Re: Baka - Selected Quotes Part II

In article
<3KCdnT2Lqc7wCMTbnZ2dnUVZ_rrinZ2d@xxxxxxxxxxxxx>,
jim beam <spamvortex@xxxxxxxxxxxxxxx> wrote:

Michael Press wrote:
In article
<iZCdnZdkQvL_7crbnZ2dnUVZ_jydnZ2d@xxxxxxxxxxxxx>,
jim beam <spamvortex@xxxxxxxxxxxxxxx> wrote:

Michael Press wrote:
In article
<gbKdnTI0UeWF2srbnZ2dnUVZ_sSmnZ2d@xxxxxxxxxxxxx>,
jim beam <spamvortex@xxxxxxxxxxxxxxx> wrote:

Michael Press wrote:
In article <CI2dnb0Jg9ldj8rbnZ2dnUVZ_sWdnZ2d@xxxxxxxxxxxxx>,
jim beam <spamvortex@xxxxxxxxxxxxxxx> wrote:

Michael Press wrote:
In article <rv2dnVBHrPl5EMjbnZ2dnUVZ_r_inZ2d@xxxxxxxxxxxxx>,
jim beam <spamvortex@xxxxxxxxxxxxxxx> wrote:

R Brickston wrote:
On Wed, 23 May 2007 22:28:29 -0700, jim beam
<spamvortex@xxxxxxxxxxxxxxx> wrote:

Michael Press wrote:
<snip>
2. A gram of water is super-cooled to -40 C. It
spontaneously undergoes a phase change to a mixture of
water-ice. What is the change in entropy?
what's the proportion of water-ice in this mixture?
Please, let Dr. Baka the Physics major answer it.
but the question is incomplete, dumb-ass.
No, it is not.
yes it is. you think 10% solid is the same /_\S as 90% solid?

Sure, you have to look up the properties of water.
Not too difficult, even if it is not in a book
on your shelf.

unless you want an algebraic answer, you need proportions.
Water supercooled to -40 C is a fully determined configuration.
(assuming it is at 1 atomosphere and insulated from the
environment, and why not?)
it's not "determined", it's subject to a number of preconditions, not
least of which is the means to contain and cool without triggering
nucleation.

It undergoes a spontaneous phase change to solid water and liquid
water. The proportion is determined by the initial conditions.
see above. you need proportion of solid to determine /_\S, your
original question.
Liquid water at -40 C, 1 atmosphere is fully determined.
With this and the known properties of water, the ratio
of liquid to solid water after the phase change can be
calculated.

satisfy my curiosity - what is it?

I posted these problems for fun, thinking some would
enjoy thinking about them. Did not intend to do much
else, or post solutions. But here goes.

Latent heat of fusion for water = h = 3.34x10^5 J/kg.
Specific heat of water = k = 4.200x10^3 J/(kg.K).

Denote by T_0 the initial temperature, -40 C in our case.
Some fracttion, f, of the super-cooled water will freeze.
0 <= f <= 1.
As it freezes it releases heat,
raising the temperature of the water-ice mixture.
It eventually equilibratetes at some temperature T_1.
If there is some liquid water, then T_1 = 0 C.
Otherwise T_0 <= T_1 <= 0 C.
Denote by m the mass of water.

m.f.h = (T_1 - T_0).m.k
so
f = (T_1 - T_0).k/h
= (T_1 - T_0).k/h
= (T_1 - T_0) x 0.0126

When T_1 - T_0 = 40 K, then f = 0.503.

thanks michael. i haven't looked at that stuff in decades - it's fun!

what's the entropy? i vaguely remember the second law but am too rusty
to apply it.

Consider the following reversible process.
* Raise the temperature of the supercooled water to 0 C.
The amount of heat added is (T_1 - T_0).m.k
* Remove heat from the 0 C water as ice forms until fm ice is formed.
The amount of heat added is -m.f.h = -(T_1 - T_0).m.k

The entropy change for the first stage of this process is
int_{T_0}^{T_1} m.k dT/T = m.k.log(T_1/T_0)

The entropy change for the second stage of this process is
-(T_1 - T_0).m.k/T_1

The total entropy change is
/_\S = m.k(log(T_1/T_0) - (T_1 - T_0)/T_1)
= m.k(log(273/233) - 40/273)
= 0.011913 m.k
= 0.011913 . 0.001 kg . 4200 J/(kg.K)
= 0.05004 J/K

--
Michael Press
.

Relevant Pages

  • Re: Baka - Selected Quotes Part II
    ... jim beam wrote: ... Liquid water at -40 C, ... of liquid to solid water after the phase change can be ... Denote by T_0 the initial temperature, ...
    (rec.bicycles.tech)
  • Re: Baka - Selected Quotes Part II
    ... jim beam wrote: ... Liquid water at -40 C, ... of liquid to solid water after the phase change can be ... Denote by T_0 the initial temperature, ...
    (rec.bicycles.tech)
  • Re: OT physical chemistry
    ... jim beam wrote: ... With this and the known properties of water, ... of liquid to solid water after the phase change can be ... Denote by T_0 the initial temperature, ...
    (rec.bicycles.tech)
  • how does God defy the second law of thermodynamics
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    (sci.physics.relativity)
  • Re: Baka - Selected Quotes Part II
    ... jim beam wrote: ... The proportion is determined by the initial conditions. ... Liquid water at -40 C, ... of liquid to solid water after the phase change can be ...
    (rec.bicycles.tech)
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