- #1

fancyterm

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## Homework Statement

A tank of water has been outdoors in cold weather, and a slab of ice 5 cm thick has formed on its surface. The air above the ice is -10C. Calculate the rate of ice formation ( in cm / h) on the ice slab. Take the thermal conductivity of ice to be 0.004 cal / (s.cm.C) and its density to be 0.92 g/cm^3. Assume no energy transfer through the tank walls or bottom.

## Homework Equations

P(conducted) = Q/t = kA (T2-T1)/ Length

p(density) = m/v

Q = Lm heat of transformation

## The Attempt at a Solution

Some thoughts:

When they say on the slab, they must mean underneath it right?

Also they provide standard info the in the question like the conductivity of ice which I could look up in a table. I could similarly look up density and conductivity of water but since they didnt explicidly mention that too, I wonder if I'm not supposed to? Like maybe that's variable in terms of the problem?

Since they arent just asking the rate of conductivity through ice I probably need to use the heat of fusion formulae somewhere too.

So I need to relate the density somehow, in order to cancel out the Area which I don't have.

Pluggin in info for the conduction rate equation I get

P = 0.004A (0--10) / 5

P = A x 0.008

now for density

p = m/v = m/Ah (using A and h because I have more information about that, than just V)

0.92 = mA5

I don't see an obvious relation.

(heat of fusion from water to ice = 333)

Q = Lm

Q = 333m

Q = 333 x 0.92/5A

Uhm I don't know where I am going with this. I could say both equations are equal to A, but then I still have P and m as variables