A pond has a layer of ice of thickness 0.25 m on its surface the temperature of the atmosphere is `10^(@) C`. Find out the time required to increase the thickness of the layer of ice `0.5 "mm"`. K of ice `= 2 "Wm"^(-2) "K"^(-1)`. Density of ice `= 900 "kg m"^(-3)` latent heat of fusion of ice `= 336 "KJ kg"^(-1)`.

Explanation

To find the time required to increase the thickness of the layer of ice by 0.5 mm, we can use the principles of thermal conduction and the properties of ice. Here’s a step-by-step solution:

### Step 1: Convert the thickness increase from mm to meters
We need to convert 0.5 mm to meters for consistency in units.

\[ 0.5 \text{ mm} = 0.5 \times 10^{-3} \text{ m} = 0.0005 \text{ m} \]

**Hint:** Always ensure that all units are consistent when performing calculations.

### Step 2: Identify the parameters given in the problem
We have the following parameters:
- Thickness of ice layer, \( x = 0.25 \text{ m} \)
- Temperature of the atmosphere, \( T_a = 10^\circ C \)
- Thermal conductivity of ice, \( k = 2 \text{ W m}^{-1} \text{K}^{-1} \)
- Density of ice, \( \rho = 900 \text{ kg m}^{-3} \)
- Latent heat of fusion of ice, \( L = 336 \text{ kJ kg}^{-1} = 336 \times 10^3 \text{ J kg}^{-1} \)
- Increase in thickness of ice, \( \Delta x = 0.0005 \text{ m} \)

**Hint:** Write down all the known values clearly to avoid confusion later.

### Step 3: Calculate the temperature difference
The temperature difference between the ice and the atmosphere is:

\[ \Delta T = T_a - T_{ice} = 10^\circ C - 0^\circ C = 10 \text{ K} \]

**Hint:** Remember that the temperature of the ice is assumed to be 0°C since it is at the freezing point.

### Step 4: Use the formula for heat transfer
The rate of heat transfer through the ice can be expressed using Fourier's law of heat conduction:

\[ \frac{dq}{dt} = k \cdot A \cdot \frac{\Delta T}{x} \]

Where:
- \( A \) is the area (we can assume it to be 1 m² for simplicity).
- \( x \) is the thickness of the ice layer.

**Hint:** Understand that the area cancels out if we are looking for a rate per unit area.

### Step 5: Substitute the values into the formula
Substituting the known values into the equation:

\[ \frac{dq}{dt} = 2 \cdot 1 \cdot \frac{10}{0.25} = 80 \text{ W} \]

**Hint:** Make sure to double-check your calculations for accuracy.

### Step 6: Calculate the mass of ice that needs to be formed
The mass of ice that corresponds to the increase in thickness can be calculated as:

\[ \Delta m = \rho \cdot A \cdot \Delta x = 900 \cdot 1 \cdot 0.0005 = 0.45 \text{ kg} \]

**Hint:** Remember that the mass is directly related to the volume and density.

### Step 7: Calculate the heat required to freeze this mass of ice
The heat required to freeze this mass of ice is given by:

\[ Q = \Delta m \cdot L = 0.45 \cdot 336 \times 10^3 = 151200 \text{ J} \]

**Hint:** Ensure you are using the correct units for energy (Joules).

### Step 8: Calculate the time required
Now, we can find the time required for this heat transfer using:

\[ t = \frac{Q}{\frac{dq}{dt}} = \frac{151200}{80} = 1890 \text{ seconds} \]

**Hint:** Time can be calculated by dividing the total energy required by the rate of energy transfer.

### Final Answer
The time required to increase the thickness of the layer of ice by 0.5 mm is approximately **1890 seconds** or **31.5 minutes**.

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