Re: There's work, and then there's work

[David Rutherford <drutherford@SOFTCOM.NET>]

Bob LaMontagne wrote:
> David Rutherford wrote:
>
> > > We are talking about quasi steady state currents. There is no net
> > > Transverse current.
> >
> > I never said there was. The net motion in the transverse direction is
> > zero, but the net heat rise in the water due to the transverse
> > components of the motion is not zero.
>
> The focus is on measuring the energy stored in the capacitor. This is done by
> discharging it - giving a dq/dt, which is the current. This is the current used
> in calculating the i^2 R heating in the resistor. When the experiment is over,
> the energy in the capacitor is zero, and there has been a rise in the
> temperature of the water. The mass of the water times the specific heat of water
> times the temperature rise is numerically a measure of the energy previously
> associated with the capacitor. It's that simple.

Talk to Ludwik about it, he agrees with me that doing both experiments
at the same time on the same circuit is valid. Check his response in
this thread at

http://mailgate1.nau.edu/cgi-bin/wa?A2=ind0301&L=phys-l&F=&S=&P=38796

Maybe he can explain it better to you than I can. I'm fresh out of new
ways to explain it, sorry.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555/newtransform.pdf

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