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Time needed for a lake to freeze over.

by adrian116
Tags: freeze, lake, time
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adrian116
#1
May20-06, 10:11 AM
P: 43
a). Show that the thickness of the ice sheet formed on the surface of a lake is proportional to the square root of the time if the heat of fusion of the water freezing on the underside of the ice sheet is conducted through the sheet.
b). Assuming that the upper surface of the ice sheet is at -10 degree celsius and that the bottom surface is at 0 degree celcius, calculate the time it will take to form an ice sheet 25 cm thick.
c). If the lake in part (b) is uniformly 40m deep, how long would it take to freeze all the water in the lake?

I have the difficulty on part a that how to work out the relationship.

thx for attention
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maverick280857
#2
May20-06, 11:12 AM
P: 1,780
Show us your work.

Can you perform an energy balance for part (a)? What is the physical process involved in ice formation here? What are the assumptions you have to make?

Hint: If you do part (a) you can do the others as well.
adrian116
#3
May20-06, 09:22 PM
P: 43
I have applied the forumla (dQ/dt)=kA(T2 - T1)/l,

(T2-higher temperture,T1-lower temperture
k- thermal conductivity, A- area , l-length)

then move dt to the right hand side, take the integral,

I would get the formula like this Q=tkA(T2 - T1)/l,

then applied the latent heat formula that Q=mL (L is latent heat),

such that mL=tkA(T2 - T1)/l , then

p(l^3) L =tkA(T2-T1)/l, (p-density)

i have stopped here..........was my assumption of Q wrong?
shouldn't i apply the latent heat forumla?

maverick280857
#4
May20-06, 11:22 PM
P: 1,780
Time needed for a lake to freeze over.

Quote Quote by adrian116
I have applied the forumla (dQ/dt)=kA(T2 - T1)/l,

(T2-higher temperture,T1-lower temperture
k- thermal conductivity, A- area , l-length)
So far its correct. You're on the right track.

then move dt to the right hand side, take the integral,

I would get the formula like this Q=tkA(T2 - T1)/l,
No, this isn't correct. You have assumed that the length (l) of the developing ice layer is independent of time and so Q is linearly varying with time t. You have to use the general form of Fourier's heat conduction equation here:

[tex]\frac{dQ}{dt} = -kA\frac{dT}{dy}[/tex]

Assume that the atmosphere (the air) above the water surface is at a temperature -T degrees (so T > 0). If water is to freeze, it must loose "latent heat" to the surroundings and come to a temperature of 0 degrees.

Consider a differential layer of freshly formed ice at a depth y from the water surface. This layer is of width dy and is at 0 degrees. At this stage, you will have to assume that heat conduction is taking place at steady state and hence [itex]dQ/dt[/itex] is the same everywhere in the medium. The heat conduction equation gives

[tex]\frac{dQ}{dt} = kA\frac{(0-(-T)}{y} = kA\frac{T}{y}[/tex]

Now we cannot integrate at this stage to get Q because we do not know y as a function of time t (which is what we want to find out). The trick here is to observe that [itex]dQ[/itex] is also the "latent heat lost" by the differential layer which enables it to freeze without change in temperature (from 0 degrees).

Hence [tex]dQ = dmL[/tex]. Also observe that the mass of the differential layer is [itex]dm = \rho A dy[/itex] where [itex]\rho[/itex] as the density of water. It is also the density of ice. This is an unrealistic assumption here: density of water = density of ice. Hence,

[tex]\frac{dQ}{dt} = \rho A \frac{dy}{dt}[/tex]

Now you can equate the two expressions for [itex]dQ/dt[/itex] to get your expression.

The assumptions made are:

1. Conduction of heat to the atmosphere from any differential layer takes place in steady state.
2. Density of ice is equal to density of water (as we are assuming that A and dy are both constant as the ice forms out of an equal mass of water).

EDIT: Armed with y as a function of time t, you can also find [itex]dQ/dt = \rho A dy/dt[/itex] just to see how it varies with time.
adrian116
#5
May21-06, 08:50 AM
P: 43
I have got the answer , thx so much


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