Heat transfer

luis20

I found the following:

We have already seen that the flow of heat through conduction occurs when fast (hot) atoms collide with slow (cool) atoms, transferring some of their kinetic energy in the process. One might wonder why the fast atoms don't collide with the cool ones and subsequently speed up, thereby gaining kinetic energy as the cool ones lose kinetic energy - this would involve the spontaneous transfer of heat from a cool object to a hot one, in violation of the 2nd law. The answer lies in energy and momentum conservation in a collision - one can show, using these two principles, that in a collision between two objects which conserves energy (called an elastic collision) the faster object slows down and the slower object speeds up.

I believe there are some configurations in which the hot atoms speed up and the colder atoms slow down.
The energy flows from hot to cold, but it's possible that some energy also flow from cold to hot, naturally. But in the end, the flow is like 99% from hot to cold.

Am I right or is the statement in bold true, and if so, how can you prove it?

mfb

If you have an elastic collision between a fast object and a slow object of the same mass, the fast object will always lose energy and the slow object will gain energy. With different masses, I would assume that you can make the same statement with kinetic energy, but I did not check this.
Note that individual atoms do not have a temperature - there are no "not atoms", just atoms with a high kinetic energy.

Yes. This is related to the velocity distribution of the atoms - some atoms in a cold object are quite fast, too, and some atoms in a hot object are slow. If those hit each other, you transfer some energy from the cold object to the hot one. However, this is rare, the common process is energy transfer hot -> cold.

In this post, I neglected other energy types which can be present, but the concept is similar for them.

luis20

Thanks a lot for the answer.

But look at this example (in the attachment):

Atom A is slower and atom B is faster. The configuration in the left side is when the atoms start to really feel the repelling forces from each other. If this is the real configuration, then atom B will gain |Vx| and |Vy| so it will get even faster than atom A.

So, according to this example, hot atoms can speed up and colder atoms slow down. What do you think now?

I really need someone to support this idea, or explain me why I'm possibly wrong !! :)

Attached Thumbnails

 

luis20

Anyone? :o

Look at the picture. It seems there are collisions in which the faster object gets even faster than the slower object.

mfb

Let's see, all units are m/s:
The center of mass system moves with 5 to the right, 0.5 down. In this, both objects have a velocity of sqrt(5^2+0.5^2). This velocity is the same after the impact, but the direction can be different. In the ideal case, the fast object moves in the same direction as the center of mass system in our lab system. Transformed back, it gets a velocity of 2sqrt(5^2+0.5^2)=~10.05. Which means you are right. Even more: One object can stop completely (maybe a feature of the orthongonal velocity vectors). Unless I made a mistake here.

However, this is quite unlikely - on average, there is an transfer energy to the slower particle.

256bits

elastic collisions have conservation of kinetic energy as well as conservation of momentum.

You can play around with the equations on this site here, with different velocities and masses of particles.
http://chemistrydaily.com/chemistry/Elastic_collision

luis20

I didn't quite understand how did you make the calculations but I could see that the faster atom most speed up! If the slower atom comes perpendicularly or from behind, the faster object will speed up.
I think this could be likely in a system with few atoms. In the real situation, atoms are surrounded in many directions, and so the entropy goes up. Energy flows from hot to cold.

Another point: You said atom B gained 0.05 m/s, but we could come up with another situation in which the faster atom (B) would gain even more velocity, right?

luis20

Thanks. The problem is: I'm talking about collisions in at least 2 dimensions. Calculations become harder. :\

mikeph

I'm looking at your picture, and wondering how two "billiard balls" can collide in such a way. Specifically, you're requiring that the slow moving ball strikes the rear of the fast moving ball. If they are to collide with those precise velocities, would they not collide in such a way that the contact point is on the forward-moving face of the faster ball?

At the point when they are perfectly aligned vertically, if they have not yet collided, then I suggest that they never will, because the receding ball is travelling faster than the approaching ball, so the separation between the centres will increase, and since they are both circular, this means the minimum separation between the edges will also increase.

mfb

The whole kinetic energy is transferred, there is no way to make it faster.

The contact point is the uppermost point of the lower ball. The horizontal velocity remains unchanged, and the vertical velocity is as in a 1-dimensional collision.


That is not right, as the movement of the slower ball is directly in the direction of the ball separation, while the other ball moves perpendicular to that. Those collisions are possible, too, but the energy of the faster ball is a bit lower there.

luis20

Right! One last question

The equation of conservation of energy is:

Ma*sqrt(Viax^2 + Viay^2) + Mb*sqrt(Vibx^2 + Viby^2) = Ma*sqrt(Vfax^2 + Vfay^2) + Mb*sqrt(Vfbx^2 + Vfby^2)

and not separated like this:


Ma*Viax^2 + Mb*Vibx^2 = Ma*Vfax^2 + Mb*Vfbx^2 , for direction x

Ma*Viay^2 + Mb*Viby^2 = Ma*Vfay^2 + Mb*Vfby^2 , for direction y


Right? It's a stupid question, I'm just not sure x_x


Note: Via= initial velocity of body A; Vfb= final velocity of body B...