# Thermal Conductivity - Forming ice under water

1. May 21, 2014

### unscientific

1. The problem statement, all variables and given/known data

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.

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

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

2. May 21, 2014

### Staff: Mentor

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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