the initial water temperature.  In fact, $C$ is scaled by a smaller factor so it is less clear how high $T_0$ can be for energy to decrease with power.  Additionally, I only estimated the energy for constant applied power.  I think that there is probably a more efficient way if you let the power be time dependent.  For example, if you slowly decreased power as the temperature of the water increased, that may be more efficient.  It would also be a good exercise for using the variational principle, if anyone is interested.

## 24 thoughts on “Energy efficiency and boiling water”

1. Hey, think of how much energy could be saved if everyone cooked pasta in the most efficient way possible.

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2. Cian says:

Added bonus: max heat is quickest.

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3. One of the few instances of having your cake and eating it too!

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4. Daniel says:

In theory you could save a lot of energy, but how much energy would have to be wasted estimating the unknown parameters?

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5. I would guess that you could do all the necessary experiments in a few days.

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6. romain says:

In your model, the transfer of energy from the burner to the water does not depend on the power of the burner or of the temperature of the water. The only thing that makes your system non-optimal is the loss of energy through air cooling. It is therefore straightforward to say that the most efficient strategy is to reduce this loss by making the water boil as fast as possible. This amounts to setting the burner at its maximum power.

Am I wrong? or is this a very nice example (that all physics teachers should tell their students) to show that sometimes you don’t need mathematics to solve that kind of idealized problem?

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7. That is what I thought initially but when I did the calculation I found that if the initial temperature is high enough then there is a nontrivial minimum in the energy use. I think it is because if the distance to threshold is small enough then there is a smallest burner temperature that makes it effectively non-lossy. Also, if the burner temperature is nonlinear in the input power then your intuition may also not work. I think the bottom line is that intuition is very useful but sometimes doesn’t get the whole picture.

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8. How about some math/physics on the hissing/buzzing noises water makes as it’s coming to the boil?

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9. The mechanism for the sounds during boiling are well known: it is due to the nucleation and bursting of bubbles. Modeling precisely how the noise is made quantitatively is a hard problem as it involves the nucleation and collapse of a three dimensional boundary layer between liquid and vapour immersed in a convective background of water and other bubbles.

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10. Is it really that the bubbles are bursting, or that they’re collapsing as they move up into cooler water? Is there an on-line reference for this?

Do we know why the pitch changes as the temperature rises? Does each bit of sound come from the collapse of a single bubble (micro-bubble)?

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11. I think it is that they are collapsing as they move into cooler water and as the temperature rises the point of collapse changes. When it reaches boiling, they make it to the surface. This is all conjecture though. I couldn’t find a reference although I vaguely remember it being covered on Quirks and Quarks recently. Good question though.

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12. Cloe says:

cool experiment

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13. Is it more efficient to micro wave water to boiling? And what would that equation look like?

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14. Depending on what you are comparing it to, a microwave could be more efficient. In a microwave, energy goes into the magnetron to make microwaves, which then heat the water by dieletctric heating (i.e. rotating the dipoles in water on each cycle of the microwave). Thus the microwave heats the water on all surfaces up to some penetration depth. This could potentially allow for faster heating, especially small volumes. My guess is that a stove top with a stainless steel kettle and a microwave would go head to head depending on the exact circumstances but I really don’t know. As for the equations, they should roughly look the same, i.e. a constant heat source with Newton’s law of cooling, although the parameters will be different.

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16. water coolers…

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17. Thank you for the math.
Love it.

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18. @Linden Not sure what you mean. You are free to use anything in the post in any way you like. There is no other paper associated with it though.

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I am not a math guy, but I have been in the kitchen learning how to cook since I was a young child. Standing on a chair watching my grandmother cook eagerly awaiting any task she would let me tend to sparked my love of preparing meals. With that being said, your calculated formulas are simply astonoshing. Furthermore, brilliant. I learned, without the math, that minimal water needed + a stainless steel pot with a lid + maximum heat, then at boiling point, reduce the heat to about half, seemed to be the most efficient and quickest boiling procedure Ive found. Thats about the extent of my formulas fir this equation, and I apologize for any grammar mistakes as well. I hope this was useful and a different spin on the subject. Enjoy and thank you for your knowledge as well.

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20. Roberta says:

What cooks really want to know is: Does it PULL more electricity to use an electric burner on high for a short time or on medium for a long time? If say, you are boiling water. Or is the electricity PULLED (and money cost) constant as relates to the time of use regardless of the electric stove setting?

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21. Medium for a long time uses more electricity to boil water.

Electricity use (or energy which equals dollars) is a function of setting, i.e. high vs low and how long. In fact, it is given by the formula of V times I times T, where V is the voltage, I is the current, and T is the time. Given that for an electric stove, V is fixed, then it is all about current and time. The thermostat on your stove controls the amount of current running through the coils. So, if medium uses half as much current as high (and I don’t know if this is true for your stove) then a stove on medium can be run for twice as long as high and use the same amount of energy. In boiling water, it will generally take longer than twice the amount of time to boil on medium vs high and thus high is more efficient. This may not be true for other tasks.

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22. KACY says:

It seems that the ways of heat loss are not all taken into consideration. All these hidden parameters should not ignore before making conclusion. Or, at least some experiments should be done. e.g. Boil some water, record the gas used, and repeat.

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23. @KACY All the assumptions are there and you can improve it in any way you see fit. As for other forms of heat loss, they can all be characterized in terms of dependence on temperature. So, there could be a term independent of temperature and those that are nonlinearly dependent. An experiment would always be good to do. It would be a good science project.

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