Re: Baka - Selected Quotes Part II

[Michael Press]

In article <[email protected]>,
Bill <[email protected]> wrote:

> Michael Press wrote:
> > In article <[email protected]>,
> > R Brickston <rb20170REMOVE.yahoo.com@> wrote:
> >> If so, then Baka should be the one to figure out that it's not. In
> >> fact, I wonder if the OP asked the question in this way for that
> >> specific reason.
> >
> > Serious students of secondary school chemistry remember
> > the phrase `latent heat of fusion.'
> >
> OK,
> The amount of heat required to change a standard weight from solid, to
> liquid, to vapor, with water being among the highest naturally occurring
> compound.
> Also if it's flash frozen to -40C/F crossover point, it should all be
> solid, unless a certain amount remains super-cooled but not frozen.
> There would also be a noticeable lack of crystalline spurs, since sperm
> can be frozen and revived but not a whole body since the heat doesn't
> 'flash' out.
> Wrong group, and the math could get a bit much, some of which I hated 40
> years ago.

Not flash frozen; rather super-cooled water.
Question is what is the change in entropy upon
phase change from super-cooled water
to ice and liquid water mixture.

--
Michael Press

 

[jim beam]

Michael Press wrote:
> In article <[email protected]>,
> jim beam <[email protected]> wrote:
>
>> R Brickston wrote:
>>> On Wed, 23 May 2007 22:28:29 -0700, jim beam
>>> <[email protected]> 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.

 

[Bill]

Michael Press wrote:
> In article <[email protected]>,
> Bill <[email protected]> wrote:
>
>> Michael Press wrote:
>>> In article <[email protected]>,
>>> R Brickston <rb20170REMOVE.yahoo.com@> wrote:
>>>> If so, then Baka should be the one to figure out that it's not. In
>>>> fact, I wonder if the OP asked the question in this way for that
>>>> specific reason.
>>> Serious students of secondary school chemistry remember
>>> the phrase `latent heat of fusion.'
>>>
>> OK,
>> The amount of heat required to change a standard weight from solid, to
>> liquid, to vapor, with water being among the highest naturally occurring
>> compound.
>> Also if it's flash frozen to -40C/F crossover point, it should all be
>> solid, unless a certain amount remains super-cooled but not frozen.
>> There would also be a noticeable lack of crystalline spurs, since sperm
>> can be frozen and revived but not a whole body since the heat doesn't
>> 'flash' out.
>> Wrong group, and the math could get a bit much, some of which I hated 40
>> years ago.
>
> Not flash frozen; rather super-cooled water.
> Question is what is the change in entropy upon
> phase change from super-cooled water
> to ice and liquid water mixture.
>
If it's all at -40 and stays that way the entropy is temperature
dependent. If it figures out that it should be frozen then the state
change should pop the ice up to a higher temperature but with the same
overall entropy.
Super heated water is where I have a problem with my microwave.
Hot dogs sometimes get to way more than the boiling point in a microwave
and then explode 5 or ten seconds after being taken out.
Fun with physics.
Bill Baka

 

[Bill]

jim beam wrote:
> Michael Press wrote:
>> In article <[email protected]>,
>> jim beam <[email protected]> wrote:
>>
>>> R Brickston wrote:
>>>> On Wed, 23 May 2007 22:28:29 -0700, jim beam
>>>> <[email protected]> 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.

This is somewhat amusing but not enough to actually look up the calories
per gram per degree (C,F,K) pick one.
Bill Baka

 

[Michael Press]

In article <[email protected]>,
jim beam <[email protected]> wrote:

> Michael Press wrote:
> > In article <[email protected]>,
> > jim beam <[email protected]> wrote:
> >
> >> R Brickston wrote:
> >>> On Wed, 23 May 2007 22:28:29 -0700, jim beam
> >>> <[email protected]> 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 undergoes a spontaneous phase change to solid water and liquid
water. The proportion is determined by the initial conditions.

--
Michael Press

 

[jim beam]

Michael Press wrote:
> In article <[email protected]>,
> jim beam <[email protected]> wrote:
>
>> Michael Press wrote:
>>> In article <[email protected]>,
>>> jim beam <[email protected]> wrote:
>>>
>>>> R Brickston wrote:
>>>>> On Wed, 23 May 2007 22:28:29 -0700, jim beam
>>>>> <[email protected]> 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.

 

[Michael Press]

In article
<[email protected]>,
jim beam <[email protected]> wrote:

> Michael Press wrote:
> > In article <[email protected]>,
> > jim beam <[email protected]> wrote:
> >
> >> Michael Press wrote:
> >>> In article <[email protected]>,
> >>> jim beam <[email protected]> wrote:
> >>>
> >>>> R Brickston wrote:
> >>>>> On Wed, 23 May 2007 22:28:29 -0700, jim beam
> >>>>> <[email protected]> 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.

--
Michael Press

 

[jim beam]

Michael Press wrote:
> In article
> <[email protected]>,
> jim beam <[email protected]> wrote:
>
>> Michael Press wrote:
>>> In article <[email protected]>,
>>> jim beam <[email protected]> wrote:
>>>
>>>> Michael Press wrote:
>>>>> In article <[email protected]>,
>>>>> jim beam <[email protected]> wrote:
>>>>>
>>>>>> R Brickston wrote:
>>>>>>> On Wed, 23 May 2007 22:28:29 -0700, jim beam
>>>>>>> <[email protected]> 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?

 

[Bill]

jim beam wrote:
> Michael Press wrote:
>> 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?

Be sure to give the phase of the moon and the alignment of the planets,
and of course, how much is Deuterium, etc.
I having too much fun watching you guys now to even think about
answering the question.
Bill Baka

 

[Michael Press]

In article
<[email protected]>,
jim beam <[email protected]> wrote:

> Michael Press wrote:
> > In article
> > <[email protected]>,
> > jim beam <[email protected]> wrote:
> >
> >> Michael Press wrote:
> >>> In article <[email protected]>,
> >>> jim beam <[email protected]> wrote:
> >>>
> >>>> Michael Press wrote:
> >>>>> In article <[email protected]>,
> >>>>> jim beam <[email protected]> wrote:
> >>>>>
> >>>>>> R Brickston wrote:
> >>>>>>> On Wed, 23 May 2007 22:28:29 -0700, jim beam
> >>>>>>> <[email protected]> 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.

--
Michael Press

 

[DI]

"Michael Press" <[email protected]> wrote in message
news:[email protected]...
> In article
> <[email protected]>,
> jim beam <[email protected]> wrote:
>
>> Michael Press wrote:
>> > In article
>> > <[email protected]>,
>> > jim beam <[email protected]> wrote:
>> >
>> >> Michael Press wrote:
>> >>> In article <[email protected]>,
>> >>> jim beam <[email protected]> wrote:
>> >>>
>> >>>> Michael Press wrote:
>> >>>>> In article <[email protected]>,
>> >>>>> jim beam <[email protected]> wrote:
>> >>>>>
>> >>>>>> R Brickston wrote:
>> >>>>>>> On Wed, 23 May 2007 22:28:29 -0700, jim beam
>> >>>>>>> <[email protected]> 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.
>
> --
> Michael Press

That's just what I thought it was :<) (honestly I have no idea what was
ever being talked about)

 

[Bill]

Michael Press wrote:
> 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.
>
You made it a bit of a trick question by saying flash 'freezing' to -40
and not mentioning a liquid state at -40. The release of heat at the
freezing point is what causes the plateau effect at both the freezing
point and the boiling point with a constant energy input or output.
The trick would be to have all the water in a liquid state at -40 and
watch the released heat from the phase change to solid release heat back
up to 0C. I missed that point so there is the ambiguity problem with
data not being totally provided.
Welcome to the 'Off Topic' nature of news groups.
Bill Baka

 

[R Brickston]

On Fri, 25 May 2007 18:15:12 -0700, Bill <[email protected]> wrote:

>Michael Press wrote:
>> 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.
>>
>You made it a bit of a trick question by saying flash 'freezing' to -40
>and not mentioning a liquid state at -40. The release of heat at the
>freezing point is what causes the plateau effect at both the freezing
>point and the boiling point with a constant energy input or output.
>The trick would be to have all the water in a liquid state at -40 and
>watch the released heat from the phase change to solid release heat back
>up to 0C. I missed that point so there is the ambiguity problem with
>data not being totally provided.
>Welcome to the 'Off Topic' nature of news groups.
>Bill Baka

Fact is you didn't answer the question.

 

[Michael Press]

In article <[email protected]>,
"DI" <[email protected]> wrote:
> That's just what I thought it was :<) (honestly I have no idea what was
> ever being talked about)

Super cooled water. Water freezes at 0 C.
But if it is prepared with very little impurities and
kept still, then it can be cooled below 0 C and
remain liquid. Dropping in a speck
of material will form a nucleus for crystallization of ice.
That is all.

<URL:http://www.newton.dep.anl.gov/askasci/gen01/gen01672.htm>

--
Michael Press

 

[Michael Press]

In article
<[email protected]>,
Bill <[email protected]> wrote:

> Michael Press wrote:
> > 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.
> >
> You made it a bit of a trick question by saying flash 'freezing' to -40
> and not mentioning a liquid state at -40. The release of heat at the
> freezing point is what causes the plateau effect at both the freezing
> point and the boiling point with a constant energy input or output.
> The trick would be to have all the water in a liquid state at -40 and
> watch the released heat from the phase change to solid release heat back
> up to 0C. I missed that point so there is the ambiguity problem with
> data not being totally provided.
> Welcome to the 'Off Topic' nature of news groups.

I did not say flash `freezing'.

--
Michael Press

 

[jim beam]

Michael Press wrote:
> In article
> <[email protected]>,
> jim beam <[email protected]> wrote:
>
>> Michael Press wrote:
>>> In article
>>> <[email protected]>,
>>> jim beam <[email protected]> wrote:
>>>
>>>> Michael Press wrote:
>>>>> In article <[email protected]>,
>>>>> jim beam <[email protected]> wrote:
>>>>>
>>>>>> Michael Press wrote:
>>>>>>> In article <[email protected]>,
>>>>>>> jim beam <[email protected]> wrote:
>>>>>>>
>>>>>>>> R Brickston wrote:
>>>>>>>>> On Wed, 23 May 2007 22:28:29 -0700, jim beam
>>>>>>>>> <[email protected]> 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.

 

[Michael Press]

In article
<[email protected]>,
jim beam <[email protected]> wrote:

> Michael Press wrote:
> > In article
> > <[email protected]>,
> > jim beam <[email protected]> wrote:
> >
> >> Michael Press wrote:
> >>> In article
> >>> <[email protected]>,
> >>> jim beam <[email protected]> wrote:
> >>>
> >>>> Michael Press wrote:
> >>>>> In article <[email protected]>,
> >>>>> jim beam <[email protected]> wrote:
> >>>>>
> >>>>>> Michael Press wrote:
> >>>>>>> In article <[email protected]>,
> >>>>>>> jim beam <[email protected]> wrote:
> >>>>>>>
> >>>>>>>> R Brickston wrote:
> >>>>>>>>> On Wed, 23 May 2007 22:28:29 -0700, jim beam
> >>>>>>>>> <[email protected]> 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.

I'll write it up soon.

--
Michael Press

 

[still me]

On Sun, 27 May 2007 07:27:57 -0700, jim beam
<[email protected]> wrote:

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

Entropy is the habit of a (every) newsgroup to degenerate into a state
of inert posts that have nothing to do with the original purpose.

 

[Michael Press]

In article
<[email protected]>,
jim beam <[email protected]> wrote:

> Michael Press wrote:
> > In article
> > <[email protected]>,
> > jim beam <[email protected]> wrote:
> >
> >> Michael Press wrote:
> >>> In article
> >>> <[email protected]>,
> >>> jim beam <[email protected]> wrote:
> >>>
> >>>> Michael Press wrote:
> >>>>> In article <[email protected]>,
> >>>>> jim beam <[email protected]> wrote:
> >>>>>
> >>>>>> Michael Press wrote:
> >>>>>>> In article <[email protected]>,
> >>>>>>> jim beam <[email protected]> wrote:
> >>>>>>>
> >>>>>>>> R Brickston wrote:
> >>>>>>>>> On Wed, 23 May 2007 22:28:29 -0700, jim beam
> >>>>>>>>> <[email protected]> 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