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The air above a large lake is at –2°C, while the water of the lake is at 0°C. Assuming that only thermal conduction is important and using relevant data selected from the given below find how many hours it would take for a layer of ice 10 cm thick to form on the lake’s surface.
Data(where $w$=water and $i$=ice):
$K_w=5.6Wm^{-1}k^{-1},$ $K_i=2.3Wm^{-1}k^{-1}$
$L_i=3.3*10^5J{kg}^{-1},$ $\rho_w=10^3kg m^{-3}$ $\rho_i=920kgm^{-3}$

This is how I solved.
Let the area of crossection of lake be A and same throught its depth.
Let the layer of ice formed at time $t$ be $x$ where $t$ is in seconds and $x$ is in meters.
enter image description here

Now in time $dt$, $dm$ mass of water will be converted to ice but $dm=Adx\rho_w$.
So heat lost by water is $$dmL_i=A\rho_wL_idx$$
That amount of heat is transferred to the atmosphere through ice.
But heat transfer through ice is $$\frac{dq}{dt}=\frac{K_iA\Delta\theta}{x}$$
So heat lost by water is equal to heat transferred by ice. $$A\rho_wL_idx=\frac{K_iA\Delta\theta}{x}dt$$
Now on solving the differential equatio we get $$t=\frac{x^2\rho_wL_i}{2K_i\Delta\theta}$$
On substituing the values answer comes out to be 99.6 hours.
Where as the solution of the question has taken $\rho_i$ instead of $\rho_w$ which I took to solve the question.
This is here where I am stuck and unable to comprehend and understand why. After giving it an enough thought both seems to be correct.
Hence the final answer comes out to be 92 hours.

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2 Answers 2

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Assuming that only thermal conduction is important [...]

That's a very dicy assumption. There certainly is convection at the ice/air interface but assuming the water and ice are at the same temperature, convection at that boundary is $0$.

We can deal with that with a so-called overall heat transfer coefficient $U$:

$$\frac{\mathrm{d}q}{\mathrm{d}t}=UA\Delta\theta$$

$U$ is found using the principle of parallel heat resistors. $h_{ia}$ is the ice-to-air convection heat transfer coefficient:

$$\frac{1}{U}=\frac{1}{h_{ia}}+\frac{x}{K_i}$$

$$A\rho_iL_i\mathrm{d}x=UA\Delta\theta \mathrm{d}t$$

$$\frac1U=\frac{K_i+h_{ia}x}{h_{ia}K_i}$$ $$U=\frac{h_{ia}K_i}{{K_i+h_{ia}x}}$$ It then follows that: $$\rho_iL_i\mathrm{d}x=\frac{h_{ia}K_i}{{K_i+h_{ia}x}} \Delta\theta \mathrm{d}t$$ $$\rho_iL_i\left(K_i+h_{ia}x\right)\mathrm{d}x=h_{ia}K_i\Delta \theta \mathrm{d}t$$ Integrate as appropriate.

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  • $\begingroup$ This site is about physics and not homework. Being a cautious amateur of skating on natural ice, thanks for your answer. So what is the time required for 10 cm of ice with the correct formula? And how good is the assumption of no convection at the water ice boundary? $\endgroup$
    – my2cts
    May 21, 2021 at 8:11
  • $\begingroup$ That time depends on the air temperature and the convection coefficient $h_{ia}$, which you'll have to look up in various tables. The 'no convection at the water-ice boundary assumption' would be 100% fulfilled if the water and ice are in thermal equilibrium (same temperature) because then there simply is no heat flux crossing that boundary. It's a reasonable assumption, IMO. $\endgroup$
    – Gert
    May 21, 2021 at 18:23
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Note that the distance $x$ that you considered is the thickness of ice, not water. Hence $dx$ will be the elemental increase in thickness of ice, so we will use $\rho_i$ and not $\rho_w$. The rest of your calculation is correct.

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  • $\begingroup$ The rest of your calculation is correct. No, it's not. It's mostly nonsensical rubbish, actually. $\endgroup$
    – Gert
    May 16, 2021 at 9:28
  • $\begingroup$ Most certainly not. OP has equated heat current passing through ice to the rate at which water loses heat. This is most definitely correct. What are you talking about? $\endgroup$ May 16, 2021 at 9:32
  • $\begingroup$ The calculation of the heat flux through the ice is wrong. You cannot simply brush aside convection layers at the boundaries (water/ice and ice/air). One needs to calculate an overall heat transfer coefficient, not simply use thermal conductivity of ice. Failing to do so will lead to gross errors. $\endgroup$
    – Gert
    May 16, 2021 at 9:38
  • $\begingroup$ While I realize that, I am aware that the OP is a high school student, and the problem is a homework problem. In fact I had faced the same problem a few years back. Convection currents are to be ignored here, and that has been stated in the problem. OP should have stated that and tagged this under "homework and exercises". $\endgroup$ May 16, 2021 at 9:57
  • $\begingroup$ I was simply writing in the context of the problem here. The objective here was never to create an accurate model of the freezing of the lake. $\endgroup$ May 16, 2021 at 9:58

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