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the problem is outside temperature is -10 C and on the lake there is 5 cm of ice. After what time will 15 cm be reached. $Q=\frac{kA(T_2-T_1)}{H}$, where $A$ is area, $H$ thickness of ice, $T$ temperatures on both sides. thermal conductivity coefficients $k$: $k_{ice}=0.56$ W/m*C, $k_{water}=1.7$ (same units), $\rho_{ice}=920$ kg/m$^3$, $\rho_{water}= 1000$, latent heat for water $L=333$ kJ/kg

Looking at units $Q$ is watts. So $Q=\frac E t$, where $E=Lm$ energy needed for the freezing of ice, $m=\rho A \Delta H$, where $\Delta H$ is height of column which will freeze. So $$\Delta t=\frac{L \rho H \cdot H}{k(T_2-T_1)}$$

So if I sum all the $\Delta H$ up and take the limit as $\Delta H$ goes to $0$. I am left with an integral

$$t= \int_{H_{initial}}^{H_{final}} \frac{L \rho H}{k(T_2-T_1)} dH.$$

Solving the integral putting in $H_{final}= 0.15$ m and $H_{initial}= 0.05$, and $T_{water}=0$ C, I get that it will take $1.959 \cdot 10^5$ seconds which is 136 days

It does not seem right and I do not know really how to interpret the change in volume from water to ice. Can somebody tell me where I went wrong

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  • $\begingroup$ What is the initial rate of growth? $\endgroup$ – Pieter Mar 27 at 21:09
  • $\begingroup$ the first part of the exercise was about finding how long will the ice grow to 6 cm, and I calculated (without calculus so with a slight mistake probably) that it would take just 177 minutes. Otherwise there was not anything given about initial rate of growth $\endgroup$ – ata4444 Mar 27 at 21:14
  • $\begingroup$ Please use the MathJax syntax to type mathematical expressions. $\endgroup$ – flaudemus Mar 27 at 21:25
  • $\begingroup$ I am new to here, and dont really know how to do that but will try to do that in future $\endgroup$ – ata4444 Mar 27 at 21:28
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    $\begingroup$ no it is a previous year exercise, haha $\endgroup$ – ata4444 Mar 27 at 21:52
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If h is the ice thickness at time t, the rate of heat flow through area A from the ice to the air is $$q(t)=\frac{k_{ice}(0-(-10))A}{h}$$This is also equal to the heat required for freezing: $$q(t)=\rho A L\frac{dh}{dt}$$If you set these equal and integrate the resulting differential equation, you can get h as a function of t. Note that the area A cancels.

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  • $\begingroup$ Oh i understand, but is that$\rho$ of water or of ice, and how or if the difference between density of ice and water changes anything? $\endgroup$ – ata4444 Mar 27 at 21:56
  • $\begingroup$ I suppose using the density of ice would be a slightly better approximation, although the difference would probably not be significant. $\endgroup$ – Chet Miller Mar 27 at 22:05
  • $\begingroup$ okay, anyway thanks for the help appreciated $\endgroup$ – ata4444 Mar 27 at 22:07

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