Does Boiling Water Faster Depend on Amount or Incremental Additions?

[Jotty]

Hi guys
First time posting on here.
Family and I were having dinner and a discussion came up.
If I need to boil 1000ml of water, would it boil faster if I boil all 1000 ml at once or if I boil say 200ml, wait for it to boil, add another 200 , let it boil and add another 200 till there's 1000ml in the pot.

Thanks !

 

[ShayanJ]

I think its better t boil all of the water at once.
Because when you add cold water to boiled water,they reach to an equilibrium temperature which is below the temperature of the boiled water and above the temperature of the cold water.So the oven should boil the mixture again and the fact that part of it was boiled before mixing doesn't matter.

 

[Jotty]

I think its better t boil all of the water at once.
Because when you add cold water to boiled water,they reach to an equilibrium temperature which is below the temperature of the boiled water and above the temperature of the cold water.So the oven should boil the mixture again and the fact that part of it was boiled before mixing doesn't matter.

Thanks for the quick reply!

 

[TumblingDice]

The textbook, purist answer: It would take the same amount of energy, so it would take the same amount of time.

In the real world, wasted energy would be the major influence for any difference (i.e., energy transferred to the pot, or escaped to the external environment). The amount of wasted energy depends on the heating procedure, but it would seem reasonable to suggest that the extra steps of adding water would result in extra ways that energy could escape from the system.

What does the family think?

 

[voko]

This is a difficult question, because much depends on the way you conduct the experiment.

In a typical setup, boiling the whole bit at once should be faster, and that is because the ratio of surface area to volume is less for bigger volumes (unless the vessel has some weird shape), so you lose less heat through evaporation, head conduction and radiation.

If no heat were lost, and you could account for the time lost on "topping up" properly, you should get identical results.

 

[TumblingDice]

I think its better t boil all of the water at once.
Because when you add cold water to boiled water,they reach to an equilibrium temperature which is below the temperature of the boiled water and above the temperature of the cold water.So the oven should boil the mixture again and the fact that part of it was boiled before mixing doesn't matter.

As long as energy doesn't escape the system, fluctuation of temperature should not make a difference.

 

[ShayanJ]

As long as energy doesn't escape the system, fluctuation of temperature should not make a difference.

Well,because both cases need the same amount of energy,if we take the powers to be equal,then it seems they should need equal amounts of time.
But consider to following:
In the case of boiling the whole water at once,we have for only reaching to the boiling point:

Q=mc(T_f-300) \Rightarrow P\tau=mc(T_f-300) \Rightarrow \tau=\frac{mc}{P}(T_f-300)

Now consider the other case:
<br /> Q=\frac{m}{n}c(T_f-300)+\frac{2m}{n}c(T_f-T_e^1)+\frac{3m}{n}c(T_f-T_e^2)+...+mc(T_f-T_e^{n-1})=mc \sum_1^n \frac{k}{n} (T_f-T_e^{k-1}) \Rightarrow \tau=\frac{mc}{P} \sum_1^n \frac{k}{n} (T_f-T_e^{k-1})<br />
Where T_e^{k-1} is the equilibrium temperature of \frac{(k-1)m}{n} and \frac{m}{n} amounts of water and T_e^0=300

The two expressions for \tau don't seem to be equal!

 

[TumblingDice]

Well,because both cases need the same amount of energy,if we take the powers to be equal,then it seems they should need equal amounts of time.

Right.

But consider to following:
In the case of boiling the whole water at once,we have for only reaching to the boiling point:

Q=mc(T_f-300) \Rightarrow P\tau=mc(T_f-300) \Rightarrow \tau=\frac{mc}{P}(T_f-300)

Now consider the other case:
<br /> Q=\frac{m}{n}c(T_f-300)+\frac{2m}{n}c(T_f-T_e^1)+\frac{3m}{n}c(T_f-T_e^2)+...+mc(T_f-T_e^{n-1})=mc \sum_1^n \frac{k}{n} (T_f-T_e^{k-1}) \Rightarrow \tau=\frac{mc}{P} \sum_1^n \frac{k}{n} (T_f-T_e^{k-1})<br />
Where T_e^{k-1} is the equilibrium temperature of \frac{(k-1)m}{n} and \frac{m}{n} amounts of water and T_e^0=300

The two expressions for \tau don't seem to be equal!

In the real world scenario, liquid water will state change to gas at the surface as well as in bubbles within the liquid if the system is not contained (pressure sealed). Boiling water in an open system doesn't represent an equilibrium condition. The gaseous water is no longer in an equal forward/reverse state for equilibrium condition with liquid water.

I don't want to get lost in a forest of details. I thought it was important to stress to the OP that there were no surprises here, like making ice faster with hot water than cold. But I am realizing that setting up a 'conservation of energy' example in the real world for the OP is a lot crazier than I even initially thought.

So because the OP specifically said, "wait for it to boil", I'll switch to "it will take longer" to boil by adding water incrementally. Only if we added water before liquid water transitioned to gas could we come closer to a utopian result.

 

[sophiecentaur]

Calculations involving 'the method of mixtures' (beloved of 1960s A level Physics) should always give the same answer, whichever order you do them and if heat loss is not included.
Heat loss should be minimised and this could be achieved by keeping surface temperature low so the best deal may be to delay mixing the outer layers and waiting for the central core to heat up first. (Some intelligent baffles followed by stirring).