Re: [Phys-L] work versus mechanical transfer of energy

[Diego Saravia <dsa@unsa.edu.ar>]

> Feynman famously defined equilibrium as the situation "When all
> the fast things have happened, and the slow things have not."
> That is a slightly smart-alecky way of saying it, but the idea
> is sound:  You can't talk about equilibrium unless you have a
> particular timescale in mind, not too long and not too short.
beatiful


   My model

> corresponds to two capacitors connected by very high value
> resistor (not by a short circuit).
I agree



> For any halfway reasonable value of the resistor, the total
> energy dissipated in the C-R-C circuit is independent of the
> resistor value ...


ok


> so you can ignore its inner workings.
mmm, if you are not interested in the entropy generation, ok

but what was the original question?

if we have an irreversible system wich obeys that equation.

so we know the we have combined irreversible systems in wich the reversible
parts obeys that equation,

 ok we can say that the irreversible part its stationary/ciclic/small and
ignore it.

but what happens anyway in the part that generates entropy?

 for example you can divide the filament in small parts each with one T,
but ... irreversibility is in the borders!

so you are not appling the equation to the irrevesible part.

As I know, the thermodynamic way of studing a system implies

a) the system have a U(X....Z) function  1st principle, experimental
b) you have a mechanical/electrical/etc definiton of work (process
quantity))
c) heat is defined as the diference in the process.

meassuring what happens in the interface (work,heat) is important, you
losse info if you only studies what happens in each of the systems

force, distance, are easy to meassure/visualize, so work

One system for study the problem could be a constant adiabatic volume box
with a mixing pad,  U(S) exists and could be seen as continuous. You have
work, but is not pdV, is torque * modulus of angle displacement

You dont have heat.

Its irreversible, and obeys dU=TdS, even simper one than dU=TdS-pdV

but i haves a problem, work is not reflected directly in variables

You need the first principle in the form d U= dd Q - dd W

Ok you can say the external system loose energy, the systems gain its. But
the form of calculating external loose is by work definition


-- 
Diego Saravia
Diego.Saravia@gmail.com
NO FUNCIONA->dsa@unsa.edu.ar