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Freezing a gallon of water

by Taiki_Kazuma
Tags: freezing, gallon, water
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Taiki_Kazuma
#1
Mar15-07, 06:29 PM
P: 23
I have a Gallon of water at room temperature at Sea Level.
I want that gallon completely frozen in 5 seconds.


Supposing you don't account for impurities (ie. it's pure water)...

1) Is this possible?
2) How cold would it have to be?

Let's say you add 50% salt...

1) Is this possible?
2) Howcold would it have to be?



Also...how long would it take to be completely freeze if we were at the freezing point.


Also...what equations are being used? I've been out of School for a bit...but I want to know.
Thanks
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russ_watters
#2
Mar15-07, 10:46 PM
Mentor
P: 22,303
This is such a vague question, that the answers are:

1. Yes.
2. Depends on how you do it.

1. Yes.
2. Depends on how you do it.

This snowmaker, for example, freezes ten gallons every five seconds: http://www.hedco.com/
Taiki_Kazuma
#3
Mar16-07, 07:18 AM
P: 23
That snow maker is taking something that is around freezing anyways and freezing it from what it looks like.

I want the minimal temperature to freeze 1 gallon of pure water from room temperature in 5 seconds.

There is a number for that. What I'm thinking that needs to be done for this problem has something to do with enthalpy. (i think.)
Basically I was thinking of figuring how much energy needs to be added to the system to decrease it to freezing, then how much energy needs to be added to freeze the system. Add the two energies together. Convert Work into time.

I just don't have any equations. My one Chem book from college...is well...I returned it. I just don't know where to begin. There has to be an exact number for this to occur in a timely manner. I'm pretty sure I accounted for all the variables in the system:

Time = 5 seconds
Pressure = 1 atm
Current Temperature = Room Temperature (98 F ~ is that room temp?)
Freezing Point = 0 C
Temperature Outside = ? ~ this is what we are solving for

To solve for temperature outside I believe we need:

Energy needed for the system to freeze....Beyond this my knowledge is cloudy.

I need equations...




I just noticed the previous reply stated this is vague...How can I make this concrete?

I want to basically instantly freeze something...How cold would the surrounding area be for this to happen?

wind chill is not a factor...It's a closed system. What variables am I not thinking of?

(Note: This is not actual homework. This is the search for personal knowledge. I'm curious and want to know how to figure this out.)

eaboujaoudeh
#4
Mar16-07, 07:31 AM
P: 183
Freezing a gallon of water

since its not a homework : u should apply Q = m c deltaT
Q is the heat needed, m is the mass of material, c 4.182 Kj/Kg for pure water, deltaT is your temperature differential..this is the amount ur looking for. physically its possible, but how to do it,well good luck
Taiki_Kazuma
#5
Mar16-07, 07:44 AM
P: 23
Ok...I'm a little rusty since I've been out of school for awhile.

M would be the mass of 1 gallon of water...which I'm pretty should I can figure out by googling it.

C...thanks for the constant ^_^

DeltaT is what I'm trying to find....

What's Q? How do I figure this out?

The problem with this equation...is I don't see how time is a factor. How do I specify...I want this to happen in 5 seconds?


-----

Unless... I'm actually suppose to be solving for Q. In which case the initial equation will give the ........ok I thought I had something. I'm lost once again ;-;

I thought that maybe delatT would be room temp minus freezing point. But I don't know how that works. Is Q's unit in Joules?


------------------

Ok I think I have it:

I need 1 more equation.
An equation that states how much "heat" or energy is needed to freeze a system in 5 seconds.

There needs to be an equation with Q and time in it.

What would this equation be?
Once I have that I can solve the problem.....ooooo I'm soooo close ^___^
eaboujaoudeh
#6
Mar16-07, 08:10 AM
P: 183
yes Q is in Joules, if u divide by the time in the right side of the equation , u get Q in Watts, or J/s
translation would be : from T1 to T2 over a certain amount of time needs this much watts
to get m is simple : density x volume
Taiki_Kazuma
#7
Mar16-07, 08:20 AM
P: 23
Ok, How do I figure out how much energy is needed for the system?

Energy is Joules...which doesn't matter for time.

Watts would be the amount of energy per time.

So I need x amount of energy divided by 5 seconds will equal how cold it would need to be.

I have 2 variables from what I see.

Q and deltaT

I know one of the temperatures which is room temperature.
But I'm kinda at a lost of where to begin for Q.

Will I have to solve the energy required to take the gallon to freezing....and then how long it will take to completely freeze?
I'm thinking that's my next step.

I think I need 2 equations.

the first one to solve for Q.
In this equation would Surface Area be a factor? Or is Volume a factor?


================

Actually I think I might be going about this wrong. Should I first solve the equation for Q? Knowing that the deltaT will be Room temp minus freezing? Then solve it for temp? @_@

-----------------

http://www.svgs.k12.va.us/Courses/Ph...RR2_Chap14.pdf

That pdf mentions something about latent heat. How do I find out the Q for that latent heat? If I found that and added it to the previous energy..wouldn't that be the energy to instantly jump it to the solid phase?


OH MY what's Q = m L?????????
(also in the same pdf) should I use that? o.o
eaboujaoudeh
#8
Mar16-07, 08:31 AM
P: 183
mass is the factor, neither surface area nor volume (well volume through mass anyway). deltaT is the difference of temperature !! its room temperature minus the temperature ur planning to reach which is zero or the freezing temperature. then after that u need to solve how much energy needed to freeze the mass u have, all that in 5 mins. i forgot how to calculate the freezing Q after we get to the freezing temperature line, mayb if u look it on the net i'm sure u'll find it
Taiki_Kazuma
#9
Mar16-07, 08:38 AM
P: 23
Thank you for directing me to the internet.....that's what I've been doing =P

* What are the states (or phases) of matter?

* What can cause a change in the state of matter?

During a phase change (e.g. ice melting) the temperature of the object will be constant even though energy is entering or leaving the object. The equation Q = m c D T cannot be used if the D T range includes a phase change. e.g. water T1 = 20o C T2 = -5o C.

The energy required to change the phase of a substance is called the latent heat. The latent heat of fusion refers to the energy per kg required to melt or freeze a substance. The latent heat of fusion for water is 3.33 x 105 J/kg or 79.7 kcal/kg. The latent heat of vaporization refers to the energy per kg required to vaporize or condense a substance. The latent heat of vaporization for water at 100o C is 22.6 x 105 J/kg or 539 kcal/kg. Table 14.3 gives values for the latent heat, L. Q = mL

http://physicspointers.com/download/giancoli/g14.html


Ok I found Laten Heat...as you can see. This is how you solve for that energy.

My problem is now....after reading that. How do I combine the two equations. I'm trying to change the phase of something. Which means I need to include that. Is this saying that no matter how cold it is outside, the water will freeze at a constant rate?

I know that I can add the 2 energies together which will give me the Q total.
After that...how do I figure the temperature needed to freeze in 5 seconds? I'm not understanding how you can just snap your fingers and say there it's now the temperature needed. What is that equation?

If you have been given me the answer to my question...can you please rephrase it. I'm not understanding and I would like to understand. Thank you so very much for your help thus far. I've gotten farther than I would have without you.
eaboujaoudeh
#10
Mar16-07, 08:44 AM
P: 183
ok the only way u can go on by this is to add the 2 Q's. then u divide them by the time period u want, and u get the value in kW. the reason we look for the kW value in engineering domains is because in all heat applications we are given the power ratings in kW, like heaters, AC's, heat exchangers, radiators.... alll are rated in kW. so we search to find the kW value of our need then we look at what device gives us the amount we are looking for. hope its clear enough..but that's how i understood ur final question
Taiki_Kazuma
#11
Mar16-07, 08:46 AM
P: 23
Ok...Let me try solving for kW so we have a number to work with. Then we'll go from there. It'll probably take me a few minutes. I will have a reply before 15 mintues are up from this time. Thank you so far for your help.
eaboujaoudeh
#12
Mar16-07, 08:57 AM
P: 183
ur welcome..but i have to go,,i'll reply to u as soon as possible. c you around
Taiki_Kazuma
#13
Mar16-07, 09:00 AM
P: 23
http://www.newton.dep.anl.gov/askasc.../chem00005.htm
Based off of this
Density of pure water at room temperature (saying room temperature equals 25 C) is 997.07 g/L

3.79 L = 1 Gallon

m = D/V = (997.07 g/L) * (3.79 L) * (1 kg / 1000 g) = 3.78 kg

c = 4.182 KJ / Kg

DelT = T2 - T1 = 0 C - 25 C = -25 C

Q1 = m (c) delT = (3.78 kg) (4.182 kJ/ Kg) (-25 C) = -395 KJ*C ???

-----

L = 3.33 x 10^5 J/kg * (1 KJ / 1000 J) = 3.33 x 10^3 KJ/Kg
m = 3.78 kg

Q2 = m (L) = 3.78 kg (3.33 x 10^3 KJ/KG) = 126 KJ

-----------

Q3 = Q1 + Q2 = |-395KJ???| + 126 KJ = 521 KJ???


How does my work look so far? I think Q1 is screwed up. I have temperature...and it didn't cancel out.
Taiki_Kazuma
#14
Mar16-07, 09:54 AM
P: 23
----

http://www.engineeringtoolbox.com/wa...ies-d_162.html

This states that rougly the specific heat (I think that's what C is) for the range we are working with is roughly 4.182 kJ/(kg*K)

4.182 KJ/(kg*K)

So let's resolve for Q1 again....

3.79 L = 1 Gallon

m = D/V = (997.07 g/L) * (3.79 L) * (1 kg / 1000 g) = 3.78 kg

c = 4.182 KJ / Kg

DelT = T2 - T1 = 0 C - 25 C = -25 C + 273 K = 248 K

Q1 = m (c) delT = (3.78 kg) (4.182 KJ/(kg*K)) (248 K) = 3920 KJ

-----

Q1 = 3920 KJ
Q2 = 126 KJ

Q3 = Q1 + Q2 = 3920 KJ + 126 KJ = 4046 KJ

-------------

Q3 = 4046 KJ
t = 5 seconds

kW = Q3/t = 4046 KJ / 5 s = 809 KJ/s = 809 kW

--------

Now if all of this is correct.
How to I find the temperature required to freeze a gallon of water in 5 seconds? starting from room temperature.

Or ...if you can't think on how to do that......how about this.
How long would it take for this gallon to freeze? or what equation would be used to figure that out?
chemisttree
#15
Mar17-07, 01:14 AM
Sci Advisor
HW Helper
PF Gold
chemisttree's Avatar
P: 3,724
At any temperature less than 0C, a thin sheet of 1 gallon of water will freeze much quicker than a sphere of the same volume.

You can see that your OP is missing key information.
Russ was right...
eaboujaoudeh
#16
Mar17-07, 02:46 AM
P: 183
no 1st trial was correct.. C is the same in Kelvins or Celsius, cause delta T in kelvin or celsius is the same: (sorry i forgot the degrees in C)
deltaT in celsius: 25-0 = 25
deltaT in Kelvin: (25+273)-(0+273) = 25...same
the first one wass correct
Taiki_Kazuma
#17
Mar19-07, 07:24 AM
P: 23
Ok, so the shape of the liquid matters....

Then let's say it's a cube of liquid. Not a gallon.

1 m^3 will be the size of liquid.
Now how do I figure the time frame? I want this to freeze in 5 seconds....so how cold would it need to be outside for a cube of water to freeze starting from 25 Celsuis?

what equation do I need now to find the time frame? That's what I wanted from the beginning =\

If you guys would have said the equation for time requires surface area...or volume...I would have added that information to it. =\

So now the problem has changed a little.

Now I'm trying to have a Meter Cube of water to freeze in 5 seconds starting from Room temperature at sea level. @_@

where do I go from here?...Let me change the first equation a little.

-------------------------

-------------------------

D = 997.07 g/L
V = 1 m^3 = 1 L

m = D/V = (997.07 g/L) * (1 L) * (1 kg / 1000 g) = 0.99707 kg

c = 4.182 KJ / Kg

DelT = T2 - T1 = 25 C - 0 C = 25 C = 25 K

Q1 = m (c) delT = (0.99707 kg) (4.182 kJ/ Kg*K) (25 K) = 104.2 KJ

-----

L = 3.33 x 10^5 J/kg * (1 KJ / 1000 J) = 3.33 x 10^3 KJ/Kg
m = 3.78 kg

Q2 = m (L) = 3.78 kg (3.33 x 10^3 KJ/KG) = 126 KJ

-----------

Q3 = Q1 + Q2 = 104.2KJ + 126 KJ = 230 KJ



Now how do I figure the time it will take to freeze this....or how do I figure the temperature I need to freeze this in 5 seconds?

Is there some other variable of this problem I'm missing?



Here's a different question.
Let's say you have a 2-dimensional graph.
Y-axis is amount of solid (0% on bottom, 100% on top)
X-axis is time (0 seconds on the left, up to infinity on the right)

We are going to graph the rate at which a mass freezes, would this graph look like a parabola?
If so...wouldn't there be an easier equation to figure out how fast something completely freezes or how cold it would be to freeze something....or how frozen it is at a certain time frame?


Please note this is not actually homework for a particular class. I just want to know this information to apply it to something else I'm doing.
eaboujaoudeh
#18
Mar19-07, 09:23 PM
P: 183
well if it was a H.W they wouldn't have asked u the ambient temperature to freeze this because its another subject.
The Q u are finding is the Q needed to move the temperature from a certain temperature to another, in your case you are moving the object from 25 degrees to 0 degrees, the energy needed to do that is 230 KJ, if u need to do it in 5 seconods then Q = 230/5 = 46 kW.
Now to find how much the temperature is outside then u need a mechanical engineer to calculate the heat convection/conduction from the outside to the inside...a whole different complex subject.


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