Time Needed for a Lake to Freeze Over

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  • #1
Aeonic333
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I have been rather unfortunate to attend a school with a physics department that uses Sears and Zemansky's University Physics textbook. I have been working unsuccessfully on the following problem (Chapter 15, #103 for those who have the book) for 2 days...

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 ice sheet

Here is what I know... or think I know...

heat current = dQ/dt = (kA(Th - Tc))/L
That is the only equation I have that involves time; however I was always under the impression that I cannot separate dQ/dt since dt is part of an expression representing the instantaneous heat flow, and not a single variable.
 

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  • #2
quantumdude
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Originally posted by Aeonic333
heat current = dQ/dt = (kA(Th - Tc))/L
That is the only equation I have that involves time; however I was always under the impression that I cannot separate dQ/dt since dt is part of an expression representing the instantaneous heat flow, and not a single variable.

Sorry, but I still don't understand why you think it can't be separated. In the equation dv/dt=-g, dt "is part of an expression representing the instantaneous" velocity, and it sure can be separated.
 
  • #3
Aeonic333
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That makes sense. I was wondering whether or not that was possible. I just remember my Calculus book saying to treat it as a single expression... but if it is bounded by a constant like -g, or in this case, heat flow (which should be constant for a given thickness), then I should be able to treat the dt as a single variable. Is it possible to do that in other situations? For instance, something without constant acceleration, heat flow, etc?
 
  • #4
quantumdude
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Originally posted by Aeonic333
Is it possible to do that in other situations? For instance, something without constant acceleration, heat flow, etc?

Yes, it is possible. Start with constant acceleration:

dv/dt=-g
dv=-g dt
∫dv=-g∫dt
v-v0=-gt
v=v0-gt

OK, now recall that v=dx/dt.

dx/dt=v0-gt

Note that on the right hand side we have not a constant, but a function of time. We can still separate the variables.

dx=(v0-gt)dt

and you continue with the integration.

The same rule (separation of variables) is valid for your problem.
 
  • #5
Aeonic333
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I think I got it...

dQ/dt = k((Th-Tc)/x) Q = Hf*m dQ/dm = dQ/dx dm = dx
dQ = [del]Hf*dx dQ/dt = [del]Hf*(dx/dt)
dQ/dt = k((Th-Tc)/x) = [del]Hf(dx/dt)
xdx = k(Th-Tc)dt/[del]Hf
[inte]xdx = [inte]k(Th-Tc)dt/[del]Hf
(x^2)/2 = k(Th-Tc)t/[del]Hf + C
x = [squ]2k(Th-Tc)t/[del]Hf
x is proportional to [squ]t

Does this look good?
 
  • #6
quantumdude
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Originally posted by Aeonic333
I think I got it...

Well, you arrived at the correct answer but it is not clear to me that your solution is correct.

dQ/dt = k((Th-Tc)/x) Q = Hf*m dQ/dm = dQ/dx dm = dx

How is it that dQ/dt=dx?
 
  • #7
Aeonic333
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dQ/dx = dQ/dm
dQ/dt [x=] dx

Is it ok to assume that dx = dm ?
 
  • #8
quantumdude
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Originally posted by Aeonic333
Is it ok to assume that dx = dm ?

What is dm?
 
  • #9
Aeonic333
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dm is the change in mass, and dx is the change in length. Since a change in length results in a change in mass I set dQ/dm = dQ/dx
 
  • #10
quantumdude
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Originally posted by Aeonic333
dm is the change in mass, and dx is the change in length. Since a change in length results in a change in mass I set dQ/dm = dQ/dx

In that case, you can't do it like that because dm and dx have different units. However, dm/dx is the linear density (call it λ), so you could write dm=λdx.
 

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