A deep rectangular pond of surface area A,con | vedclass

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(A) The heat conducted through the ice layer of thickness $x$ in time $dt$ is given by $dQ = \frac{KA(T_2 - T_1)}{x} dt$.Here,$T_2 = 0^{\circ}C$ (temp

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$26 K / (\rho x L)$

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(A) The heat conducted through the ice layer of thickness $x$ in time $dt$ is given by $dQ = \frac{KA(T_2 - T_1)}{x} dt$.Here,$T_2 = 0^{\circ}C$ (temperature at the water-ice interface) and $T_1 = -26^{\circ}C$ (outside air temperature).So,$dQ = \frac{KA(0 - (-26))}{x} dt = \frac{26KA}{x} dt$.This heat causes an additional thickness $dx$ of water to freeze. The mass of this new ice layer is $dm = A \cdot dx \cdot \rho$.The heat released during this phase change is $dQ = dm \cdot L = A \cdot dx \cdot \rho \cdot L$.Equating the two expressions for $dQ$:$\frac{26KA}{x} dt = A \cdot dx \cdot \rho \cdot L$.Rearranging to find the rate of increase of thickness $\frac{dx}{dt}$:$\frac{dx}{dt} = \frac{26K}{\rho x L}$.

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