Entropy change and reversible/irreversible processes

  • #1
Erik Schildt
1
0

Homework Statement


A mass m is thrown from a height of h. Find the entropy change of the mass, the surroundings and the universe. The temperature of the surroundings is T.

Homework Equations


dS=dQ/T
ΔU=Q+W

The Attempt at a Solution


The main thing I don't understand in this question is how we even can apply dS=dQ/T to the surroundings and the mass since all processes are irreversible. I know that entropy is a state function and you can choose a reversible process between two states and get the same entropy change as in a irreversible process, provided that all quantities we use to calculate the entropy change are state functions. Take the free expansion as an example, we can't use dS=dQ/T to calculate the entropy change since it's a irreversible process and Q isn't a state function, we have to use the central equation which only dependes on state functions. From the solution to this problem, which I've seen, it seems that dS=dQ/T was used. The entropy change of the surroundings is mgh/T, but the first law gives us Q=0 since the change in internal energy and the work the ground does is the same. I don't understand how both are true. The entropy change of the mass is zero, which I can understand if dS=dQ/T can be applied somehow.

Thanks in advance!
 

Answers and Replies

  • #2
22,231
5,134
Here's what I get out of this problem statement. The potential energy of the mass is dissipated in this irreversible collision (or collisions if the mass bounces until is stops), and is converted to internal energy of the mass and the surroundings. Actually, the internal energy of the mass changes virtually not at all, and essentially all the energy is transferred to the surroundings. The surroundings is being treated as an ideal reservoir, with infinite capacity to absorb heat without its temperature changing. So the change in entropy of the surroundings is Q/T, where Q = mgh.
 

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