Timeline for Thermodynamics of a single spring in gravitational potential
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 25, 2019 at 21:56 | comment | added | Chet Miller | For a simple cookbook primer (with worked examples) on how to determine the change in entropy for a system that has experienced an irreversible process, see the following: physicsforums.com/insights/grandpa-chets-entropy-recipe | |
| Sep 25, 2019 at 21:54 | comment | added | Chet Miller | It is much simpler than that. Since entropy is a physical property of the materials comprising the system (i.e, the mass and the air in the room), all you need to do for the reversible process is to take these from their initial thermodynamic equilibrium state to their final thermodynamic equilibrium state by putting them in contact with a constant temperature reservoir at a slightly higher temperature than the initial temperature until the total heat transferred matches the change in their internal energy that they experienced in the irreversible damping. No mechanical effects are necessary | |
| Sep 25, 2019 at 19:00 | comment | added | user8736288 | I understand you "devise" a reversible process in order to obtain $dS= \delta Q_{rev} / T$. If we were to write something from the standpoint of the system "mass+ spring": the work done by the conservative gravity forces should go into $\Delta Ep$ in the lhs. I would still write $\Delta U$ as the variation of strain energy. The quantity computed then corresponds to the sum of the work done by non conservative (drag) forces + some heat exchange $Q$, but this cannot further be specified, without knowing the specifics of the dissipation mechanisms. Would that be correct? Many thanks. | |
| Sep 25, 2019 at 16:42 | comment | added | Chet Miller | Notice that I didn't say anything about heat flow in my answer. The heat flow in the actual process is irrelevant to the actual entropy change, which must be established by devising an alternative reversible process, say, by separating the mass and the surrounding air, and transferring reversible heat to each of them separately, consistent with their changes in internal energy in the actual process (i.e., between their initial and final thermodynamic equilibrium states). | |
| Sep 25, 2019 at 16:36 | comment | added | Chet Miller | The amount of heat transfer between the spring/mass and its surroundings is way less than this. Plus, mechanistically, all the entropy change is due to viscous dissipation of mechanical energy (resulting from air drag) in the surrounding air boundary layer adjacent to the mass. This is the region that first increases in temperature. A small amount of heat then flows from the air boundary layer to the mass, and the rest is transported by convection and conduction throughout the surrounding air. So the heat flow from the mass to the surroundings is very small, and actually negative. | |
| Sep 25, 2019 at 15:22 | comment | added | user8736288 | The second method appears to yield a consistent result with your answer, and looks correct to me. So does your answer. It would be interesting if you could specifically point to what is wrong in the proposed second method. | |
| Sep 25, 2019 at 3:32 | history | answered | Chet Miller | CC BY-SA 4.0 |