Thermodynamics: Change in entropy of a falling ball

  • #1

Homework Statement


In a closed isolated system, a small rigid ball is held at height h1, which we will call state A. By remote control a pin is pulled and the ball is released. It falls to height h2, called state B. Calculate the change of entropy S of the system. Is it a reversible or irreversible process?


Homework Equations



Delta S = Delta Q/T

The Attempt at a Solution


My attempt is that since the system is isolated that means no heat is transferred and the entropy is not changed at all.

But again, since the position of the ball is now lowered, it also means it has lost some potential energy, so change in Q can be related to change in energy which can be mg(h1-h2), with T being constant. So am not sure which one is the right answer.

Also sometimes I think it is an irreversible process because external work will be needed to bring the ball back to its original height but then I also get confused that it might be reversible process because it's an closed isolated system.

Thanks for your help in advance!
 

Answers and Replies

  • #2
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The ball seems to be still falling at H2. In this case it would appear the since the ball can still do the same work it could at H1, I'd say the entropy is unchanged. If you said the ball bounced off the bottom, there'd be a little heat there and the ball would loose the ability to recoil back to H1. In that case you'd have an entropy increase.

It's reversible if the ball can bounce back up to H1. It's rigid so that might work. If the ball was made of clay-filled leather you'd imagine it would be quite irreversible.
 
  • #3
The ball seems to be still falling at H2. In this case it would appear the since the ball can still do the same work it could at H1, I'd say the entropy is unchanged. If you said the ball bounced off the bottom, there'd be a little heat there and the ball would loose the ability to recoil back to H1. In that case you'd have an entropy increase.

It's reversible if the ball can bounce back up to H1. It's rigid so that might work. If the ball was made of clay-filled leather you'd imagine it would be quite irreversible.

Well I thought, the ball kind of stays at the second position. I agree to your theory but again this notion of mechanical work done in this process bugs me a lot. What about the loss in potential energy of the ball?

Note: Note that height h1>h2, the initial height is greater than the final height.
 
  • #4
Andrew Mason
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Well I thought, the ball kind of stays at the second position. I agree to your theory but again this notion of mechanical work done in this process bugs me a lot. What about the loss in potential energy of the ball?

Note: Note that height h1>h2, the initial height is greater than the final height.
It is not a very good question because we don't know how the ball loses its energy. If it landed on a spring and the spring energy held by another pin at its maximum point, there would be no heat produced and no change in entropy. The ball could go back to its original height by releasing the spring. So it is reversible.

Having said that, I suspect that you are supposed to assume that the change in potential energy is lost as heat.

AM
 
  • #5
It is not a very good question because we don't know how the ball loses its energy. If it landed on a spring and the spring energy held by another pin at its maximum point, there would be no heat produced and no change in entropy. The ball could go back to its original height by releasing the spring. So it is reversible.

Having said that, I suspect that you are supposed to assume that the change in potential energy is lost as heat.

AM

Thanks Andrew!

So you suggest the answer would be that "Entropy doesn't change" and the process is "reversible"

Right? I am going to confirm that with the Teaching Assistant or the professor himself soon, and if I get something from them, I'll share here :)
 
  • #6
Andrew Mason
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Thanks Andrew!

So you suggest the answer would be that "Entropy doesn't change" and the process is "reversible"

Right? I am going to confirm that with the Teaching Assistant or the professor himself soon, and if I get something from them, I'll share here :)
No. I am not sure. I think the person who posed the question may want you to assume that the change in potential energy is transferred to the surroundings as heat, in which case, the entropy increases by the amount mg(h1-h2)/T where T is the ambient temperature.

AM
 
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