Timeline for Internal energy, Enthalpy, and Heat
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 6, 2016 at 16:15 | comment | added | garyp | I suppose it does. If the boundary can conduct thermal energy, then if you do work the thermal energy will increase, increasing the temperature, after which transfer of thermal energy because of temperature gradient (heat) is inevitable. | |
| Mar 6, 2016 at 15:57 | vote | accept | Jose Lopez Garcia | ||
| Mar 6, 2016 at 15:57 | comment | added | Jose Lopez Garcia | Thank you both. Excuse my insistence, but I have yet another question: does that mean that (for an ideal gas) Work implies Heat transfer? (assuming a non-adiabatic boundary and a "non-slow" process) | |
| Mar 6, 2016 at 15:42 | comment | added | hyportnex | @garyp is absolutely right. Even better, do not use the word "heat" as a noun, but only as a verb, that is in the sense of "to heat" and then no mistake will be made. And on occasions when you see it used as a noun substitute in your mind immediately the word "heat" with the words "internal energy exchanged because of temperature gradient". | |
| Mar 6, 2016 at 15:01 | comment | added | garyp | Yes, if at the same time work is done. (For ideal gases only. The situation can be different for real gases. If a phase transition is possible, then $Q$ can be transferred and the temperature remains constant even in the absence of work.) | |
| Mar 6, 2016 at 14:59 | comment | added | Jose Lopez Garcia | Does this mean that a gas can exchange heat, $Q$, without changing its own temperature $T$? How? Thanks. (Can't upvote because I don't have enough reputation). | |
| Mar 6, 2016 at 14:54 | history | answered | garyp | CC BY-SA 3.0 |