Why is boiling water bubbling?

rogerk8

This is probably one of the most stupid questions you'll ever get.

But I was looking at my boiling potatoes the other day and tried really hard to understand why the water was bubbling.

I did understand that closest to the plate the water is hotter due to the stove than higher up in the "can".

This did however not explain the bubbling phenomena because higher temperature just makes the molecules close to the stove's plate move faster.

Something like

$$Ek=\frac{mv^2}{2}\propto {kT}$$

should then apply.

But this just gives the speed of the water molecule, not the bubbling effect.

So what makes these higher Ek water molecules rise to the top of the water level?

They are not lighter than "solid" water.

It seems like more kinetic water molecules rise to the top level of water and then it penetrates the surface tension of water in the form of bubbles.

Most of all I do not get what makes the hotter molecules rise to the surface.

Why do they move to the surface?

It is clear that lighter gases/liquids move above denser ones.

And maybe it is as simple as that (while hotter liquid somehow is less dense than colder liquid)?

Best regards, Roger

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mfb

Mentor
The bubbles are water vapour - a gas. The gas is significantly less dense than the liquid.
A small difference in average energy leads to a large difference in the state, that is the point of phase transitions.

• rogerk8

rogerk8

So closest to the stove, water enters the gas state (within the liquid, observe) and is then less dense than the liquid and therefore rises to the surface?

But how about my asumption that the gas then has to penetrate the surface tension of the water which gives the bubbles, is this right?

And why does less dense gases/liquids lie above eachother?

Does it have to do with gravity, or?

Best regards, Roger

mfb

Mentor
So closest to the stove, water enters the gas state (within the liquid, observe) and is then less dense than the liquid and therefore rises to the surface?
Right.
But how about my asumption that the gas then has to penetrate the surface tension of the water which gives the bubbles, is this right?
Surface tension "tries" to reduce the surface, and releasing the gas to the atmosphere reduces the surface significantly. There is nothing to penetrate, surface tension is helping.
And why does less dense gases/liquids lie above eachother?
What do you mean with "lie"? The heat is coming from the bottom so water there boils. The resulting gas goes up.
There is also constant convection that transports hotter, less dense water to the top and colder, denser water to the bottom.

Chestermiller

Mentor
And why does less dense gases/liquids lie above eachother?

Does it have to do with gravity, or?
Yes. It has to do with gravity. Have you studied Archimedes principle?

Chet

rogerk8

Surface tension "tries" to reduce the surface, and releasing the gas to the atmosphere reduces the surface significantly. There is nothing to penetrate, surface tension is helping.
What do you mean by this?
Especially interested in the bold parts.
What do you mean with "lie"? The heat is coming from the bottom so water there boils. The resulting gas goes up.
There is also constant convection that transports hotter, less dense water to the top and colder, denser water to the bottom.
This was a very good explanation. Thanks!
But, why is there a constant convection?
My interpretation of what you say is that less dense boiling water (gas) moves upwards and this makes room for denser non-boiling water to "fall down" and take it's place continously (=convection?).

Best regards, Roger

mfb

Mentor
"Surface with bubble on it" has a larger surface area than a flat surface (because the bubble wall went away).
My interpretation of what you say is that less dense boiling water (gas) moves upwards and this makes room for denser non-boiling water to "fall down" and take it's place continously (=convection?).
Convection happens in the part that is still liquid. The gas moves up much more rapidly, that part is not called convection.

rogerk8

"Surface with bubble on it" has a larger surface area than a flat surface (because the bubble wall went away).
And surface tension tries to keep the surface small, as I think you said, so surface tension is actually "pushing away" the bubbles, right?
Very interesting to know!
Convection happens in the part that is still liquid. The gas moves up much more rapidly, that part is not called convection.
I will have to look convection up then.
I do however know that most tubes are cooled by convection (i.e air).

Finally, say that I wish to calculate the size of the bubbles (within some probability) while the water is just about boiling, is this possible?

Best regards, Roger

mfb

Mentor
Finally, say that I wish to calculate the size of the bubbles (within some probability) while the water is just about boiling, is this possible?
I guess it is possible to make some model, but it will depend on various details of the boiling process.

rogerk8

Yes. It has to do with gravity. Have you studied Archimedes principle?

Chet
Hi Chet!

Correct me if I'm wrong but I think Archimedes principle goes something like this:

Say that you have a bucket. Let's say that you fill that bucket with some known litres of water while noting the level. If you then take your newly found irregular meteor and submerges it (either by force, due to lower density, or by itself) and measures the increase of water level you then have its volume. Weighing it thus gives its density.

Best regards, Roger

Khashishi

Boiling happens when the vapor pressure exceeds the (mechanical) pressure in the liquid. This is when a bubble can grow with enough force to lift up the weight of the water and air on top of it. However, the entire liquid doesn't evaporate at once because it is difficult for an initial bubble to form, so the rate of boiling depends very much on the roughness of the pot and the presence of impurities to act as nucleation sites (and, of course, the heating power). The vapor bubbles rise because they are less dense than the liquid.

rogerk8

I guess it is possible to make some model, but it will depend on various details of the boiling process.
Is there any chance you would like to elaborate?

I can think of some variables:

1) The actual temperature in the bottom of the cannister (i.e what level the stove is set to)
2) Considering pure water only (without salt, "pure" does however vary from city to city but I think the differences may be omitted for our experiment).
3) Volume is not an issue (just takes time...)
4) Surface tension of water (still a variable, I think)
5) Viscosity

I think the most critical thing is how close to above 100C the stove plate may be set to.

Best regards, Roger

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rogerk8

Boiling happens when the vapor pressure exceeds the (mechanical) pressure in the liquid. This is when a bubble can grow with enough force to lift up the weight of the water and air on top of it. However, the entire liquid doesn't evaporate at once because it is difficult for an initial bubble to form, so the rate of boiling depends very much on the roughness of the pot and the presence of impurities to act as nucleation sites (and, of course, the heating power). The vapor bubbles rise because they are less dense than the liquid.
Hi Khashishi!

This was very interesting to read!

I did however not understand a single thing, except for the last senrtence.

Let's start with the first sentence only.

What do you mean by the bold part, more precicely (I like equations).

May

$$p=nkT$$

be relevant?

Best regards, Roger

Khashishi

The ideal gas law won't work here because a boiling vapor is not ideal, since there are intermolecular attractive forces which cause it to condense into a liquid. You can try a Van der Waals equation of state http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/waal.html. There are some values for water vapor in the table. If you solve the equation, I think you will find two solutions for the density which correspond to a liquid phase and a vapor phase.

Take a look at the phase diagram for water here. https://en.wikipedia.org/wiki/Phase_diagram
The vapor pressure is the line between the vapor and liquid phases at some given temperature. If you are in a liquid region, you can move into the vapor region by reducing the pressure or increasing the temperature.

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Chestermiller

Mentor
Hi Chet!

Correct me if I'm wrong but I think Archimedes principle goes something like this:

Say that you have a bucket. Let's say that you fill that bucket with some known litres of water while noting the level. If you then take your newly found irregular meteor and submerges it (either by force, due to lower density, or by itself) and measures the increase of water level you then have its volume. Weighing it thus gives its density.

Best regards, Roger
This is not Archimedes principle in any form that I can relate to. Kashishi captured what I was referring to when he wrote: "The vapor bubbles rise because they are less dense than the liquid."

Chet

rogerk8

I loved this link!

I have only read it once and think I will have to read it at least two times more to maybe understand.

I liked the way they explained, using formulas and taking it step by step.

I did not get

$$P=P_{IG}-a/V^2$$

which stipulates that the "real pressure would be the ideal gas (IG) pressure minus the contracting forces per unit area due to the intermolecular attractions".

What I do not get is why V^2 is involved because what has this to do with intermolucular forces?

I do however get

$$V=V_{IG}+b$$

where b is the addition of volume due to the molecules themselves.

I find it extremely interesting that there is a "Two-Phase" region for many compounds(?).

This seems to be a region where, at a certain temperature, pressure decreasement stops and is constant for a range of different volumes.

At the left end in the PV-diagram the compound is in its liquid phase while at the right end of this "dome" the compound is at its vapor phase.

This means that at one certain temperature, which seems to determine the with of this dome, a compound can switch from liquid to vapor just by changing the volume.

The interesting thing being that pressure is not affected, only volume while this makes the change of state.

Best regards, Roger
PS
What is pressure, really? I know it has the unit N/m^2 or even more remarkable, J/m^3. The last one is kind of funny and hard to understand because how can there be energy within a volume? What is this energy? Its not speed of the particles because that relates to temperature. So what is it? May you perhaps view it as fast moving prticles (with a certain temperature/speed) that you contract so they can move within a smaller and smaller volume and thereby hit a virtual cannister more often yielding a higher pressure? I really, really do not grasp pressure. Please help me :)

rogerk8

Pressure:

I think you may write

$$F=m\frac{dv}{dt}=\frac{dp}{dt}$$

i.e it's how fast the impulse is changing that gives the force.

The rate of this is probably very hard to actually know but you may begin with a totally elastic (or is it non-elastic? I never seem to learn this but what I mean is that no energy is wasted due to "softness") situation

Now, let's look at what is common in movies, that is considering a rifle with a bullet of 10 gram mass.

Conservation of impulse gives

$$m_1v_1=m_2v_2$$

if 1 denotes the bullet we have

$$v_2=\frac{m_1}{m_2}v_1$$

Now a bullet from a rifle is just about supersonic (I think) so we may set v1 to at least 340m/s.

Considering holding the rifle close to your shoulder while weighing 100kg you then have

$$v_2=\frac{0,01}{100}340m/s=0,034m/s$$

Which means that when you see people fly in movies while shooting bullets, that is just nonsense.

Getting back to pressure.

The impulse will have to change with a certain rate to give a force on the cannister wall.

But how do you know this rate?

And what determines this rate?

Worst case estimations (i.e totally elastic) will of course work but does this agree with reality?

My thought is that the force on the (virtual) cannister always is less.

So pressure (force over area) is always less than calculated.

It is less because the molecules hits the cannister wall in a "softer" way than in the totally elastic case.

But what about when we don't have any cannister walls?

What is pressure then?

I really do not have any idea.

The only idea I have is for gravitational pressure i.e

$$p=\rho gh$$

I get this one but not the plasma pressure I'm talking about.

Would anyone like to help me?

Best regards, Roger

rogerk8

Actually, the conservation of impulse is nothing more than Newton's first law.

I think the first law stipulates that every action has a reaction (i.e the forces are opposite in direction).

Now, considering

$$p_1=p_2$$

and

$$F=\frac{dp}{dt}$$

You just need to integrate F over the same time span to get the first equation.

So the conservation of impulse is actually Newton's first law.

Am I wrong?

Best regards, Roger

rogerk8

Boiling happens when the vapor pressure exceeds the (mechanical) pressure in the liquid. This is when a bubble can grow with enough force to lift up the weight of the water and air on top of it. However, the entire liquid doesn't evaporate at once because it is difficult for an initial bubble to form, so the rate of boiling depends very much on the roughness of the pot and the presence of impurities to act as nucleation sites (and, of course, the heating power). The vapor bubbles rise because they are less dense than the liquid.
Now I'm curious of the second sentence.

How do you come to this conclusion?

Best regards, Roger

Khashishi

I don't know if I got that from somewhere, but it seems reasonable to me. The pressure in the water is essentially equal to the weight per unit area of everything on top of it. If you consider an infinitesimal displacement of this column of fluid upward, it is going to cost an energy of $mg dA = P$. When the vapor pressure is above this, then the fluid will try to expand.

rogerk8

Hi Khashishi!

I'm sorry, but I have stared at this equation for two hours now and I simply don't get it.

Two versions (neither accurate, probably):

1) mgd*A=P[J/m^3=N/m^2]
2) mg*dA=P

The left side of equation 1) gives [Ep*Area] as result (d being the hight) and the unit of this is [Jm^2] which is unrecognizable.

The left side of equation 2) gives [F*Area] as result (dA being the infinitesimal Area) and the unit of this is [Nm^2] which also is unrecognizable.

I like this homework very much but in this case I simply give up.

Best regards, Roger

sophiecentaur

Gold Member
The normal process of boiling can be looked upon as an exaggerated form of convection. It's something that can't happen in zero g. If you have a heating element immersed in water in a space ship, you will just get a volume of vapour round the element (effectively insulating it, thermally) and the pressure can increase catastrophically without heating the liquid that's quite near to the element.

It is common for bubbles to collapse as they rise (normal g) because the surrounding liquid is cooler and the vapour can condense. That can be a noisy process and is sometimes referred to as Bumping. They put 'anti bumping granules' in the water in laboratory distillation equipment (many nuclei will prevent single, large bubbles forming).

Khashishi

Sorry! I should have divided by area instead of multiplying. It should be $mg/A = P$

rogerk8

Ok, Khashishi!

Now it looks like the expected

$$P=F/A=mg/A$$

Let me try to understand this with regard to a fluid column.

Mass may then be replaced with

$$m=h*A*\rho$$

which gives us the familiar

$$P=\rho gh$$

But this last formula is not trivial.

If we use it on a liquid, yes there is some possillity we may view it as a column.

But a column of gas, what is that?

Best regards, Roger

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