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I am trying to clarify some of the concepts that I learned so far in my thermodynamic adventure.

For that purpose I will set up a salad with those concepts, so I can see not only their isolated definition but the practical relationship between them all.

SCENARIO:

Suppose I am in a room and I have a ball in my left hand.

Suppose the ball is my thermodynamic system because it's the thing I want to analyze. Hence the room is the environment.

Suppose I let the ball fall free.

THERMODYNAMIC PERSPECTIVE:

Potential energy is gravity pulling the ball downwards. All this potential energy is part of the internal energy of the system.

Part of it, because another part is made of the kinetic energy of the system's particles (be it a gas or be it a whatever).

It will hence swim down across the air, bounce on the ground, and swim back up...

...A few times, while losing its internal energy as heat flow that goes into the surroundings.

Is there any work flow while the ball is swimming down in contact with the air? - I say swim because it represents better that there is an opposition with the air, falling sounds too easy -

There is a positive work flow when the ball is touching the ground, because it is compressing its volume, and hence there is work being done on the system.

There is a negative work flow when the ball is bouncing back from the ground because it is expanding again, and hence the system is doing the work.

Being entropy the freedom of the system to have the possibility to set itself into "entropy" amount of different micro-states, where can I see this entropy?

While falling the room gets the heat flow, so the room increases its entropy while the ball itself is losing entropy?

But while "swimming up" against gravity, the ball gaining entropy again (not as much as the previous bounce because it doesn't get so high).

What if the room is very hot and the ball is very cold? If heat is not flowing out of the ball, is it still losing internal energy while falling? If the room was exaggeratedly hot in contrast to the ball, would that help the ball gain more energy and hence bounce higher than before?

In the end, the ball will have lost all its kinetic energy and will be in thermal balance with the room. It has lost all the entropy it could have lost, and that is the reason why it doesn't keep acting. If it had some entropy to lose, it would definitely keep doing something (everything happens because everything wants to provide entropy to the universe, and it cannot say no until it has completely been robbed of its entropy?).

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    $\begingroup$ Looks like you have typed in a great hurry. You must make your questions clearer. Potential and kinetic energy are not part of internal energy of the ball. They appear as separate terms along with internal energy and work, i.e. on right hand side of $Q=\Delta U+W$. $\endgroup$
    – Deep
    Commented Aug 18, 2016 at 10:43
  • $\begingroup$ @Zero Thank you for your comment. But seriously, I took my time and re-read it at least ten times. I even edited it a few times. The thing is, all those afirmations are there to be corrected. We're all just learning. But to what you say, and extracting the first sentence from Wikipedia: "In thermodynamics, the internal energy of a system is the energy contained within the system, including the kinetic and potential energy as a whole." I still see that internal energy is composed of kinetic and potential energy, and hence those are part of internal energy. Not correcting you, just wondering. :) $\endgroup$
    – Álvaro N. Franz
    Commented Aug 18, 2016 at 10:53
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    $\begingroup$ Is this a thermodynamics question? I assumed the reason the ball stopped bouncing was because of Newton's three laws. $\endgroup$
    – user126513
    Commented Aug 18, 2016 at 16:51
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    $\begingroup$ @ÁlvaroN.Franz Your effort to study thermodynamics on your own is commendable. I would suggest that you stay away from Wikipedia, at least let it be only your first step. Read some popular science books, and I would recommend "Laws of thermodynamics" by Peter Atkins. $\endgroup$
    – Deep
    Commented Aug 19, 2016 at 4:31

2 Answers 2

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We can consider that the whole system {ball + gas + room} is isolated so that the total energy is constant through time. Even in that case where the total energy remains the same, the system can evolve towards a macrostate of higher entropy than initially. Entropy is to be understood here in the sense of the number of microstates of the system {ball + gas + room} compatible with a given macrostate. Now, the macrostate for such a system will be characterised by the velocity distribution of the center of mass of the ball and the temperature of the whole system (temperature of the ball and temperature of the gas and possibly temperature of the walls of the room).

Upon colliding with the molecules of the gas and the walls, the ball will give up energy to its surrounding roughly until it has about the same kinetic energy as one molecule. As it gives up energy, the ball does not really lose entropy (first because its own entropy does not change much and second because the motion of its center of mass does not contribute much to the entropy of the system as a whole) but does contribute to increase the entropy of the gas that has now more energy. Since the gas has more kinetic energy, it has more microstates compatible with this new kinetic energy state and thus the entropy of the gas is greater than before.

Now to comment on the point

In the end, the ball will have lost all its kinetic energy and will be in thermal balance with the room. It has lost all the entropy it could have lost, and that is the reason why it doesn't keep acting. If it had some entropy to lose, it would definitely keep doing something (everything happens because everything wants to provide entropy to the universe, and it cannot say no until it has completely been robbed of its entropy?).

I am not sure this is the right way of formulating it. The ball on its own does not have to lose entropy for the entropy of the universe to increase. Entropy is not something that is conserved but something that gets created. It just so happens that a ball with mass $m_B$ going at velocity $\vec{v}_B$ such that $m_B ||\vec{v}_B|| \gg \sqrt{m k_B T}$ (where $m$ is the mass of a molecule) is bound to be a situation that generates more opportunities for the gas and walls molecules to occupy new states of motions unaccessible before the ball released its energy; that's how entropy is created in this case.

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When the room is very hot and the ball is very cold, Ball will gain some energy in form of heat from the room.

But that heat energy will not add to its potential energy. in other words heat energy will not be transform in kinetic energy while ball is falling down.

So, in this scenario, The ball never bounces back higher than earlier bounce.

Assumption: gaining heat does not change the characteristics of the material of the ball.

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