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Thermal Conductivity Ice Question

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Homework Statement



Ice has formed on a shallow pond, and a steady state has been reached, with the air above the ice at -9.6°C and the bottom of the pond at 5.5°C. If the total depth of ice + water is 2.1 m, how thick is the ice? (Assume that the thermal conductivities of ice and water are 0.40 and 0.12 cal/m·C°·s, respectively.)

Homework Equations



P = KA(Th - Tc)/L

The Attempt at a Solution



Do I have the right equation for this question? I'm not too sure because I don't have any information related to the area besides depth.
 

Answers and Replies

  • #2
gneill
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Pick a representative area (say one square meter) and assume that the rest of the pond area will behave similarly.
 
  • #3
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The hint says this: Do you realize that the energy conducted through the water must equal that conducted through the ice? You know the temperature of the ice-water interface (don't you?). If you equate the two expressions, the ice thickness is the only unknown left.

Not sure what expressions to use though, that's my problem.
 
  • #4
gneill
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The Relevant Equation that you gave looks promising. Try it.
 
  • #5
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KA(Th - Tc)/L = KA(Th - Tc)/L

But what I'm not understanding here is, that I'll only have one unknown when I do this. One side is Ice and the other is Ice + Water correct? So L for ice is going to be unknown. Area is unknown as well for both. Not sure if I should have the temp change on the ice side either. Should it just be Temperature of ice multiplied by KA and divided by L?
 
  • #6
gneill
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If you think about it, you should be able to identify two distinct thermal gradients that you need to consider. Think in terms of interfaces.
 
  • #7
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I'm not quite sure what you mean by thermal gradients but this is what I would have if I plugged in the values.

(0.12)A1(5.5 - (-9.6))/2.1 = (0.40)A2(5.5 - (-9.6))/L

Is this even remotely close to what I need to solve this for L?
 
  • #8
gneill
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I'm not quite sure what you mean by thermal gradients but this is what I would have if I plugged in the values.

(0.12)A1(5.5 - (-9.6))/2.1 = (0.40)A2(5.5 - (-9.6))/L

Is this even remotely close to what I need to solve this for L?
A thermal gradient means that the temperature changes gradually from one place to another through a medium. In this case you have the water and the ice as the two media that you're interested in.

The water has two surfaces: One at the pond bottom, the other at the ice bottom. The ice has two surfaces: One at the top of the water, and one at the "bottom" of the air.

You need to consider how you'll work the total depth of the water+ice into the scheme.
 
  • #9
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But that doesn't help me find out what the Area's are for those surfaces. Or does it?
 
  • #10
gneill
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Just assume a representative area; The heat flow should be uniform over the whole pond. In fact, you should probably find that the area variable cancels out on both sides of the equation.
 
  • #11
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If you have an ice+water system. How do you utilize the given K values on that expression? Do you add them or multiply them? Jeez I'm about to give up on this problem, I'm just not seeing what you're saying.
 
  • #12
gneill
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In an ice-water system in equilibrium, the interface of the two are at the freezing/melting temperature; As many molecules of liquid freeze to the surface of the ice as do molecules of ice melt to join the liquid. That fixes the water/ice interface temperature.

The Relevant Equation gives the rate of heat passing through a given thickness and area of a material. You have two materials to worry about, and you've already determined that you want the heat flow through each to be equal (so for a given area of ice surface, as much heat flows through the ice volume below it as flows through the corresponding volume of water below it, and in turn, from the area of pond bottom that the water makes contact with). There's one parameter to play with for the conductivities: the thickness of the material. You are also given that the sum of the thicknesses is fixed (water depth + ice thickness).

If you think of it in terms of an electrical circuit, you've got two two resistors in series, with a given potential difference across each, and a total potential difference across the two. The current through them is the same. Given the current and fixed potential differences, the individual conductivities (inverse of resistance) is determined.
 

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  • #13
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You have to make an equation like this right?

(K1A(Th - Tx))/L1 = (K2A(Tx - Tc)/L2)

Am I on the right track here? The Area will cancel out. Th = 5.5 C ; Tx = 0 C ; Tc = -9.6 C. I'm confused about L1 and L2 though.
 
  • #14
gneill
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Yes, you're on the right track. The Areas cancel (and were not required in the first place, since you can just as easily deal with heat flow in terms of watts per square meter as with a total value of watts).

Why are you confused about L1 and L2? What is the relationship between them?
 
  • #15
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Yes, you're on the right track. The Areas cancel (and were not required in the first place, since you can just as easily deal with heat flow in terms of watts per square meter as with a total value of watts).

Why are you confused about L1 and L2? What is the relationship between them?
2.1 - L1 = L2
2.1 - L2 = L1

(K1A(Th - Tx))/(2.1 - L2)) = (K2A(Tx - Tc)/(2.1 - L1))

That's why I'm confused. It said in the hint when I set up the equation I'll find that only the thickness is left as unknown. But I have 2 unknowns here. Which side is correct?
 
  • #16
gneill
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I don't know why you substituted for *both* L1 and L2, since that just left the same two unknowns in your equation, only in different places. Only substitute for one of them. Since you want to find the thickness of the ice, I suggest leaving that one alone!
 
  • #17
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I don't know why you substituted for *both* L1 and L2, since that just left the same two unknowns in your equation, only in different places. Only substitute for one of them. Since you want to find the thickness of the ice, I suggest leaving that one alone!
Ohhh, my bad I was reading your post wrong.

(K1A(Th - Tx))/(2.1 - L2)) = (K2A(Tx - Tc)/(L2))

Is that basically the set up there? I saw in my text book that in some problems you divide the thickness by the k value but I guess that doesn't really apply here does it?
 
  • #18
gneill
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I've lost track of what you've assigned for all the variables and constants. Can you provide the equation once more with them identified and given values?
 
  • #19
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thermal conductive for ice = .4 cal/m*C
thermal conductive for water = .12 cal/m*C

Air above the ice = -9.6 C
Water at bottom of pond = 5.5 C

Th = 5.5 C
Tx = 0 C
Tc = -9.6 C

(.4 cal/m*C)(5.5 C - 0 C)/(2.1 - L,ice) = (.12 cal/m*C)(0 C - (-9.6 C))/(L,ice)

L,ice = 1.323 m
 
Last edited:
  • #20
gneill
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On the left hand side of your equation, you are using the thermal conductivity for ice where there is liquid water. Similarly, on the right hand side, you're using the conductivity for liquid water where there's ice.
 
  • #21
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On the left hand side of your equation, you are using the thermal conductivity for ice where there is liquid water. Similarly, on the right hand side, you're using the conductivity for liquid water where there's ice.
When I put the answer into the system though, it said it was correct.
 
  • #22
gneill
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Well, you can't argue with success.:biggrin:
 
  • #23
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Well, you can't argue with success.:biggrin:
Hmmm it is strange that it said it was right because I just looked at my work again and I did indeed have K for ice on the side of the water. Now I get a smaller value. The tolerance for the answers is only +/- 2% too.

Edit: I tried setting it up so that I have to divide the thickness by k of the substance as well. I got closer to the answer I originally got. Still the question is, what method is right.
 
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  • #24
gneill
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Hmmm it is strange that it said it was right because I just looked at my work again and I did indeed have K for ice on the side of the water. Now I get a smaller value. The tolerance for the answers is only +/- 2% too.
I agree. It's a mystery. :confused:
 
  • #25
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I agree. It's a mystery. :confused:
Wow, nevermind I made a mistake. I get the exact answer as what they get now. So yeah your method is right. I just got extremely lucky before.
 

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