A deep rectangular pond of surface area A, containing water (density = `rho` ), specific heat capacity = s), is located in a region where the outside air temperature is at a steady value of `-26^@C` . The thickness of the frozen ice layer in this pond, at a certain instant is x. Taking the thermal conductivity of ice as K, and its specific latent heat of fusion as L, the rate of increase of the thickness of ice layer, at this instant, would be given by

Options

A. `26K//rho x (L+4s)`
B. `(26K // rho x (L - 4s)`
C. `26K//(rho x^2 L)`
D. `26K//(rho x L)`

Explanation

To find the rate of increase of the thickness of the ice layer in the pond, we can use the principles of heat transfer and the properties of ice and water. Here's a step-by-step solution:

### Step 1: Understand the Heat Transfer Process
The heat transfer occurs from the water below the ice to the outside air. The heat lost by the water is used to freeze the water into ice. The rate of heat loss through the ice layer can be described using Fourier's law of heat conduction.

### Step 2: Apply Fourier's Law of Heat Conduction
According to Fourier's law, the rate of heat transfer (Q) through a material is given by:
\[ Q = K \cdot A \cdot \frac{\Delta T}{x} \]
where:
- \( K \) is the thermal conductivity of ice,
- \( A \) is the surface area of the pond,
- \( \Delta T \) is the temperature difference between the water and the outside air,
- \( x \) is the thickness of the ice layer.

### Step 3: Determine the Temperature Difference
The temperature of the water just below the ice is at 0°C (assuming the ice is at the freezing point), and the outside air temperature is -26°C. Therefore, the temperature difference \( \Delta T \) is:
\[ \Delta T = 0 - (-26) = 26 \, \text{°C} \]

### Step 4: Substitute \( \Delta T \) into the Heat Transfer Equation
Now, substituting \( \Delta T \) into the equation:
\[ Q = K \cdot A \cdot \frac{26}{x} \]

### Step 5: Relate Heat Transfer to Ice Formation
The heat lost by the water is used to freeze a certain volume of water into ice. The amount of heat required to freeze a mass \( m \) of water is given by:
\[ Q = m \cdot L \]
where \( L \) is the specific latent heat of fusion.

### Step 6: Express Mass in Terms of Volume and Density
The mass of the ice formed can be expressed in terms of the volume of ice formed (which is the increase in thickness \( dx \)):
\[ m = \rho \cdot A \cdot dx \]
where \( \rho \) is the density of ice.

### Step 7: Set the Heat Lost Equal to the Heat Gained
Equating the two expressions for \( Q \):
\[ K \cdot A \cdot \frac{26}{x} = \rho \cdot A \cdot dx \cdot L \]

### Step 8: Simplify the Equation
Cancel \( A \) from both sides:
\[ K \cdot \frac{26}{x} = \rho \cdot dx \cdot L \]

### Step 9: Solve for \( \frac{dx}{dt} \)
Rearranging gives:
\[ \frac{dx}{dt} = \frac{K \cdot 26}{\rho \cdot L \cdot x} \]

### Final Result
Thus, the rate of increase of the thickness of the ice layer is given by:
\[ \frac{dx}{dt} = \frac{K \cdot 26}{\rho \cdot L \cdot x} \]

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