- #1
nonequilibrium
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So I bought myself the Feynman lectures and was reading up on his discussion on Statistical Mechanics and Thermodynamics (in volume I). The following is regarding Thermodynamics.
So he defined entropy (at least the change in it from a to b) as delta S = int(Q/T) over a reversible path from a to b. He then proved that all reversible paths from a to b give the same result (at least in the framework of gases). However:
"First, suppose that we do irreversible work on an object by friction, generating a heat Q on some object at temperature T. The entropy is increased by Q/T. The heat Q is equal to the work, and thus when we do a certain amount of work by friction against an object whose temperature is T, the entropy of the whole world increases by W/T."
This makes me wonder about two things:
1) I thought that for calculating delta S, you had to find a reversible path from start to end. Is it perhaps because of the following: to calculate the change in entropy for the floor alone, we can say heating it is in itself reversible (floors can cool down), so delta S_floor = Q_floor/T; identically for the object delta S_object = Q_object/T. This would give S = S_floor + S_object = Q/T. That does seem like a big jump to ignore in your reasoning (especially since it was his first example). But even if this is how it works, then that leads me to question 2
2) How do you know that it is well-defined then? For example, imagine a model for friction force where if you were to pull an object faster or slower, it would result in a different Q (for example because if you were to pull really slowly, you'd continously have to fight the larger static friction, then if you were to pull it reasonably fast, just giving the normal (and lower) kinetic friction). Following the reasoning from 1, we get different entropies for the same change from a to b.
All replies welcome,
mr. vodka
So he defined entropy (at least the change in it from a to b) as delta S = int(Q/T) over a reversible path from a to b. He then proved that all reversible paths from a to b give the same result (at least in the framework of gases). However:
"First, suppose that we do irreversible work on an object by friction, generating a heat Q on some object at temperature T. The entropy is increased by Q/T. The heat Q is equal to the work, and thus when we do a certain amount of work by friction against an object whose temperature is T, the entropy of the whole world increases by W/T."
This makes me wonder about two things:
1) I thought that for calculating delta S, you had to find a reversible path from start to end. Is it perhaps because of the following: to calculate the change in entropy for the floor alone, we can say heating it is in itself reversible (floors can cool down), so delta S_floor = Q_floor/T; identically for the object delta S_object = Q_object/T. This would give S = S_floor + S_object = Q/T. That does seem like a big jump to ignore in your reasoning (especially since it was his first example). But even if this is how it works, then that leads me to question 2
2) How do you know that it is well-defined then? For example, imagine a model for friction force where if you were to pull an object faster or slower, it would result in a different Q (for example because if you were to pull really slowly, you'd continously have to fight the larger static friction, then if you were to pull it reasonably fast, just giving the normal (and lower) kinetic friction). Following the reasoning from 1, we get different entropies for the same change from a to b.
All replies welcome,
mr. vodka