Ice thickness in a pond with temperature gradient

[Rockdog]

During a cold Chicago winter, a steady thermodynamic state has been reached in a small pond with a layer of ice on the top. The air directly above the pond is at -7 oC and the ground at the bottom of the pond is held at 2 oC. The pond holds a total of 342.1 m3 of water and has a relatively uniform depth (water and ice) of 1.1 m. Assume the thermal conductivities of ice and water are 0.40 cal/m/s/oC and 0.12 cal/m/s/oC, respectively.
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What is the thickness of the ice?

Ok, this problem appears tough. How do I start this ??

 

[quantumdude]

You know that the water-ice boundary is going to be 0^o, because below that the temperature will be too high and the ice will melt. Using the data on thermal conductivities, you need to set up an expression for the temperature distribution in both the water and the ice. If you let y=0 be at the bottom of the lake, then your expression for the temperature distribution in the water will be in terms of y, and that for the ice will be in terms of 1.1-y.

You can equate the temperatures at the boundary (because you know they are both zero), and solve for y.

 

[M98Ranger]

To calculate the thickness of the ice, we can use the heat transfer equation: Q = kAΔT/Δx, where Q is the heat transfer rate, k is the thermal conductivity, A is the surface area, ΔT is the temperature difference, and Δx is the thickness of the material.

First, we need to calculate the heat transfer rate from the air to the water and ice in the pond. We can use the given temperatures of -7 oC and 2 oC to find the temperature difference, which is ΔT = 2 oC - (-7 oC) = 9 oC.

Next, we need to find the surface area of the pond. Since the pond has a uniform depth of 1.1 m, we can use the total volume of 342.1 m3 to find the surface area. The surface area can be calculated as A = V/Δx = 342.1 m3/1.1 m = 311 m2.

Now, we can plug in the values into the heat transfer equation to find the heat transfer rate: Q = (0.12 cal/m/s/oC)(311 m2)(9 oC)/Δx.

Since we are looking for the thickness of the ice, we can rearrange the equation to solve for Δx: Δx = (0.12 cal/m/s/oC)(311 m2)(9 oC)/Q.

Finally, we need to find the thermal conductivity of ice. The problem states that the thermal conductivity of ice is 0.40 cal/m/s/oC.

Plugging in all the values, we get: Δx = (0.12 cal/m/s/oC)(311 m2)(9 oC)/(0.40 cal/m/s/oC) = 83.9 m.

Therefore, the thickness of the ice on the pond is approximately 83.9 m.