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

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Discussion Overview

The discussion revolves around the freezing and thawing times of a gallon of milk, specifically why it freezes solid in 12 hours at 20°F and takes 3 days to thaw back to 45°F in a refrigerator. Participants explore various factors affecting these processes, including temperature differentials, heat transfer mechanisms, and the properties of milk.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest that wind could significantly affect the heat transfer rate when the milk is outside, while noting that there is little wind in a refrigerator.
  • One participant proposes that the temperature differential asymmetry between freezing and thawing could explain the longer thawing time, mentioning that the outside temperature may drop below 20°F overnight.
  • Another participant discusses the latent heat of fusion and its potential role in requiring more energy to thaw the milk than to freeze it, speculating on the molecular structure differences between frozen and liquid milk.
  • A participant presents calculations involving specific heat capacities, latent heat, and thermal conductivity, indicating that the heat transfer coefficient would be crucial for determining thawing time.
  • Concerns are raised about the accuracy of the initial conditions, particularly the refrigerator temperature, which some argue is typically lower than 45°F.
  • Some participants express frustration over the lack of resolution to the problem, questioning the interest level among others in solving it.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the reasons behind the differing freezing and thawing times. Multiple competing views and hypotheses are presented, indicating that the discussion remains unresolved.

Contextual Notes

Participants note various assumptions, such as the refrigerator temperature and the influence of external conditions like wind and humidity, which may not have been fully accounted for in the initial scenario.

gary350
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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?
 
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If the temp stayed at 20 F, the answer is probably just wind...though 45F seems a little high for a fridge.
 
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).
 
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.
 
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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.
 
D H said:
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.
 
gary350 said:
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?

dacruick said:
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?
 
dacruick said:
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.
 
Changing temperature outside, humidity differences, wind differences
 
  • #10
oh, well in the case, you should do what i said haha
 
  • #11
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
 
  • #12
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.
 
  • #13
cronxeh said:
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.
 
  • #14
cronxeh said:
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.
 
  • #15
DaveC426913 said:
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!
 
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  • #16
cronxeh said:
As milks thaws the ice will stay on the bottom
Are you sure about this?
 
  • #17
DaveC426913 said:
Are you sure about this?

Why won't I be sure about this? I said it, I must be somewhat sure of it, right?
 
  • #18
cronxeh said:
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?
 
  • #19
DaveC426913 said:
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?
 
  • #20
cronxeh said:
Yes. Oh, I'm sorry, you want my reason?
Sure.
 
  • #21
DaveC426913 said:
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.
 
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  • #22
cronxeh said:
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...

cronxeh said:
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. :-p
 
  • #23
DaveC426913 said:
But fat floats in water...


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

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.
 
  • #24
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.
 
  • #25
cronxeh said:
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. :biggrin:

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.
 
  • #26
cronxeh said:
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 disappoint... (maybe this could be one of those reasons why there aren't 270 posts coupled with 270 views?)
 
  • #27
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.
 
  • #28
D H said:
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.
 
  • #29
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.
 
  • #30
What is this 45 degrees nonsense? No refrigerator is kept at 45 degrees.
 

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