Re: ENERGY WITH Q

[kowalskil <kowalskil@MAIL.MONTCLAIR.EDU>]

For the time being deltaQ is due to friction and is equal
to Emech which is lost in the process. The way in which
I introduced deltaQ makes it path dependent. I plan to
addres the first law in Model 3. The corrected draft
for Models 1 and 2 will be posted tonight. Thanks for
constructive comments.
Ludwik Kowalski

"RAUBER, JOEL" wrote:
> I'm a little worried about the Delta Q notation, in the
> sense of trying to avoid definitions (and notation, that
> isn't consistent with later usage.) The Delta Q notation
> implies to me, that the calculation only depends on the
> endpoints of the process.  Is that what you intend to convey?
>
> ps, this is the old "d q" versus "dbar q" issue.