There is ice formation on a tank of water of thickness 10 cm.How much time it will take to have a layer of 0.1 cm below it ?The outer temperature is `-5^(@)C`, the thermal conductivity of ice is ` 0.005 calcm^(-1)``""^(@)C^(-1)` and latent heat of ice is `80cal//g` and the density of ice is `0.91 gcm^(-3)`:

Options

A. 46.39 minute
B. 47.63 minute
C. 48.78 minute
D. 49.31 minute

Explanation

To solve the problem of how much time it will take to have a layer of 0.1 cm of ice below the existing 10 cm layer, we can follow these steps:

### Step 1: Understand the Problem
We need to calculate the time required for a 0.1 cm layer of ice to form beneath an existing 10 cm layer of ice, given the outer temperature, thermal conductivity, latent heat of fusion, and density of ice.

### Step 2: Identify Given Data
- Thickness of existing ice layer, \( x = 10 \, \text{cm} \)
- Thickness of ice to be formed, \( dx = 0.1 \, \text{cm} \)
- Outer temperature, \( \theta_1 = -5^\circ C \)
- Temperature at the ice-water interface, \( \theta_2 = 0^\circ C \) (since water is at 0°C)
- Thermal conductivity of ice, \( k = 0.005 \, \text{cal cm}^{-1} \text{°C}^{-1} \)
- Latent heat of fusion of ice, \( L = 80 \, \text{cal/g} \)
- Density of ice, \( \rho = 0.91 \, \text{g/cm}^3 \)

### Step 3: Calculate the Heat Required to Form Ice
The heat required to form a mass \( dm \) of ice is given by:
\[ dQ = dm \cdot L \]
Where \( dm = \text{Volume} \times \text{Density} = A \cdot dx \cdot \rho \).

Thus,
\[ dQ = A \cdot dx \cdot \rho \cdot L \]

### Step 4: Calculate the Heat Flow Rate
The heat flow rate through the ice can be expressed using Fourier's law:
\[ \frac{dQ}{dt} = k \cdot A \cdot \frac{\theta_2 - \theta_1}{x} \]
Where \( \theta_2 - \theta_1 = 0 - (-5) = 5 \, \text{°C} \).

### Step 5: Set Up the Equation
Equating the heat required to the heat flow rate:
\[ A \cdot dx \cdot \rho \cdot L = k \cdot A \cdot \frac{5}{x} \cdot dt \]
We can cancel \( A \) from both sides:
\[ dx \cdot \rho \cdot L = k \cdot \frac{5}{x} \cdot dt \]

### Step 6: Rearrange the Equation
Rearranging gives:
\[ dt = \frac{dx \cdot \rho \cdot L \cdot x}{k \cdot 5} \]

### Step 7: Substitute Values
Substituting the known values:
- \( dx = 0.1 \, \text{cm} \)
- \( \rho = 0.91 \, \text{g/cm}^3 \)
- \( L = 80 \, \text{cal/g} \)
- \( k = 0.005 \, \text{cal cm}^{-1} \text{°C}^{-1} \)
- \( x = 10 \, \text{cm} \)

\[ dt = \frac{0.1 \cdot 0.91 \cdot 80 \cdot 10}{0.005 \cdot 5} \]

### Step 8: Calculate \( dt \)
Calculating the right-hand side:
\[ dt = \frac{0.1 \cdot 0.91 \cdot 80 \cdot 10}{0.025} \]
\[ dt = \frac{72.8}{0.025} = 2912 \, \text{seconds} \]

### Step 9: Convert Seconds to Minutes
To convert seconds to minutes:
\[ \text{Time in minutes} = \frac{2912}{60} \approx 48.53 \, \text{minutes} \]

### Final Answer
The time it will take to have a layer of 0.1 cm of ice below the existing layer is approximately **48.53 minutes**.