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Thermal Conductivity - Forming ice under water

  1. May 21, 2014 #1
    1. The problem statement, all variables and given/known data

    2f06g76.png

    Part (a): Derive Clausius-Clapeyron Equation. Find latent heat of fusion of ice.
    Part (b): Find rate of formation of ice
    Part (c): What is the maximum thickness of ice formed?

    2. Relevant equations



    3. The attempt at a solution

    Part (a)
    I have derived the relation. Using the values, I found that the latent heat of fusion of ice is ##L = 3.44 \times 10^5 J kg^{-1}## which seems right.

    Part (b)

    Using heat flux ##J = k (\frac{\partial T}{\partial z})##, consider in time ##\delta t## amount of ice ##\Delta m## is formed.

    [tex]\Delta Q = (\Delta m)L = (JA) \Delta t[/tex]
    [tex] \rho (\Delta z)L = \frac{k(T_2-T_1)}{z} \Delta t[/tex]
    [tex]\frac{dz}{dt} = \frac{k(T_2-T_1)}{\rho L z}[/tex]

    Taking ##k = 2.3##, ##T_2-T_1 = 0.5##, ##\rho = 1000##, ##L = 3.44 \times 10^5##:

    I find that ##\frac{dz}{dt} = 3.34 \times 10^{-7} m s^{-1}##, so ice forms awfully slow.

    Part (c)
    4hr4pk.png

    In steady state, ## \frac{dQ_1}{dt} = \frac{dQ_2}{dt}##.

    Therefore, we equate the heat fluxes:

    [tex]k_{water} \left(\frac{2}{1-z_f}\right)A = k_{ice}\frac{0.5}{z_f}A[/tex]

    [tex]z_f = \frac{k_{ice}}{4k_{water} + k_{ice}} = 0.51 m [/tex]

    It's hard to believe that over half the lake would be frozen, when the rate of formation of ice above (without the interference of the bottom of the lake) is merely ##3.34 \times 10^{-7} m s^{-1}##.
     
  2. jcsd
  3. May 21, 2014 #2
    That's about 3 cm/day. That doesn't seem extremely low to me, considering the very low temperature driving force of 0.5 C.

    Chet
     
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