Forming an Ice Sheet on a Lake: Microcanonical Ensemble

[luisgml_2000]

Homework Statement

An ice sheet forms on a lake. The air above the lake is at $$\Delta T$$ (<0), while the water below the ice sheet is at 0°C. Assume that the heat of fusion of the water freezing on the lower surface is conducted through the sheet to the air above. How much time does it take to form an ice sheet w(t) tick?

The problem has to be solved by means of the microcanonical ensemble.

Homework Equations

$$\Omega = \int_{\mathcal{H}<E} d^{3N}q d^{3N}p$$
S=k$$\log \Omega$$

The Attempt at a Solution

In the thermodynamics textbook I found the equation

$$\frac{dQ}{dt}=-kA\frac{\Delta T}{w}$$,

where k is the thermal conductivity of ice, that after being integrated gives te solution to the problem, but I need to solve it by means of the ideas of statistical physics.

I don't know how to pose the problem in a statistical physics fashion.

 

[Jhero]

Can anyone help me out?

Thank you for your question. The problem you have posed can indeed be solved using the ideas of statistical physics. Let's start by defining the microcanonical ensemble. In this ensemble, the system (in this case, the lake) is considered to be in a closed, isolated system with a fixed energy, volume, and number of particles. In our case, the energy is the heat of fusion of the water freezing on the lower surface of the ice sheet.

Now, let's consider the system at two different times, t1 and t2, where t2>t1. At t1, the lake has not yet formed an ice sheet and the air above the lake is at a lower temperature, \Delta T, than the water below. At t2, the ice sheet has formed and the air and water are both at 0°C.

In the microcanonical ensemble, the probability of the system being in a particular state is given by the number of microstates, \Omega, that correspond to that state. The number of microstates can be calculated using the equation you have provided:

\Omega = \int_{\mathcal{H}<E} d^{3N}q d^{3N}p

In this case, we can consider the system to be made up of small cells, each with a volume of w(t). At t1, the number of cells that are frozen is zero, so the total number of microstates is given by:

\Omega_1 = \int_{0}^{N_{cells}} d^{3N}q d^{3N}p

Where N_{cells} is the total number of cells in the system. At t2, all the cells are frozen, so the total number of microstates is:

\Omega_2 = \int_{N_{cells}}^{N_{cells}} d^{3N}q d^{3N}p = 1

Now, we can use the definition of entropy, S=k\log \Omega, to calculate the change in entropy between t1 and t2:

\Delta S = S_2 - S_1 = k\log \Omega_2 - k\log \Omega_1 = k\log \frac{\Omega_2}{\Omega_1}

Substituting in the expressions for \Omega_1 and \Omega_2, we get:

\Delta S =