a balloon filled with hottor water submerged in colder water (more dense) will rise due to bouyancy. But when we remove the balloon's wall why is there still bouyancy. I would think that the molecules disperse (diffusion): At the molecular scale there is no density value/difference so no bouyancy, only diffusion, but then how do we explain the dynamics of wheather systems?
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3 Answers
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Buoyancy on a molecular scale just doesn't make a ton of sense.
Buoyancy is driven by pressure differences, and the way we calculate the pressure differences that generate buoyancy, we are treating the fluids as a continuous medium, not as individual molecules. You might be able to find the same behaviour if you modeled it as a bunch of molecules; but that would get very complicated; and our macroscopic approach usually works very well without that.
As far as your balloon example goes, I believe you would get a bit of both. There would be some diffusion; but there would also be a strong directional component. The average interaction of the warmer water with the cooler water would still result in a net upwards movement of the warmer water, along with some diffusion of molecules and heat in all directions. The flows of heat in this situation can be modeled by convection.
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$\begingroup$ But archimedes' law is Newton's mechanics on the semi-rigid balloon. downward gravity on the balloon and the resulting upward force of the pressure over the surface of the balloon. But pressure does not make any sense on moleculare level so summing pressure over an imaginary surface wouldn't either (I think) I try to reason starting from the kinetics of the gas, i.e. brownian motion, average kinetic energy, speed and distance travelled between collisions. How to arrive at a upward force and acceleration of regions with lower density? I can only imagine the molecules a billiard balls ... $\endgroup$ Commented Jun 16, 2020 at 3:43
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But when we remove the balloon's wall why is there still buoyancy. I would think that the molecules disperse (diffusion): At the molecular scale there is no density value/difference so no buoyancy, only diffusion
when we remove the balloon, as you say, the water will diffuse in the cold one. but I think(we have to experiment it), the cold water will stay higher because of the buoyancy and after a while, it will diffuse. why?
1- when water is hot it means that it has kinetic energy.
2- when it has kinetic energy it means that molecules in a certain volume (the same amount of hot water in the balloon) can increase the inter molecule space.
3- increasing the inter molecule space means that the volume increases and by that, the density gets lower and buoyant force will hold it up.
note: the molecules still have their Brownian movement, but the force coming from the deep parts(buoyancy) will make them stay higher.
4- by the explanation of heat, the collision of hot molecules (with high kinetic energy) with cold molecules (with low kinetic energy) eventually transfers heat from the hot body to the cold, and the average kinetic energy of water will be equal. As a result, the total density of the water container will increase over time and be equal.
but then how do we explain the dynamics of weather systems?
it's exactly the same thing and the process is called convection.
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$\begingroup$ I'm not that much professional but I hope this helps. anyone who saw any mistake I'll be happy to remind it to me. $\endgroup$– AminCommented Jun 14, 2020 at 19:17
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For most of the molecules in the bulk of the balloon, all they experience are other molecules also in the bulk and of the same original states of temperature and pressure. Thus the overall density doesn't change very fast.
For molecules at the interface, they will exchange heat through collision with the surrounding cooler molecules (though this was true with the balloon present as well, just with a rubber molecule middle man). The exchange of heat will lead to a very slow dissipation of energy, but because this exchange is occurring across an interface, it takes a fairly long time for enough energy to redistribute among all of the molecule in the bulk.
