Calculating Heat Transfer: Freezing Water from -15°C to 10°C

In summary, to work out the rate of heat transfer between different states of matter, such as freezing water in a cup in a freezer, you would need to use equations such as the Biot or Fourier analysis. These equations require specific information about the materials and conditions involved, such as density, heat capacity, and convection coefficient. A quick way to calculate the heat removal rate would be to use the heat removal rate of the freezer. However, for a more accurate and detailed analysis, a longer process would be needed.
  • #1
Energize
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How do I work out the rate of heat transfer between different states of matter? Say for example I wanted to calculate how long it takes for a freezer at -15 degrees to freeze 30g of water in a cup at 10 degrees, what equations would I need to use?

I've already worked out that 11,274 joules needs to be extracted from the water.
 
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  • #2
Energize said:
How do I work out the rate of heat transfer between different states of matter? Say for example I wanted to calculate how long it takes for a freezer at -15 degrees to freeze 30g of water in a cup at 10 degrees, what equations would I need to use?

I've already worked out that 11,274 joules needs to be extracted from the water.

For a quick answer, all you would need is the heat removal rate of the freezer (watts or btu/hr). How did you arrive at 11,274 joules? Average Cp would be around 4.2 kj/kg-k, so m*Cp*dT =(.03)(4.2)(10)= 1.26kj= 1260 joules. Am I messing something up?

For a long drawn out process, I would try a biot or Fourier analysis (see a heat transfer book). Of course every little thing would be need to be known including the material the cup is made of, the cups wall thickness, cup dimensions, convection coefficient of the air moving around in the freezer..etc. It would take at least an hour to solve it this way.
 
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  • #3
Jupiter6 said:
For a quick answer, all you would need is the heat removal rate of the freezer (watts or btu/hr). How did you arrive at 11,274 joules? Average Cp would be around 4.2 kj/kg-k, so m*Cp*dT =(.03)(4.2)(10)= 1.26kj= 1260 joules.

You forgot to add the specific latent heat of fusion.
For a long drawn out process, I would try a biot or Fourier analysis.

I don't have a clue what those are. :P
 
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  • #4
Hi Energize,

Figuring out the amount of energy you need to remove is actually the easy part. The tricky part is the heat transfer analysis to figure out how long it takes. One possible equation to use is

[tex]\rho cV\frac{\partial T}{\partial t}=\dot Q[/itex]

where [itex]\rho[/itex] is the density of the object, [itex]c[/itex] is the heat capacity, [itex]T[/itex] is temperature, [itex]t[/itex] is time, and [itex]\dot Q[/itex] is the rate of heat loss in watts.

This is the Biot analysis that Jupiter6 mentioned, which assumes that the cup and water are all about the same temperature during cooling (if you have to worry about temperature variations in the cup and water, things get a lot more complicated).

Right away we run into a complication, however, because the cup and water each have different densities and heat capacities.

Next we have to decide how to calculate the heat loss rate [itex]\dot Q[/itex]. In a freezer with a fan, the dominant rate of heat loss might be convection, which is modeled as

[tex]\dot Q=hA(T_\infty-T)[/itex]

where [itex]h[/itex] is the convection coefficient (which we don't know), [itex]A[/itex] is the surface area of the cup and water, and [itex]T_\infty[/itex] is the temperature of the air in the freezer.

Now we put these two equations together to get a first-order differential equation we can solve. Sound good so far?

(Fortunately, this type of analysis is explained in every good textbook on heat transfer. I recommend Incropera and DeWitt.)
 

1. How is heat transfer calculated?

Heat transfer is calculated using the equation Q = mcΔT, where Q is the amount of heat transferred, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.

2. What is the specific heat capacity of water?

The specific heat capacity of water is 4.18 J/g°C. This means that it takes 4.18 joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.

3. How do you calculate the change in temperature?

The change in temperature, ΔT, is calculated by subtracting the initial temperature from the final temperature. In this case, it would be 10°C - (-15°C) = 25°C.

4. Can you use the same equation to calculate heat transfer for other substances?

Yes, the equation Q = mcΔT can be used to calculate heat transfer for any substance. However, the specific heat capacity (c) will vary depending on the substance.

5. What units should be used for the variables in the heat transfer equation?

The mass (m) should be in grams, the specific heat capacity (c) should be in joules per gram degree Celsius, and the change in temperature (ΔT) should be in degrees Celsius.

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