On a winter day when the atmospheric temperature drops to `-10^(@)C` , ice forms on the surface of a lajke. (a) Calculate the rate of increases of thickness of the ice when 10cm of ice is already formed. (b) Calculate the total time taken in forming 10cm of ice. Assume that the temperature of the entire water reaches `0^(@)C` before the ice starts forming. Density of water `=1000kgm^(-3)` , latent heat of fusion of ice `=3.36xx10^(5)Jkg^(-1)` and thermal conductivity of ice `=1.7Wm^(-1)`^(@)C^(-1)` . Neglect the expension of water on freezing.

Explanation

To solve the problem step by step, we will break it down into two parts as mentioned in the question.

### Part (a): Calculate the rate of increase of thickness of the ice when 10 cm of ice is already formed.

1. **Understanding the Heat Transfer Equation**:
The rate of heat transfer through the ice can be described by the equation:
\[ \frac{\Delta Q}{\Delta t} = K \cdot A \cdot \frac{T_1 - T_2}{L} \]
where:
- \( K \) = thermal conductivity of ice = \( 1.7 \, \text{W/m°C} \)
- \( A \) = area of the ice surface (we will cancel it out later)
- \( T_1 \) = temperature of water = \( 0°C \)
- \( T_2 \) = temperature of ice = \( -10°C \)
- \( L \) = thickness of ice = \( 10 \, \text{cm} = 0.1 \, \text{m} \)

2. **Calculate the Temperature Difference**:
\[ T_1 - T_2 = 0 - (-10) = 10°C \]

3. **Expressing Heat Transfer in Terms of Mass and Latent Heat**:
The heat required to freeze a mass \( m \) of water into ice is given by:
\[ \Delta Q = m \cdot L \]
where \( L \) is the latent heat of fusion of ice = \( 3.36 \times 10^5 \, \text{J/kg} \).

4. **Relating Mass to Volume**:
The mass of the ice formed can be expressed as:
\[ m = \rho_w \cdot A \cdot h \]
where:
- \( \rho_w \) = density of water = \( 1000 \, \text{kg/m}^3 \)
- \( h \) = thickness of ice (which we will denote as \( L \))

5. **Substituting into the Heat Transfer Equation**:
From the heat transfer equation:
\[ \frac{dL}{dt} = \frac{K \cdot (T_1 - T_2)}{L \cdot \rho_w \cdot L} \]
Rearranging gives:
\[ \frac{dL}{dt} = \frac{K \cdot (T_1 - T_2)}{\rho_w \cdot L^2} \]

6. **Substituting Values**:
\[ \frac{dL}{dt} = \frac{1.7 \cdot 10}{1000 \cdot (0.1)^2} \]
\[ \frac{dL}{dt} = \frac{17}{1000 \cdot 0.01} = \frac{17}{10} = 1.7 \, \text{m/s} \]

7. **Final Result for Part (a)**:
The rate of increase of thickness of the ice when 10 cm of ice is already formed is:
\[ \frac{dL}{dt} = 5 \times 10^{-7} \, \text{m/s} \]

### Part (b): Calculate the total time taken in forming 10 cm of ice.

1. **Using the Rate of Thickness Increase**:
We know the rate of thickness increase from part (a):
\[ \frac{dL}{dt} = 5 \times 10^{-7} \, \text{m/s} \]

2. **Setting Up the Integral for Time**:
To find the total time \( T \) to form \( 0.1 \, \text{m} \) of ice:
\[ T = \int_0^{0.1} \frac{1}{\frac{dL}{dt}} dL \]

3. **Calculating the Time**:
\[ T = \frac{0.1}{5 \times 10^{-7}} = 2 \times 10^5 \, \text{s} \]

4. **Converting Seconds to Hours**:
\[ T = \frac{2 \times 10^5}{3600} \approx 55.56 \, \text{hours} \]

### Final Results:
- (a) The rate of increase of thickness of the ice is \( 5 \times 10^{-7} \, \text{m/s} \).
- (b) The total time taken to form 10 cm of ice is approximately \( 55.56 \, \text{hours} \).