Re: dW turns into dQ

[Bob Sciamanda <trebor@VELOCITY.NET>]

----- Original Message -----
From: Joel Rauber <Joel_Rauber@SDSTATE.EDU>
To: <PHYS-L@lists.nau.edu>
Sent: Monday, November 01, 1999 1:43 PM
Subject: Re: dW turns into dQ


> Bob wrote:
>
> > This might be compared to asking, in F=mA, "if F includes the sum of
> > electrical, gravitational, etc forces and only their (vector)
> > sum matters
> > for calculating A, why the partitioning?"  Of course it is
> > because this
> > partitioning is our way of enabling the calculation of that
> > sum, but the
> > resulting A wouldn't care if you used a different
> > partitioning (taxonomy
> > of F) so long as you applied the same vector sum.
> >
> > Bob
>
> I wonder???? I realize that in your above analogy, it is only an
analogy;
> but I have a gut feeling that it is different.  The partitioning in the
NII
> example above, is in essence a coordinate system choice.  And the
physics
> doesn't care what coordinates you choose to describe the process with.
I am not speaking of the partitioning of the vector F into its
coordinate-resolved components, I am speaking of conceptually dividing the
causes of aceleration into electrical, gravitational, (perhaps frictional,
Normal, contact)etc.

> In thermodynamics . . .  my gut tells me the physics probably does care
> whether something is dQ or dW (of course that gut feeling may simply be
the
> brownie I just ate). . . Namely we also have the 2nd law (of
thermodynamics)
> to worry about . . . but to what degree?
Certainly in calculating dS=dQ/T  the specification of dQ is crucial, but
since S is a state function, even in this calculation, how you get from
state 1 to state 2 doesn't matter (the process here must of course be
reversible).

IE., once the states are specified, so is dS, and it can be calculated in
accordance with its definition: dS= dQ/T calculated for any reversible
process joining the states

> . . .
> Joel

To sum my position:  In applying dE=dW+dQ, as far as the system under
consideration is concerned, it is important that:
1) the correct total dE is used and applied so that
2) the correct final state of the system is achieved.

That done, it matters not how, or if, you choose to say to yourself, as
you sequentially write down and add the two quantities on the RHS, "this
is dW and that is dQ".  Note again that if 2) above is obeyed then the
correct change in all state variables and state functions will have been
uniquely determined including the entropy.

The choice you make will matter only (as Leigh is wont to quip) in grading
foolish multiple choice questions like "How much work did _____ do in the
described situation?"

Bob

Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor