Thermal Conductivity - Forming ice under water

[unscientific]

Homework Statement

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?

Homework Equations

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.

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

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)

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

Therefore, we equate the heat fluxes:

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

z_f = \frac{k_{ice}}{4k_{water} + k_{ice}} = 0.51 m

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 $3.34 \times 10^{-7} m s^{-1}$.

 

[Chestermiller]

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