# How Long Does It Take for Ice to Double Its Thickness in Sub-Zero Temperatures?

• NASSAfellow
In summary: K = m.../s deg Ctemperature deg C (same as K)thickness mIn summary, the problem asks how long it will take for a layer of ice to increase from 4.50 cm to 9.00 cm on top of a pond of water at 0°C, given that the air temperature is -11.0°C. The solution involves using the heat conduction equation and taking into account the latent heat of fusion and density of ice. After converting all units to SI, the answer is approximately 2.24 hours.
NASSAfellow

## Homework Statement

A pond of water at 0°C is covered with a layer of ice 4.50 cm thick. If the air temperature stays constant at -11.0°C, how much time does it take for the thickness of the ice to increase to 9.00 cm?

____________hours

Hint: To solve this problem, use the heat conduction equation,

dQ/dt = kA (change in)T/x

and note that the incremental energy dQ extracted from the water through the thickness x is the amount required to freeze a thickness dx of ice. That is, dQ = LpA dx, where p is the density of the ice, A is the area, and L is the latent heat of fusion. (The specific gravity and thermal conductivity for ice are, respectively, 0.917 is 2.0 W/m/°C.)

## Homework Equations

Heat conduction equation

## The Attempt at a Solution

I am not sure how to do this, I have done extensive research but am still confused... please help. Also I don't study maths, so please help me with a word equation if possible.

Many Thanks,
Nasser

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This is a fun problem.

Put your expressions together (I'm not doing anything but math), and you get the following.
$$LA \,dx = \frac{kA \Delta T}{x} dt$$

Separate variables, and integrate. You've already done all the physics.

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I am not sure what this all means- I am doing AS level physics and I don't do maths, would it be possible to give me a word equation?

Is A assumed to be constant? If so, you should be able to plug dQ=LA dx into dQ/dt = kA T/x and integrate.

I don't do maths... i.e. Integration means nothing to me... sorry

What do the "d"s mean in your equation then, if you are not using calculus? Are they meant to be $\Delta$; denoting "change in"?

edit: you didn't answer my question re A being constant.

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Yes, I think so, Webassign just gave me the question to do, and the formula to use.

$$LA \,dx = \frac{kA \Delta T}{x} dt$$
Someone gave me this^^, but still confused

Ok, I've seen your other thread-- you're studying AS Physics, so I presume you did moderately well in GCSE maths. Can you see how the equation you give above comes from your two equations in your original post? (simply substitute dQ from the second equation into the first).

Now, dx means "change in x", so rewrite it as dx=x-x0 where x0 is the value of x when t=t0=0 (since we will take the original time as t=0). Thus we write dt=t (since t0 is zero).

This should make your equation easier to handle. Try to rearrange it to make t the subject of the equation.

okay... I have got this far...

dt=Ldx^2/kT

Is this correct?

No, dx is not the same as x. Let's replace the d's in the equations in post #6, as I suggested in #7. This will give $$LA(x-x_0) = \frac{kA T}{x} t$$. Can you rearrange this?

I have confused myself with density and difference in etc...

Is it this? Where is density?

Change in time = Latent heat of fusion*change in x^2 / thermal conductivity of ice * 11

There should be a term for density in your equation for dQ = LAdx in your original post. You sort of mentioned it in the sentence, but left it out of your equation.

NASSAfellow said:
I have confused myself with density and difference in etc...

Is it this? Where is density?

You said this:
dQ = LA dx, where is the density of the ice, A is the area, and L is the latent heat of fusion.

Now, have you missed out the density term from the equation? My guess is that you copied the question incorrectly, and there should be a ρ somewhere.

Of course! I have seen it now! Sorry, when I pressed copy on the pc it did not copy:

dQ= LpA , where p is density of ice!

Also the delta sign did not copy for the heat conduction equation

dQ/dt = kA (Change in) T/x

So does this mean I am lost?

Cristo...?

Help needed urgently

Have you done anything with the equation I gave in post #9 (with the density put in)? If you have, why are you lost? If not, I suggest you do. I'm not going to do this for you!

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Okay, here is what I have got

LpAdx = kA (change in) T dt / x

So sub in values first then rearrange (it is easier that way);

79.72 (latent heat of fusion for ice) x 0.917 (density) x A (Area) x 4.5 (change in x) = 2 (thermal conductivity for ice) x A (Area) x 11 (change in temperature) x dt / 9 (final thickness)

So, this gives:
328.96458 x A = 22 A x dt / 9

Take 9 to other side...

328.96458 A x 9 = 22A x dt

A cancels out...

2960.68122 = 22 x dt

rearrange to get dt on its own...

2960.68122/22 = dt

so

dt = 134.58 (2 d.p.)

This could be in seconds, minutes or hours, but more likely minutes...

So Ans divided by 60 to find hours...

2.24 hours

Webassign says it is wrong.

Last edited:
NASSAfellow said:
This could be in seconds, minutes or hours, but more likely minutes...

why? what units are your other measurements in?

Density of ice is in g/ cm^3 at 0 degrees C
Heat of fusion of water is in Calories per gram
Conductivity for ice is in W/m/ degrees C
Temperature in degrees C
x (thickness) is in cm

Well, I'd change everything to SI units, then the time will be in seconds.

In that case:

I will try but I don't really know them that well...
Density of ice is in ?will find out?
Heat of fusion of water is in ?will find out?
Conductivity for ice is in ?will find out?
Temperature in ?will find out from the web? Kelvin perhaps?
x (thickness) is in m

heat of fusion... joules per mole
density of ice... kg per m^3
conductivity for ice W/m K = m kg/s3

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