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Re: "heat" is NOT a form of "energy"



Bob, this is just too convoluted for my simple mind. BUT if you are
willing to call "Q" "heat" and treat it as a noun which describes an
action, that is close enough for me.

Jim

At 02:00 AM 9/9/99 -0400, you wrote:
Hi Jim,
A statement is simply an assertion and can take a wide variety of forms,
qualitative and/or quantitative; eg.: "I perform this action and that
property increases". A mathematical equation is a very restricted type of
statement:

An equation asserts an equality between (combinations of) numerical
quantities (eg.: position, velocity, force, mass, energy, etc), not
qualitative concepts (eg.: action, process, thought, fear, etc)

The ACTION of heating is a qualitative concept and cannot perform as a
quantity in a mathematical equation. The result of the action of heating
can be an increment in the energy of a system (the numerical quantity U);
this increment in U, due to the action of heating, is Q (the noun heat).

Your statement ("I do F to the system . . .") does not refer to the
quantifiable vector F in Newton's second law, but to some qualitative
notion - the antecedent to quantitative science.

Bob

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

----- Original Message -----
From: Jim Green <JMGreen@SISNA.COM>
To: <PHYS-L@lists.nau.edu>
Sent: Thursday, September 09, 1999 1:15 AM
Subject: Re: heat is a form of energy


> At 11:38 PM 9/8/99 -0400, Bob Sciamanda <trebor@VELOCITY.NET>wrote:
> > > LOOK:
> > >
> > > dU (energy over here on the left side) = Q + W (actions over here on
the
> >right)
> > > Change in a property of the system = Actions DONE to the system
> > >. . .
> >
> >This doesn't make sense dimensionally.
> >The symbol Q stands for the increase in the system property U caused by
> >the action of heating. Q does not stand for the action; it stands for
the
> >increase in U caused by that action!
>
> Bob, I don't think that I understand what you are saying here.
> F=ma: are you saying that F does not stand for an action but for the
> acceleration caused by that action?
>
> N#2: I do F to the system and its velocity increases or I do F (an
action)
> to the system and its KE (a property) increases.
>
> T#1: I do Q (an action) to the system and its U (a property) increases.
>
> I truly wish that someone would explain to me what is so difficult to
> understand about this. There may well be something that I am missing
here.
>
> Jim Green
> mailto:JMGreen@sisna.com
> http://users.sisna.com/jmgreen

Jim Green
mailto:JMGreen@sisna.com
http://users.sisna.com/jmgreen