A layer of ice of thickness y is on the surface of a lake. The air is at a constant temperature `-theta^@C` and the ice water interface is at `0^@C`. Show that the rate at which the thickness increases is given by
` (dy)/(dt) = (Ktheta)/(Lrhoy)`
where, K is the thermal conductivity of the ice, L the latent heat of fusion and `rho` is the density of the ice.

Explanation

To solve the problem, we need to analyze the situation where a layer of ice is forming on the surface of a lake due to the heat loss to the colder air above it. We will derive the expression for the rate of increase of the thickness of the ice layer.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a layer of ice of thickness \( y \) on the surface of a lake.
- The temperature of the air above the ice is \( -\theta \) °C, while the ice-water interface is at \( 0 \) °C.

2. **Heat Transfer**:
- The heat loss from the ice to the air is conducted through the ice layer. This heat loss leads to the formation of more ice at the interface.
- The rate of heat loss \( \frac{dQ}{dt} \) through the ice can be expressed using Fourier's law of heat conduction:
\[ \frac{dQ}{dt} = K \cdot A \cdot \frac{\Delta T}{y} \]
where:
- \( K \) is the thermal conductivity of ice,
- \( A \) is the area of the ice surface,
- \( \Delta T = 0 - (-\theta) = \theta \) is the temperature difference across the ice layer,
- \( y \) is the thickness of the ice.

3. **Latent Heat of Fusion**:
- As the ice forms, it releases latent heat. The amount of heat released when a mass \( m \) of ice is formed is given by:
\[ dQ = mL \]
where \( L \) is the latent heat of fusion of ice.

4. **Mass of Ice Formed**:
- The mass of the ice formed in a time interval \( dt \) can be expressed as:
\[ m = \rho \cdot V = \rho \cdot A \cdot dy \]
where \( \rho \) is the density of ice, and \( dy \) is the increase in thickness of the ice.

5. **Equating Heat Loss and Heat Released**:
- From the heat loss and heat released, we can write:
\[ \frac{dQ}{dt} = \rho A dy \cdot L \]
- Setting the two expressions for \( \frac{dQ}{dt} \) equal gives:
\[ K \cdot A \cdot \frac{\theta}{y} = \rho \cdot A \cdot dy \cdot L \]

6. **Canceling Common Terms**:
- We can cancel \( A \) from both sides:
\[ K \cdot \frac{\theta}{y} = \rho \cdot dy \cdot L \]

7. **Rearranging for Rate of Thickness Increase**:
- Rearranging the equation to solve for \( \frac{dy}{dt} \):
\[ \frac{dy}{dt} = \frac{K \theta}{L \rho y} \]

### Final Result:
Thus, we have shown that the rate at which the thickness of the ice increases is given by:
\[ \frac{dy}{dt} = \frac{K \theta}{L \rho y} \]