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Problem 14.79: Time for a lake to freeze over.

When the air temperature is below 0 C, the water at the surface of a lake freezes to form a sheet of ice.

Part A

If the upper surface of an ice sheet 25.5 cm thick is at -10.8 C and the bottom surface is at 0.00  C, calculate the time it will take to add 2.00 mm to the thickness of this sheet.

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Explanation / Answer

given data

d(0)=0.255 m
delta T =10.8 C
d(t) = 0.255+0.002=0.257 m

Assume that the freezing of an additional layer is slow enough. Then the temperature profile can be assumed to be linear across the layer of ice. This quasistatic assumptions allows us to solve this problem without the need of solving the much more difficult moving boundary condition.

The amount of heat that leaves the top surface has two origins: the latent heat of freezing of the additional sheet of water at the bottom plus the heat from additional cooling of the existing ice.
This will give a heat balance equation from which a "thickness goes as sqrt(time) " formula emerges.

First, the linear temperature profile assumed is (meauring thickness x from the top surface downwards)

T = T0 + x delta(T)/d

The energy balance is for infinitesimal time delta(t) is

k (delta(T)/d) delta(t) = rho L delta(d) + integral_0^d rho sigma delta(T) dx

d is the thickness of the sheet (distance of growth)
With the linear profile this gives

d(t) ^2 = d(0)^2 + t * 2 k delta(T)/( rho L + 1/2 rho sigma delta(T) )

Here k is the conductivity of ice,

L the latent heat of melting, rho the density of ice

and delta(T) the temperature difference between top and bottom.

Thermal conductivity for ice is k= 2.18 W/(m·K) at 0 °C
The heat of fusion for ice is L= 334 J/g = 334000 J/kg
The density of ice is rho= 0.9167 g/cm^3 = 916.7 kg/m^3

d(t) ^2 = d(0)^2 + t * 2 k delta(T)/( rho L + 1/2 rho sigma delta(T) )

0.257^2=0.255^2+(t*2*2.18*10.8)/(916.7*334000+0.5*916.7*10.8)

t =6658.4

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