Timeline for Irreversible process involving temperature change
Current License: CC BY-SA 4.0
13 events
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| Jun 5, 2018 at 10:35 | comment | added | Chet Miller | You do realize that, when I was talking about "finite temperature gradients," what I was referring to was non-infinitesimal, and not non-infinite, correct? When two bodies with different temperatures are placed in contact, the temperature gradient at their interface will be infinite for an instant of time, after which it will decrease with time. So initially, at that exact location, the rate of entropy generation will be infinite. But, of course, when integrated over time and the volume of the bodies, the total amount of entropy generated will, in the end, be finite. | |
| Jun 5, 2018 at 5:47 | comment | added | UVCatastrophe | Since you said that heat flow in the presence of finite temperature gradient produces entropy, I began to think about what might happen in the case of an infinite one. I suppose there has to be an entropy change in that situation as well. I guess its not even that hard to create physically. Right? Any two bodies with different temperatures placed exactly next to each other will have an infinite temperature gradient at the junction. | |
| Jun 4, 2018 at 20:04 | comment | added | Chet Miller | What is your motivation for asking this? | |
| Jun 4, 2018 at 17:30 | comment | added | UVCatastrophe | Could you please give me a gist of what might happen in presence of an infinite temperature gradient? | |
| Jun 4, 2018 at 11:46 | comment | added | Chet Miller | Entropy is generated whenever and wherever heat flows in the presence of a finite temperature gradient. The entropy generation is localized to the region where this is occurring. But, overall, this results in an increase in the entropy of the universe. | |
| Jun 4, 2018 at 10:13 | comment | added | UVCatastrophe | Is it okay to say that whenever heat flows in the presence of temperature gradient , entropy of the universe increases? Of course I'm not implying that this is the only way entropy of the universe can increase. | |
| Jun 2, 2018 at 16:43 | history | edited | Chet Miller | CC BY-SA 4.0 |
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| Jun 2, 2018 at 12:30 | comment | added | Chet Miller | There is another reason. In an irreversible heat transfer process like this, there is entropy actually being generated within the gas as a result of the finite temperature gradients that are present inside the gas. The rate of entropy generation at a given location is proportional to the square of the local temperature gradient. For a detailed derivation of the equations leading to this, see Bird, Stewart, and Lightfoot, Transport Phenomena, Chapter 11, Problem 11D.1. The analysis is very enlightening. For how this applies to the present system, see the Addendum to my answer. | |
| Jun 2, 2018 at 12:29 | history | edited | Chet Miller | CC BY-SA 4.0 |
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| Jun 2, 2018 at 8:41 | comment | added | UVCatastrophe | Is this entropy change coming into the picture because during the cooling process, heat is added to surrounding while surroundings are assumed to be at a constant temperature while when the gas is being heated back up, even though the total heat added is the same (same specific heat for isochoric process and same temperature change), the temperature at which it is being added changes continuously, right? Or is there any other reason why this process should be irreversible. | |
| Jun 1, 2018 at 12:02 | comment | added | Chet Miller | The process I described is neither quasistsatic nor reversible. For a process to be reversible, it is not enough that the system can be restored to its original state. Both the system and its surroundings must be restorable to their original states, without producing any net effect on anything else in the universe. This is not possible for the irreversible process I described. | |
| Jun 1, 2018 at 2:33 | comment | added | UVCatastrophe | Any reversible process needs to be a quasi-static one, i.e. the change in the value of the variable being changed need to be 'continuous' in some way. So in the example, if we raise the temperature of a substance back up again to the original point, the net exchange of heat will become zero (work done=0 & Δ U=0), making it a reversible one. Since it is a reversible process, it must have been quasi-static, right? So do laws of thermodynamics somehow take into account that temperature change of system is a continuous process even if the external temperature change is not? | |
| May 31, 2018 at 15:50 | history | answered | Chet Miller | CC BY-SA 4.0 |