A layer of ice 0.15 m thick has formed on the surface of a deep pond. If the temperature of upper surface of ice is constant and equql to that of air which is `-12^(@)C`, determine the time it will take to increase of ice layer by 0.2 mm. take latent heat of ice =80 cal `g^(-1)`, density of ice =0.91 g `cm^(-3)` and thermal conductivity of ice =0.5 cal `s^(-1)m^(-1)K^(-1)` .

Answer

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Step 1: Understand the Problem
We need to find the time it takes to increase the thickness of the ice layer by 0.2 mm (0.02 cm) given the thickness of the ice is 0.15 m, the temperature of the upper surface is -12°C, and we have the latent heat, density, and thermal conductivity of ice.

### Step 2: Convert Units
Convert the increase in thickness from mm to cm:
- Increase in thickness = 0.2 mm = 0.02 cm

### Step 3: Calculate the Mass of Ice to be Frozen
The mass of the ice that needs to freeze can be calculated using the formula:
\[ M = A \times \text{thickness} \times \text{density} \]
Where:
- Thickness = 0.02 cm
- Density of ice = 0.91 g/cm³

Thus,
\[ M = A \times 0.02 \, \text{cm} \times 0.91 \, \text{g/cm}^3 \]
\[ M = 0.0182A \, \text{g} \]

### Step 4: Calculate the Heat Required
The heat required to freeze this mass of ice can be calculated using the formula:
\[ Q = M \times L \]
Where:
- Latent heat of ice, \( L = 80 \, \text{cal/g} \)

Thus,
\[ Q = (0.0182A) \times 80 \]
\[ Q = 1.456A \, \text{cal} \]

### Step 5: Calculate the Average Thickness of Ice
The average thickness through which heat must pass is given by:
\[ X = \frac{0.15 + 0.15 + 0.02}{2} \]
\[ X = \frac{0.32}{2} = 0.16 \, \text{m} = 16 \, \text{cm} \]

### Step 6: Use Fourier's Law of Heat Conduction
According to Fourier's law, the heat transfer can be expressed as:
\[ Q = \frac{K \cdot A \cdot (T_1 - T_2) \cdot t}{X} \]
Where:
- \( K = 0.5 \, \text{cal/s/m/K} \) = \( 0.5 \times 10^{-2} \, \text{cal/s/cm/K} \)
- \( T_1 = -12 \, \text{°C} \) (temperature of the upper surface)
- \( T_2 = 0 \, \text{°C} \) (temperature of the water below the ice)
- \( t \) is the time we want to find.

### Step 7: Rearranging the Equation for Time
Rearranging the equation to solve for \( t \):
\[ t = \frac{Q \cdot X}{K \cdot A \cdot (T_1 - T_2)} \]

### Step 8: Substitute Values
Substituting the values we calculated:
\[ t = \frac{1.456A \cdot 16 \, \text{cm}}{0.5 \times 10^{-2} \cdot A \cdot (-12 - 0)} \]
\[ t = \frac{1.456 \cdot 16}{0.5 \times 10^{-2} \cdot (-12)} \]

### Step 9: Simplify the Equation
\[ t = \frac{23.296}{-0.5 \times 10^{-2} \cdot (-12)} \]
\[ t = \frac{23.296}{0.06} \]
\[ t = 388.27 \, \text{s} \]

### Step 10: Final Result
Thus, the time it will take to increase the ice layer by 0.2 mm is approximately:
\[ t \approx 388.27 \, \text{s} \]