A lake is covered by an ice layer of thickness x having thermal...

The heat must conduct through both the ice layer and the snow layer in series.

The thermal resistance of the ice layer is $R_1 = \dfrac{x}{K_1 A}$ and that of the snow layer is $R_2 = \dfrac{y}{K_2 A}$.

The equivalent thermal resistance of the composite slab is $R_{eq} = R_1 + R_2 = \dfrac{1}{A} \left( \dfrac{x}{K_1} + \dfrac{y}{K_2} \right)$.

The temperature difference across the layers is $\Delta \theta = 0^\circ\text{C} - (-\theta^\circ\text{C}) = \theta$.

The rate of heat flow is $\dfrac{dQ}{dt} = \dfrac{\Delta \theta}{R_{eq}} = \dfrac{\theta A}{\dfrac{x}{K_1} + \dfrac{y}{K_2}}$.

The rate at which heat is released due to the freezing of water is $\dfrac{dQ}{dt} = \rho A L \dfrac{dx}{dt}$.

Equating the two expressions for the rate of heat flow, we get:

$\rho A L \dfrac{dx}{dt} = \dfrac{\theta A}{\dfrac{x}{K_1} + \dfrac{y}{K_2}}$

$\Rightarrow \dfrac{dx}{dt} = \dfrac{\theta}{\rho L \left( \dfrac{x}{K_1} + \dfrac{y}{K_2} \right)}$.

Answer: $\dfrac{\theta}{\rho L \left(\dfrac{x}{K_1} + \dfrac{y}{K_2}\right)}$