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Heat transfer question.

  1. May 26, 2008 #1
    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.
    Last edited: May 26, 2008
  2. jcsd
  3. May 26, 2008 #2
    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.
    Last edited: May 26, 2008
  4. May 26, 2008 #3
    You forgot to add the specific latent heat of fusion.

    I don't have a clue what those are. :P
    Last edited: May 26, 2008
  5. May 26, 2008 #4


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    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.)
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