1
$\begingroup$

The book I have gives the following derivation:

Let the temperature of the atmosphere be $-\theta$ and the temperature of the water be $0$. Consider unit cross sectional are of ice, if layer of thickness $dx$ forms in time $dt$ with $x$ thickness of ice above it, heat released due to its formation is $dx\rho L$ where $L$ is latent heat. If this quantity of heat is conducted upwards in time $dt$, $$dx\rho L=K\frac{\theta}{x}dt$$

Therefore, the time taken $$t=\frac{\rho L}{2K\theta}(x_{2}^2-x_{1}^2)$$

What I don't understand is why the same amount of time should be taken for the heat to be conducted and for a new layer of ice to be formed. In other words, why is it that the next layer of ice forms only after the heat is released into the atmosphere?

$\endgroup$
2
  • $\begingroup$ The 1st equation states that the time dt for a new layer of thickness dx to be formed is proportional to xdx where x is the thickness of the ice layer already formed. This time is not constant even if dx is constant. $\endgroup$ – sammy gerbil Apr 25 '16 at 11:48
  • $\begingroup$ Your 2nd question ("In other words...") is correct and in my opinion self explanatory. $\endgroup$ – sammy gerbil Apr 25 '16 at 11:55
1
$\begingroup$

The heat is continually being released to the atmosphere, and the layer is continually getting thicker. The heat has to be conducted from the water-ice interface to the ice-atmosphere interface through the layer of ice. And, as the ice gets thicker, the rate of heat being conducted slows down. And the rate of ice formation slows down. So the amount of time taken for the heat to be conducted and for a new incremental layer of ice to be formed is not the same for each incremental layer. Those are $x^2$'s in the equation, not x's.

$\endgroup$
2
  • $\begingroup$ The second equation has $x^2$ term. $\endgroup$ – Skawang Apr 24 '16 at 16:50
  • $\begingroup$ Isn't that what you were asking about? $\endgroup$ – Chet Miller Apr 24 '16 at 17:48
0
$\begingroup$

You assume that the temperature difference between the air at $-\theta^\circ C$ and the water directly under the ice $0^\circ C$ is constant.
So looking at the thermal conduction equation $\dot Q = K A \frac {\theta}{x}$ if you increase the thickness of the ice $x$ by a factor of two you reduce the rate of heat flow $\dot Q$ by a factor of two.
This is because $K, A$ and $\theta$ are constant.
So it will take twice as long to freeze a thickness of water $\Delta x$ when the thickness of ice is $2x$ than to freeze the same thickness of water when the thickness of ice is $x$.

Your analysis does not not include the additional but smaller factor of having to reduce the temperature of the water near the ice to $0^\circ C$.
If water behaved as most liquids it would not start to freeze until the temperature of all the water was $0^\circ C$ the heat being transported through the water by convection to achieve such cooling.
Since water is anomalous in that it has a maximum density at $+4 ^\circ C$, the water under the layer of ice has to be cooled to $0^\circ C$ by conduction of heat through the water and then the ice.

$\endgroup$
2
  • $\begingroup$ We assume that the temperature of water near the ice is 0 degrees. I don't see how the first part answers my question. $\endgroup$ – Skawang Apr 24 '16 at 16:49
  • $\begingroup$ The loss of heat into the air is a continuous process as is the formation of the layer of ice. So as the heat is transported through the ice so is the water freezing and heat is being lost to the air. $\endgroup$ – Farcher Apr 24 '16 at 19:11
0
$\begingroup$

A bit late maybe. A few google searches reveal that the speed at which heat propagates is infinite in the thermodynamics I am studying. So, the heat is transferred instantaneously and the time taken for a layer to form is equal to the time in which this heat is transferred.

I had thought that the heat is transferred at a finite speed as a layer is formed. So I had thought that the time taken for the heat to be transferred to the top as a layer is formed would be different from the time the layer took to form and if we take $dt$ as the time taken for the a small layer to form, there would be some delay for the heat to be conducted out of the ice and this time would obviously be greater than the the time it took for the ice to form itself. I don't know if thats what the answers are trying to say but I decided to make an answer anyway.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.