How to boil water faster

[soldar]

The question was asked in another forum but there were a lot of just plain ignorant and stupid answers and the topic died but I think it might be a good thought experiment. The question is

Is it faster to boil 10 L of water all at once (in a 10L pot) or 10 L of water in 1 L pots successively?

That was the question but, as always, the devil is in the details. If you boil 10 L of water all at once then all the water is boiling at the end whereas if you boil it in small containers then all containers were brought to a boil but at the end only one is still boiling. So let us stipulate at the start all are at room temperature and the experiment is completed when all the water was brought to a boil, even if not simultaneously.

I think bringing the water to a boil simultaneously in several pots would only be feasible with a source of heat for each pot.

I suppose we may need to specify other things like how it is heated (electric, gas, etc) but for now we can just specify the heating power is the same in both cases.

What say you?

[Andy Chee]

This sounds like a surface area/volume problem.

I'm pretty sure the 1L pots will boil quicker, even if all the pots are on the single heat source.

It's almost identical to the ice cube melting problem, which will melt quicker, a massive 10kg block of ice, or 10 x 1kg blocks of ice.

[tom66]

I would suggest that a submerged element vs a plate element at the bottom of a pot would vary the results significantly, as would stirring the fluid, but my intuition is that assuming a bottom-heated plate (i.e. a normal kettle) it is quicker to boil 10 pots than 1 larger pot.    However, intuition is often wrong!

[AndyC_772]

I'll make a case for the opposite, that the 10l container boils first.

Throughout the heating process, the heat (from whatever source) is going into a larger vessel, one with a higher ratio of volume to surface area. This means that relatively less heat is being lost from the walls of the pot, and more of it is retained within the water. If we assume the pots have similar proportions, then the larger base will also receive a greater proportion of the heat from the source, ie. less goes straight from the source to the environment, bypassing the pot entirely.

It's the opposite of the ice cube problem. Melting faster means maximising the rate of heat transfer between the ice and the air. Here we're looking to minimise that heat transfer.

[Ian.M]

Basic thermodynamics.  The energy required to heat 1 kg of a substance by 1 K (degree Kelvin) is known as its specific heat capacity.   It can (and does) vary with respect to temperature but it wouldn't be a useful concept if the energy required wasn't directly proportional to the mass.

Therefore, with the same heating power, in a perfectly thermally insulated vessel of negligible thermal mass, 1L of water will boil exactly ten times faster than 10L of water.

The only differences are the losses, and as the surface area of a container scales with the square of its linear dimensions and the volume with the cube, it is likely that the larger container will loose less heat boiling 10L than a smaller container will loose boiling 1L ten times.

The case when the container is not of negligible thermal mass is more interesting.  In real life no-one waits for the kettle to fully cool to room temperature  before refilling it to boil it again, so the first time takes longer as the kettle starts from cold, but all subsequent refills boil a little quicker due to stored energy in the kettle.

[djsb]

Do the pots mentioned have lids placed on them when filled? Are they being watched? The saying is a watched kettle (or pot) never boils 

[mendip_discovery]

10L pot.

Takes the same energy to heat the water. The mass of the pots should be similar.

As soon at a 1ltr pot is up to boiling point you need to swap for the next pot. This takes time.

But this is one of those arguments where you only need to adjust the goal to complicate the problem. Much like this 10 x 3 x 6 + 5 = ? Kind of questions go about the place as somone will adjust the parameters to meet the answer they want.

[Zero999]

This sounds like a surface area/volume problem.

I'm pretty sure the 1L pots will boil quicker, even if all the pots are on the single heat source.

It depends.

If the 1l pots were in a big oven, then they would boil quicker, since they have more surface area to absorb heat from the hotter ambient temperature. If they each had an element and surrounded by room temperature air, then they would boil slower, since more heat would be lost to ambient.

[soldar]

It depends.

If the 1l pots were in a big oven, then they would boil quicker, since they have more surface area to absorb heat from the hotter ambient temperature. If they each had an element and surrounded by room temperature air, then they would boil slower, since more heat would be lost to ambient.

"It depends" seems to be the most correct answer to most questions.

I think we all agree the amount of heat the water needs to boil is the same in both cases so the problem boils (!) down to heat gained and lost in the heating process.

To give just one example, say I have an electric hot plate that produces 250 W. It is possible that a 1 L pot will boil in X minutes but a 10 L pot will never boil because it reaches a temperature where it loses more than 250 W through the sides and top.

[soldar]

Do the pots mentioned have lids placed on them when filled? Are they being watched? The saying is a watched kettle (or pot) never boils 

We need to specify these are Schrodinger pots which are being watched and not being watched at the same time. 

[tszaboo]

It takes the same amount of energy to boil the same amount of water, that's obvious.
You are asking the wrong question though.
Why are we boiling the water?

Making tea? Desalination? Disinfection? Making soup? Energy storage/generation?
Most of these questions have a lower energy solutions.

[themadhippy]

if the pots are the same diameter then the 10 litres,cause 9 litres will be at a higher elevation and reach boiling point at a lower temperature

[tunk]

if the pots are the same diameter then the 10 litres,cause 9 litres will be at a higher elevation and reach boiling point at a lower temperature

Or at a higher temperature because the pressure at the bottom is higher.

[coppice]

What do you mean by boil? Reach 100C or bubble? Nucleation issues (e.g. surface contamination in the pots) will affect when bubbling gets going. If you are just looking to reach 100C, then if the heat is applied within the bulk of the water, that is 100% efficient, and the pots with the greatest losses will boil the slowest. Evaporation is usually a big loss, unless the pots have narrow necks, so probably the pots with the greatest total surface area will suffer the greatest loss, and boil the slowest. In a strong draught, maybe losses from the outside of the pots will dominate, and the pots with the greatest total outside surface area will be the slowest.

[IanB]

The question was asked in another forum but there were a lot of just plain ignorant and stupid answers and the topic died but I think it might be a good thought experiment. The question is

Is it faster to boil 10 L of water all at once (in a 10L pot) or 10 L of water in 1 L pots successively?

This is an example of a question that cannot be answered as given, as it will depend on so many unstated assumptions. Each person answering the question may make different assumptions, and so may arrive at as different answer.

The only way to address this question is not to answer it, but to discuss the kinds of assumptions that might need to be made to give an effective answer.

We can give a "spherical cow" answer, by making some very unrealistic assumptions:
1. In both cases we have the same mass of water (10 L = 10 kg)
2. In both cases we are heating the water from the same starting temperature to the same ending temperature (e.g. from 20°C to 100°C).
3. In both cases the rate of heating is identical (e.g. 3 kW)
4. All the heat added goes into the water and none is wasted
5. No heat escapes from the water to the surroundings (perfectly insulated containers)
6. No boiling or evaporation occurs (which will consume additional heat beyond that required for raising the temperature)

By making assumptions such as these, we can observe that the time to bring the 10 kg of water to boiling point will be identical in both cases, since heating water without losses is a path-independent process. To raise a given mass of water from one temperature to another will require the same amount of energy regardless of whether you do it in pieces or as a whole. If the power input is the same in all cases, then the time taken will also be the same.

Now, by varying or relaxing the listed assumptions above, such as by allowing for heat losses, then the answers can change, and this is where the interest may lie.

[RoGeorge]

Boiling water FAST!
MOMables - Laura Fuentes



 

[madires]

It boils down to efficiency.

[Zero999]

Boiling water FAST!
MOMables - Laura Fuentes



 

I use an electric kettle and water from the hot tap, if available. Some people say don't drink water from the hot tap, because it's from a tank and has being sitting there for a long time, but it should already be hot enough to kill bacteria and even if it isn't, you're going to boil it anyway,.

[coppice]

I use an electric kettle and water from the hot tap, if available. Some people say don't drink water from the hot tap, because it's from a tank and has being sitting there for a long time, but it should already be hot enough to kill bacteria and even if it isn't, you're going to boil it anyway,.

Most UK homes no longer have an open header tank. In those homes there is no reason not to drink the hot water, even if it hasn't been heated enough to sterilise it.

[Smokey]

Hypobaric Kitchen?
https://en.wikipedia.org/wiki/Hypobaric_chamber

[thm_w]

To give just one example, say I have an electric hot plate that produces 250 W. It is possible that a 1 L pot will boil in X minutes but a 10 L pot will never boil because it reaches a temperature where it loses more than 250 W through the sides and top.

Its an example but not really a realistic one. You can get a 2000W induction hot plate for about $50, or a 2200W kettle for less.
No one should be using a 250W heater unless its just to keep food warm.

[soldar]

Its an example but not really a realistic one. You can get a 2000W induction hot plate for about $50, or a 2200W kettle for less.
No one should be using a 250W heater unless its just to keep food warm.

It's called "proof of concept". And the price of hot plates don't enter into it.

BTW, I wonder how much spherical cows are going for these days.

[thm_w]

It's called "proof of concept". And the price of hot plates don't enter into it.

BTW, I wonder how much spherical cows are going for these days.

Correct term was "thought experiment" as per your original post, proof of concept would mean you are actually testing a product out.

The price and wattage is relevant because you didn't specify anything other than "of the same wattage" in the original post. A typical burner on an electric or gas stove is going to be 1,200-2,200W, so that is what one would expect to use when determining an answer. I think IanB put far more effort into this thought experiment, maybe consider that direction in the future. 

[SiliconWizard]

But can water boil faster than the speed of light?

[PlainName]

I use an electric kettle and water from the hot tap, if available. Some people say don't drink water from the hot tap, because it's from a tank and has being sitting there for a long time, but it should already be hot enough to kill bacteria and even if it isn't, you're going to boil it anyway,.

Most UK homes no longer have an open header tank. In those homes there is no reason not to drink the hot water, even if it hasn't been heated enough to sterilise it.

Warm water from the hot tank contains contaminants (hot water + metal = corrision). Hot water systems fur up where cold water doesn't - worth thinking about.

As for legionnaires disease, the water needs to be at least 55C to kill the bacteria, but that's too hot if you're going to put your hand in it. At 51C it will take 3 mins to scald your hand, at 56C 15secs. There is reliance on the system losing enough temperature at the tap that the tank can be hot enough to kill off stuff, yet you don't scald yourself accidentally. On the plus side, most outlets in a house get used often, so water isn't likely to be sitting in a dead spot while being warm.

[coppice]

Warm water from the hot tank contains contaminants (hot water + metal = corrision). Hot water systems fur up where cold water doesn't - worth thinking about.

Pipes fur up more rapidly with hot water, but if you live is an area with very hard water the cold pipes, taps, etc, fur up very well. All pipes and tanks are susceptible to corrosion and contaminants. There is nothing special about a hot water tank there.

[Hydro]

I didn't read the whole topic, but I remembered another task.
How to use 10 liters of water at a temperature of 100 degrees to heat another 10 liters of water at a temperature of 0 degrees to 80 degrees.
Someone will immediately claim that this violates the laws of thermodynamics.

If you know the answer, don't tell yet.

[Zero999]

Warm water from the hot tank contains contaminants (hot water + metal = corrision). Hot water systems fur up where cold water doesn't - worth thinking about.

Pipes fur up more rapidly with hot water, but if you live is an area with very hard water the cold pipes, taps, etc, fur up very well. All pipes and tanks are susceptible to corrosion and contaminants. There is nothing special about a hot water tank there.

If that's true, then hot water would be more pure, because the minerals have come out of the solution and deposited in the pipes.

[hans]

The added energy needed to boil 1x10L of water, or 10x1L of water, is obviously the same.
I assume that the question asserts that both water heaters have equal power, and therefore surface area, and the water is at an identical temperature at the start. Then only changes can be made from the boiling process and the way we treat losses.

So lets assume 2 sources of heat loss: evaporation from the hot/boiling water, and heat convection into the kettle walls.
I'm not going to speculate whatever those rates exactly are. However, if we assume they scale linearily with surface area, then we can design a kettle that has an optimal trade-off ratio between radius and height. Assuming here that a larger radius will result in more evaporation losses, and higher walls will result in more convection losses with the walls. We can find optimums for both a 1L and 10L container.

The losses of evaporation is presumably going to scale with R^2, since thats the surface area of the water body on the top of the container.
The losses in the walls is presumably going to scale with 2*R*H, as thats the surface area of the walls of the kettle.

Therefore, to balance these ratios, e.g. alpha *R^2 = beta * 2*R*H, where alpha is evaporation loss coefficient and beta is the convection loss coefficient, we can see that we must solve for R and H giving alpha and beta. Since the volume of the container is also constrained by V=R^2*H, we have a system of equation(s) that can be solved.

It means that if we go from a 1L => 10L container, we can do that either by making H 10x bigger, or R 3.16x larger. Or a mix.. whichever has the more favourable coefficient..

"Solving" the problem by having 10 containers of 1L, and boiling them one at a time (with assuming 0s switching time), is not going to help much. We have a 10x larger body of water, however, we also have 10x the losses from that smaller container.
Lets assume some extreme edge cases for such case:

Assume that alpha=0 and thus we have no evaporation losses. Having 10x 1L containers means effectively we have cut up a 10x H container into sections of 1x H. The losses into the walls should be identical, and thus it should have no influence if we boil each pot sequentially. However, since the evaporation losses is zero, we ideally would want to make R -> infinity and thus H -> 0, which makes for a non-numerical solution.

Assume that beta=0, and thus we have no losses in the walls. In this case we would want to have R -> 0, and thus H -> infinity (again non-numerical case). However, since we chop our container up in 10 sections we can reason that we also have 10x the evaporation losses. (Even where R -> 0). Ratiometrically speaking, the losses in boiling 10 separate containers is 10x worse.

The practice will probably be somewhere in between.

So my bet is on a 10L pot being more efficient, as one can set a new optimal ratio between radius and height to balance the losses of its volume.

[.RC.]

If you put the 10l in a pressure cooker, then put a vacuum pump on the pressure cooker, it will boil quicker.