# thermodynamics problem about thickness of ice

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

• What is the initial rate of growth? – Pieter Mar 27 at 21:09
• 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 – ata4444 Mar 27 at 21:14
• Please use the MathJax syntax to type mathematical expressions. – flaudemus Mar 27 at 21:25
• I am new to here, and dont really know how to do that but will try to do that in future – ata4444 Mar 27 at 21:28
• no it is a previous year exercise, haha – ata4444 Mar 27 at 21:52

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.
• 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? – ata4444 Mar 27 at 21:56