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(D) Let $A$ be the surface area of the pond and $x$ be the thickness of the ice layer at any time $t$. The temperature difference across the ice layer

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$Z^{2}=\frac{40 k}{\rho L} t$

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(D) Let $A$ be the surface area of the pond and $x$ be the thickness of the ice layer at any time $t$. The temperature difference across the ice layer is $\Delta T = 0^{\circ} C - (-20^{\circ} C) = 20^{\circ} C$.The rate of heat flow through the ice layer is given by conduction: $\frac{dQ}{dt} = \frac{kA \Delta T}{x} = \frac{kA(20)}{x}$.As the ice layer thickens by $dx$ in time $dt$, the heat released is $dQ = L dm = L(\rho A dx)$.Equating the two expressions for $\frac{dQ}{dt}$:$\rho A L \frac{dx}{dt} = \frac{20 k A}{x}$$\int_{0}^{Z} x dx = \int_{0}^{t} \frac{20 k}{\rho L} dt$$\frac{Z^2}{2} = \frac{20 k}{\rho L} t$$Z^2 = \frac{40 k}{\rho L} t$.

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