1 gal Milk freezes in 12 hrs thaws out in 3 days. WHY?

[gary350]

I take a gallon of milk out of the refrigerator the temperature of the milk is 45 deg F. I set the milk on the back porch at 6 pm. It is 20 deg F outside next morning at 6 am the milk is frozen solid.

I put the gallon of milk back into the 45 deg F refrigerator and it takes 3 days for it to thaw out.

Why?

 

[russ_watters]

If the temp stayed at 20 F, the answer is probably just wind...though 45F seems a little high for a fridge.

 

[uart]

Hi Gary. It may just be due to temperature differential asymmetry between freezing and thawing. I know you said 20F outside and 45F inside (and freezing point of milk is about 31F) but you know that the outside temperature will vary throughout the night and I suspect that it's actually getting somewhat colder than 20F in the early hours of the morning.

Also 45F sounds a little high for the inside fridge temperature, 36F to 40F would be more typical. Finally you may find the fridge internal temperature actually drops even further in response to having relatively large frozen object placed in it (especially if the fridge is infrequently opened and closed).

 

[cronxeh]

I tried finding a solution, but I'm too tired to care, someone else can give it a shot

------------------------------------------------------------------------------
density of milk = 1035 kg/m^3
mass of milk= 3924 grams

Cp= 1.97 kJ/kg deg C below 0
Cp= 3.77 kJ/kg deg C above 0
delta T = 13.88 degrees
Latent heat of Fusion = 84 kJ/kg
thermal conductivity 0.17 J m^(-1) s^(-1) K^(-1) at 20° C

Starting conditions: -6.66 deg C
End conditions: 7.22 deg C
Enthalpies:
Q1= -6.66 C -> 0
Q2= Heat of fusion
Q3= 0-> 7.22 C

Q1= m*Cp*DeltaT = 3.924 kg * 6.66 C * 1.97 Kj / kg * C = 51.48 kJoules
Q2= 84 kJ/kg * 3.924 kg = 329.61 kJoules
Q3 = 3.924 kg * 7.22 C * 3.77 kJ/kg * C = 106.80 kJoules

Qtotal = Q1 + Q2 + Q3 = 487.89 Kjoules = 116608.51 calories

Estimating surface area of a carton of 1 gallon of milk..
Surface area of cylinder = 2*pi*r^2 + 2*pi*r*h
Since 1 gallon is 231 cubic inches, suppose the diameter is 4.4 inches, and height is 12 inches

(4.4*4.4*12=232). So the radius is 2.2 inches, and height is 12 inches.. Surface area = 497.62 sq. in.

Heat applied to carton of milk by convection (in 45 deg F heat): Q=hA(Ts-Ta)

Using Newton's law of cooling, time required to heat an object: t=mCp(Ts-Tf)/Q
------------------------------------------------------------------------------

Long story short I got 5 days on one attempt, you need to get the right h for the solution (h=Heat transfer coefficient ). The h will depend on milk jug/carton material (plastic or glass or metal), plus as others have mentioned introducing a frozen block into the fridge will drop the ambient temperature.

 

[D H]

Was there a breeze blowing the night you placed the milk bottle outside? There is little, if any, wind inside a refrigerator. A breeze or a wind will significantly change the heat transfer rate.

 

[dacruick]

Was there a breeze blowing the night you placed the milk bottle outside? There is little, if any, wind inside a refrigerator. A breeze or a wind will significantly change the heat transfer rate.

He said the milk started in the 45F fridge. then he put it into the freezer, it froze, then he put it back into the fridge and it took 3 days.

My question is, did you leave it in the freezer for longer than the 12 hours, or did you measure 12 hours, leave it there, and put it in the fridge later. If you did that, it is possible for the frozen milk to be even colder.

If you measured the 12 hours, then put it back in the fridge then I have two hypotheses.

The first being something to do with the molecular structure of milk being different for frozen milk, and liquid milk.

and second, I remember from high school chemistry something about there being a certain amount of energy that during a phase change that is used up to change from phase to phase. So what I am trying to say is that, maybe more energy is required to change phases back to liquid than there is for a phase change to solid. This also might have something to do with molecular structure.

These are just guesses though. And if you want to search up that phase change thing, my memory is pushing me to say that its called latent heat of fusion or something like that. Give it a shot, and let me know if you get an answer.

 

[MotoH]

I take a gallon of milk out of the refrigerator the temperature of the milk is 45 deg F. I set the milk on the back porch at 6 pm. It is 20 deg F outside next morning at 6 am the milk is frozen solid.

I put the gallon of milk back into the 45 deg F refrigerator and it takes 3 days for it to thaw out.

Why?

He said the milk started in the 45F fridge. then he put it into the freezer, it froze, then he put it back into the fridge and it took 3 days.

wrong.

Anyways, it would have just gotten colder and colder outside, so it was probably a ways below 20 F, and putting it into the fridge slowed down the warming process by about 40 degrees or so. It does make sense that it took longer to thaw than freeze because it wasn't extremely warm in the fridge, and we all know thawing something on the counter is faster than in the fridge.

EDIT:
Why did you put the milk on the back porch?

 

[D H]

He said the milk started in the 45F fridge. then he put it into the freezer, it froze, then he put it back into the fridge and it took 3 days.

Re-read the original post. He put the milk on the back porch, not in the freezer, for 12 hours.

 

[ChmDudeCB]

Changing temperature outside, humidity differences, wind differences

 

[dacruick]

oh, well in the case, you should do what i said haha

 

[russ_watters]

Note, any cooling below freezing or heating above isn't going to have much of an impact here because the latent heat of fusion is going to be an order of magnitude larger than even the heat required to change the temp by 20F

 

[cronxeh]

Cmon you guys are PhDs.. you seriously can't solve this?

I'm lacking thermodynamics background to come up with the solution, but from the looks of it, the Newton's law of (heating) is applicable to the convection heating in the fridge, but in addition you have surface conduction on the bottom of the milk carton at the contact with the fridge. If you take h to be between 1 and 5 (for there is no wind current in the fridge) you should get your answer.

 

[DaveC426913]

Cmon you guys are PhDs.. you seriously can't solve this?

I'm lacking thermodynamics background to come up with the solution, but from the looks of it, the Newton's law of (heating) is applicable to the convection heating in the fridge, but in addition you have surface conduction on the bottom of the milk carton at the contact with the fridge. If you take h to be between 1 and 5 (for there is no wind current in the fridge) you should get your answer.

OK wait. You can't solve it but you're OK knocking other people for not solving it because you think it must be easy. And you figure it probably involves conduction through the carton's bottom...

Got it.

 

[russ_watters]

Cmon you guys are PhDs.. you seriously can't solve this?

I'm just a BS in ME...

I'm lacking thermodynamics background to come up with the solution, but from the looks of it, the Newton's law of (heating) is applicable to the convection heating in the fridge, but in addition you have surface conduction on the bottom of the milk carton at the contact with the fridge. If you take h to be between 1 and 5 (for there is no wind current in the fridge) you should get your answer.

Surface conduction is essentially nonexistent because the contact area of a milk carton is a tiny fraction of the surface area of the carton.

 

[cronxeh]

OK wait. You can't solve it but you're OK knocking other people for not solving it because you think it must be easy. And you figure it probably involves conduction through the carton's bottom...

Got it.

No, I'm knocking lack of interest in this problem. I think its a fascinating problem and I'm having a hard time solving it. I think someone with a PhD will be able to do this in 10 minutes if he just tried.

Surface conduction is essentially nonexistent because the contact area of a milk carton is a tiny fraction of the surface area of the carton.

Although it is a small area, the difference can still be hours when is taken in context of days. As milks thaws the ice will stay on the bottom in direct contact with the surface of the milk carton.

And HEY.. there are 270 views on this thread.. Who are you people?? Make yourselves known, contribute to the discussion!

 

[DaveC426913]

As milks thaws the ice will stay on the bottom

Are you sure about this?

 

[cronxeh]

Are you sure about this?

Why won't I be sure about this? I said it, I must be somewhat sure of it, right?

 

[DaveC426913]

Why won't I be sure about this? I said it, I must be somewhat sure of it, right?

Do you have reason to believe that milk-ice, being mostly water, would behave differently than water-ice?

 

[cronxeh]

Do you have reason to believe that milk-ice, being mostly water, would behave differently than water-ice?

Yes. Oh, I'm sorry, you want my reason?

 

[DaveC426913]

Yes. Oh, I'm sorry, you want my reason?

Sure.

 

[cronxeh]

Sure.

OK.. I think as milk freezes the fat and proteins and all heavy molecules float to the bottom, and as it thaws the water which is of lesser density than the milk, will float up, separating the milk and watery-milk. The iced milk with fats and proteins will be on the bottom.

I'm downloading ANSYS as we speak, should be able to model this thing soon.

 

[DaveC426913]

OK.. I think as milk freezes the fat and proteins and all heavy molecules float to the bottom, and as it floats the water which is of lesser density than the milk, will float up, separating the milk and watery-milk. The iced milk with fats and proteins will be on the bottom.

But fat floats in water...

I'm downloading ANSYS as we speak, should be able to model this thing soon.

Yes, model it. You're ten feet from the kitchen, a glass of milk and a freezer. But by all means, model it on your computer.

 

[cronxeh]

But fat floats in water...


In the time it takes you to model it, you could do the real thing in your kitchen.

When milk freezes it forms cryoglobulin with IgM and precipitates to the bottom.

I'm telling you, this is a fascinating problem! Not only you have difference in density with temperature, you have differences in temperature with temperature.

 

[dacruick]

How about you try it again only with better conditions this time. Instead of putting it on your porch. then come back if you can't figure it out. Use the fridge and the freezer only, so the temperatures are always constant. I don't see it being possible that it takes 6 times the amount of time to melt in the reverse conditions that it took 12 hours to freeze in.

 

[Redbelly98]

No, I'm knocking lack of interest in this problem. I think its a fascinating problem and I'm having a hard time solving it. I think someone with a PhD will be able to do this in 10 minutes if he just tried.

I have a PhD, but lacking your fascination with this problem I am unable to devote sufficient time to it.

As I see it, there is too much unknown or questionable information:

• We have doubts about the 45F fridge temperature.
• How does the OP know the milk was frozen solid, as in completely? Perhaps the central part of the milk did not completely freeze, but with the outer part frozen one cannot squeeze it.
• As has been pointed out, wind chill will increase the freezing rate outside. Just how fast was the wind that night? We don't know.

 

[Mentallic]

And HEY.. there are 270 views on this thread.. Who are you people?? Make yourselves known, contribute to the discussion!

Ok then...
I for one will make myself known. My contribution? I don't know why this is happening.

Sorry to dissapoint... (maybe this could be one of those reasons why there aren't 270 posts coupled with 270 views?)

 

[D H]

Even ignoring the wind chill factor, a normal refrigerator temperature of 36 F and an outside temperature of 20 F will make it take about three times as long to melt the milk as it takes to freeze it. The reason is the heat of fusion of water; see post #11.

Milk is a complex substance, a mix of water, proteins, and fats. Using water rather than milk, the difference in energy between 4.5 pounds of water (1 gallon) at 36 F and 4.5 pounds of ice at 20 F is about 690 BTU. Water has a high heat of fusion. The energy difference between 4.5 pounds of ice at 32 F and 4.5 pounds of water at 32 F is 645 BTU. Almost all (93.5%) of the heat transfer is freezing the water or melting the ice. Since 32 F - 20 F = 12 F and 36 F - 32 F = 4 F, the freezing will be three times faster than the melting process.

Add in a wind during the cooling and this factor of three can easily multiply to a factor of six.

 

[DaveC426913]

Even ignoring the wind chill factor, a normal refrigerator temperature of 36 F and an outside temperature of 20 F will make it take about three times as long to melt the milk as it takes to freeze it. The reason is the heat of fusion of water; see post #11.

Milk is a complex substance, a mix of water, proteins, and fats. Using water rather than milk, the difference in energy between 4.5 pounds of water (1 gallon) at 36 F and 4.5 pounds of ice at 20 F is about 690 BTU. Water has a high heat of fusion. The energy difference between 4.5 pounds of ice at 32 F and 4.5 pounds of water at 32 F is 645 BTU. Almost all (93.5%) of the heat transfer is freezing the water or melting the ice. Since 32 F - 20 F = 12 F and 36 F - 32 F = 4 F, the freezing will be three times faster than the melting process.

Add in a wind during the cooling and this factor of three can easily multiply to a factor of six.

I'm confused about this. If we set up an ideal experiment where we froze a volume of water, storing the dumped energy, and then later, put that stored energy back into the water to melt it, we'd have completely reversed the process, right? In this ideal setup, would the melting process still take longer than the freezing process?

Actually, now that I think about it, that does make sense. The process of heat transfer is not completely reversible. There is no convection and little conduction in a block of ice like there is in water, so melting should take longer than freezing.

 

[cronxeh]

Ok let's rephrase the problem. You guys redefining the problem too much and asking weird questions.

You have 1 gallon plastic milk carton (4.4" x 4.4" x 12") with uniform temperature of 20 deg F. You place this carton into a large box with no wind current and a constant temperature of 45 deg F. You know total energy required is 487.89 kJoules. Find time it will take to get that energy into that carton.

 

[D H]

What is this 45 degrees nonsense? No refrigerator is kept at 45 degrees.

 

[cronxeh]

What is this 45 degrees nonsense? No refrigerator is kept at 45 degrees.

I said a box. Maybe his fridge is broken or one of those redneck fridges, or perhaps its in Arizona and the freon is missing

 

[Yeti08]

Even ignoring the wind chill factor, a normal refrigerator temperature of 36 F and an outside temperature of 20 F will make it take about three times as long to melt the milk as it takes to freeze it. The reason is the heat of fusion of water; see post #11.

Milk is a complex substance, a mix of water, proteins, and fats. Using water rather than milk, the difference in energy between 4.5 pounds of water (1 gallon) at 36 F and 4.5 pounds of ice at 20 F is about 690 BTU. Water has a high heat of fusion. The energy difference between 4.5 pounds of ice at 32 F and 4.5 pounds of water at 32 F is 645 BTU. Almost all (93.5%) of the heat transfer is freezing the water or melting the ice. Since 32 F - 20 F = 12 F and 36 F - 32 F = 4 F, the freezing will be three times faster than the melting process.

Add in a wind during the cooling and this factor of three can easily multiply to a factor of six.

I'm not sure I follow that either. The process may not be reversible, but the two points at 20F and 45F (to be consistant with other posts) have defined internal energies of 30345 and -347308 J/kg, respectively assuming atmospheric pressure. So the change in energy to go from one point to the other should be 1.426 MJ (1352 BTU) for 3.78 kg, and this should be the same regardless of the direction (hot to cold or cold to hot).

If what I said above is correct, then the reason would have to be difference in heat transfer such as the convection coefficient due to wind as others have mentioned. An additional possibility is that the process of freezing creates stratified layers based on temperature. Below 4.2C, the coldest water (by the way I'm assuming water for all my calculations) will be at the top and the warmest will be at the bottom to the water freezes from the top - this effect might speed up the heat transfer, at least by increasing the natural convection inside the container. As the ice then thaws, there will be a frozen core that won't be stratified by temperature (the bulk temperature remains 0C while two phases exist) and the solid will remain fairly centered in the container due to its structure (this is what I've observerd with thawing bottles at least...). The solid core will have a liquid surrounding that could act as an insulating barrier.

All this being said, it's alright to speculate, but I don't think there is enough information to say definitively. It would be more useful if this were a controlled experiment with known parameters.

 

[cronxeh]

So the change in energy to go from one point to the other should be 1.426 MJ (1352 BTU) for 3.78 kg, and this should be the same regardless of the direction (hot to cold or cold to hot)..

How did you get this?

 

[D H]

I'm confused about this. If we set up an ideal experiment where we froze a volume of water, storing the dumped energy, and then later, put that stored energy back into the water to melt it, we'd have completely reversed the process, right? In this ideal setup, would the melting process still take longer than the freezing process?

You haven't specified the problem fully. Is the water/ice staying at 32 F, or changing temperature as well as changing phase? What are the temperatures of the surrounding environment?

Almost all of the heat transfer in this process is melting ice at 32 F / freezing water at 32 F. That is the key to the problem. Reversibility is a red herring in this case.

The process may not be reversible, but the two points at 20F and 45F (to be consistant with other posts) have defined internal energies of 30345 and -347308 J/kg, respectively assuming atmospheric pressure. So the change in energy to go from one point to the other should be 1.426 MJ (1352 BTU) for 3.78 kg, and this should be the same regardless of the direction (hot to cold or cold to hot).

First point: Those numbers are (20F and 45F) are inconsistent with the observation.

Second point: Yes, the energy change is the same (absolute) in both directions. So what? That does not mean the warming and cooling times will be the same.

 

[Yeti08]

How did you get this?

Internal energy of water at 1 atm and 20F = -1428 kJ/kg. Internal energy of water at 1 atm and 45F = 30.34 kJ/kg. Given 1 gallon to be 3.78 kg that total change in energy comes out to about 1.43 MJ. (Data taken from EES)

First point: Those numbers are (20F and 45F) are inconsistent with the observation.

Those are the numbers from the original post, so that's what I used.

Second point: Yes, the energy change is the same (absolute) in both directions. So what? That does not mean the warming and cooling times will be the same.

My point is that just saying the heat of fusion is the answer doesn't explain anything since it is the same amount of energy both ways. Also, I never said it would freeze and warm in the same time period and went on to speculate on why.