Re: Hungarian Heat

[guntbert@cut.big.ac.at (Guntbert Reiter)]

*"Re: Hungarian Heat"*
Roger A. Pruitt wrote on Thu, 17 Apr 1997 11:38:11 -0500 :

> They also probably use the same sign for Q and W. In the U.S. we use a
> positive sign for work done by the system and negative sign for "heat" =
flow
> out of the system. In Europe, they use the same sign for this.

Oh my god - not that sign-case again :-))

In ernest: I'd be glad if it were so easy. Even within one single
textbook, sometimes within one chapter, authors change the
interpretation for signs but I hopefully manage to guide the kids
through this swamp. I too use the same sign for whatever kind of
energy trasition between a system and its environment.

------
To get rid  of it: one example for that ambiguity I meant above:
(not omitting anything in between)

Chapter: "1st ?principle? of Thermodynamics"

* (DELTA)U =3D (DELTA)Q + (DELTA)W

Example A:
a) compression of an ideal gas in a cylinder (cross area A) by
moving the piston by (DELTA)x. Whats the change of 'inner energy'
(U)?
answer: volume is reduced, (DELTA)V =3D A * (DELTA)x=20
> > > !?((DELTA)V>0!!)
 -> (DELTA)x =3D (DELTA)V / A.

The work done on the system is: (DELTA)W =3D F*(DELTA)x =3D=20
=46 * (DELTA)V / A =3D p * (DELTA)V

If work is done ON a system we count it positive. By 1st
principle follows=20
(DELTA)U =3D (DELTA)Q + p*(DELTA)V
> > > I could live with this, but now to part=20
b) if the gas does work on its environment by expanding against
an external pressure p its inner energy U will decrease. This
work will be counted negative then=20
> > > still ok, BUT
(DELTA)W =3D -p*(DELTA)V
Hence:
(DELTA)U =3D (DELTA)Q - p*(DELTA)V
--------------
End of example.
those poor kids should accept that there are different formulas
to apply wether there is expansion or compression - Pfffff!



Bye, bye
	Guntbert
-------------------------------------------------
Guntbert Reiter, Graz, Austria
e-mail     : guntbert@cut.big.ac.at