Question

A layer of ice of thickness y is on the surface of a lake. The air is at a constant temperature $-{\theta }^{\circ }C$ and the ice water interface is at $0^{\circ }$. Then that the rate at which the thickness increases is given by.
$\frac{dy}{dt}=\frac{k\theta }{L\rho y}$
where k is the thermal conductivity of the ice, L the latent heat of fusion and $\rho$ is the density of the ice.Type 1 for true and 0 for false

Answer 1

Let the air temperature be $q$ and water temperature just below ice be $0^{o}C$. At a certain time let the thickness of ice b $x$ and let it increase $dy$ in time $dt$. The latent heat released on melting has to be conducted away through the ice layers as the water freezes and therefore

quantity of heat lost due to increase $dy=rLAdy$

$\rho$ - density of ice

$L$ - Specific latent heat of fusion of water

$A$ - Area of ice surface

$x$ - thickness of ice after time t

rate of loss of heat = $LA\frac{\rho dy}{dt}=\frac{k\theta A}{y}$

So, $\frac{dy}{dt}=\frac{k\theta }{L\rho y}$