adrian116 said:
I have applied the formula (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.