Re: dW turns into dQ

[Joel Rauber <Joel_Rauber@SDSTATE.EDU>]

Bob wrote in part:

> Yes, once you have chosen a reversible path for the calculation of
> dS=dQ/T, you must correctly (and uniquely) specify dQ vs. dW  (to
> correctly evaluate dS this way).  In such idealized quasi-static
> processes, however, there is (I think) never much to quibble
> about as to
> what is dW and what is dQ.

As mentioned in a reply to Jim Green, I'm wondering if we really disagree.
The above is something that I think I could've written.  In particular the
"once you have chosen a reversible path for the calculation of dS=dQ/T, you
must correctly (and uniquely) specify dQ vs. dW" part.

> But I really think that logically it goes the other way.
> Once the final
> state is determined (in part by applying the first law, with wide
> discretion in dW vs dQ), then the change in entropy is determined and
> known from its state function -  this then determines dQ=TdS. In a
> reversible process it is the entropy change which defines dQ.

It strikes me that there is no wide discretion; you must partition it in the
unique way for the path you are utilizing in calculating dS.  But this may
be a chicken/egg disagreement.

Relating all this back to the block sliding on the surface.  Here I'd
maintain that we have a specific mechanism (path) i.e. the path is given;
and there should be a unique way to partition the dW and dQ  . . .

And I guess I'd arrive at that partitioning through examing TdS somehow . .
.  Any thoughts?

Joel