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[thermodynamics] rate of ice formation, -area, +density

  1. Feb 7, 2009 #1
    1. The problem statement, all variables and given/known data

    A tank of water has been outdoors in cold weather, and a slab of ice 5 cm thick has formed on its surface. The air above the ice is -10C. Calculate the rate of ice formation ( in cm / h) on the ice slab. Take the thermal conductivity of ice to be 0.004 cal / (s.cm.C) and its density to be 0.92 g/cm^3. Assume no energy transfer through the tank walls or bottom.

    2. Relevant equations

    P(conducted) = Q/t = kA (T2-T1)/ Length
    p(density) = m/v
    Q = Lm heat of transformation

    3. The attempt at a solution

    Some thoughts:
    When they say on the slab, they must mean underneath it right?
    Also they provide standard info the in the question like the conductivity of ice which I could look up in a table. I could similarly look up density and conductivity of water but since they didnt explicidly mention that too, I wonder if I'm not supposed to? Like maybe thats variable in terms of the problem?
    Since they arent just asking the rate of conductivity through ice I probably need to use the heat of fusion formulae somewhere too.

    So I need to relate the density somehow, in order to cancel out the Area which I dont have.

    Pluggin in info for the conduction rate equation I get
    P = 0.004A (0--10) / 5
    P = A x 0.008

    now for density
    p = m/v = m/Ah (using A and h cuz I have more information about that, than just V)
    0.92 = mA5

    I don't see an obvious relation.
    (heat of fusion from water to ice = 333)

    Q = Lm
    Q = 333m
    Q = 333 x 0.92/5A

    Uhm I don't know where im going with this. I could say both equations are equal to A, but then I still have P and m as variables
     
  2. jcsd
  3. Feb 7, 2009 #2

    Mapes

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    You're relating density to the current thickness of the slab, but don't you want to relate it to the ice growth rate (i.e., connect a cm/h rate to a g/s rate)?
     
  4. Feb 12, 2009 #3
    But the density of the ice doesnt change as more is added to it, does it? The ice formed already keeps its density and as for the ice busy forming - isnt each bit formed instantly at its maximum density?
     
  5. Feb 12, 2009 #4

    Mapes

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    I don't mean that the density is variable; I mean that you've written an equation equating the ice density to the ratio of the original slab mass and the original slab volume (which is valid but irrelevant), but what you want to do is write an equation equating the ice density to the ratio of the rate of increasing ice mass and the rate of increasing ice volume.
     
  6. Dec 16, 2009 #5
    Hey, I know this post is old, but i find myself working over a close to identical problem and find myself in the same situation. I cant figure out how to get the area from the density. I have an expression Pcont = .008A and I know that density is mass over volume. However, I also know it is constant in the Ice. I have no expression for mass. The only thing I need to get is the rate of the formation of ice, which should be equal to Pcont. I just cant find an expression to manipulate the density to get the Area. The correct answer is .4 cm/s
     
  7. Dec 16, 2009 #6

    Mapes

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    Hi blackstrat, welcome to PF.

    If you know the mass rate of ice formation (in kg/s) and you know the ice density (in kg/m^3), then you can easily find the volume rate of ice formation (in m^3/s). Know what I mean?
     
  8. Sep 5, 2010 #7
    Recived the same homework and solved as follows:
    P=kA ∆T/L
    Q/t=kA ∆T/L
    Q_1=t.kA ∆T/L=3600 .0,004 .2A=28,8cal/〖cm〗^2
    Q_2=L .m=333kJ/kg=79,55cal/g
    p=m/v=0,92g/〖cm〗^3
    ∴Q_2=73,19cal/〖cm〗^3 (substitute density intoheat of transformation)
    but Q_1=Q_2 so then cm=28,8/79,55=0,4cm
     
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