Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

If you heat up water in a tank to boiling point, some of the water will turn to steam. Gas bubbles are created in the water. According to Archimedes principle, the gas bubbles will have a force equal to the weight of what they are occupying. Let's denote this force $F_g$.

$$F_g = \text{Bouyant force of a gas bubble}.$$

Let's denote the height of the tank $h$. We also assume the steam bubbles are created at the bottom of the tank and also that the viscosity of water is nearly zero. So the work done when a bubble reaches the surface is

$$\text{Energy} = (F_g-g m)h$$

$g$ is the gravitational acceleration and $m$ the mass of the bubble.

Where did this energy come from? From the surrounding water? From the boiler? From the bubble itself?

Can someone in detail explain to me this phenomena?

share|improve this question
    
Note that the size of the bubbles is also increasing as they go up because the hydrostatic pressure goes down. One has to incorporate $\int p\,dV$ to the energy conservation, too. –  Luboš Motl Aug 3 at 9:03

2 Answers 2

up vote 3 down vote accepted

Let us assume that once the bubble is created, the physics is essentially no different from what is responsible for the work making an Helium balloon rise in the air i.e. you have a macroscopic/mesoscopic object which has a density smaller than that of the fluid surrounding it.

Now, we all know that in a static fluid under gravity, pressure is decreasing with height. The reason for that is because any fluid follows an equation of state $p = f(n,T)$ where, at fixed temperature say $f$ is an increasing function of the particle density $n$.

Hence saying that the pressure decreases as we go higher in a fluid means that the density goes lower.

If the temperature is the same (buoyancy is observed on length scales much smaller than the scale over which temperature changes) then, it means that the big particle doesn't receive the same amount of collisions coming from the bottom than that coming from the top. As a consequence it will experience a net force in the upward direction.

As Pratyay Gosh has said, the energy of the system need not change here, the thing that changes however is the total entropy of the system that is related to the fact that closer to the bottom wall there are more particles that are "bothered" by the presence of the lighter bubble (because it's huge) than if you were to put this bubble at a higher position.

One way to realize that is to imagine the amount of work one needs to perform to insert a bubble somewhere at some height in the fluid. It is kind of intuitive to see that the less dense the fluid is, the more probable it is to find a "void" in which one can insert the bubble. That is what I mean when I say that is because of entropy.

share|improve this answer
    
Great answer! I just have a quick follow question. So the surrounding water pushes the bubble up? Does it means that it converts thermal energy to do this work? Does the bubble pushes itself up also? –  Woho87 Aug 3 at 9:24
    
Here energy is not the point (at least not in the simple set up I have in mind). The point is to minimize the free energy of the system which is basically $E-TS$. So even if $E$ (the total energy of the system) remains the same it is enough to provide work to simply have a final state which has higher entropy. Remember that what matters (over everything else) to predict the future of a big system is to know which state maximizes the entropy of the universe. –  gatsu Aug 3 at 9:41

When a bubble is rising up, water is filling up the space behind it. The work done by the bubble rising up is exactly same as the water coming down to fill the space behind it, but with a -ve sign. So the total energy will remain constant.

share|improve this answer
    
So the surrounding water makes the work? Does it mean that the surrounding water convert its thermal energy to lift the bubble? –  Woho87 Aug 3 at 9:03
    
Don't even think about the thermal energy part. Even if the water is not boiling, a bubble will always rise up. Thermal Energy has nothing to do with it. It only helps the creation of the bubble. It is not responsible for its upward motion.I guess gatsu has already explained you that. –  Pratyay Ghosh Aug 3 at 13:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.