On a winter day when the atmospheric temperature drops to `-10^(@)C` , ice forms on the surface of a lake. (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 expansion of water on freezing.

Explanation

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

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

1. **Identify the given values:**
- Atmospheric temperature, \( T_a = -10^\circ C \)
- Temperature of water, \( T_w = 0^\circ C \)
- Thickness of ice, \( x = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)
- Latent heat of fusion of ice, \( L = 3.36 \times 10^5 \, \text{J/kg} \)
- Thermal conductivity of ice, \( k = 1.7 \, \text{W/m} \cdot \text{C} \)

2. **Calculate the rate of heat transfer through the ice:**
The rate of heat transfer \( \frac{dQ}{dt} \) can be expressed using Fourier's law:
\[ \frac{dQ}{dt} = k \cdot A \cdot \frac{\Delta T}{x} \]
where \( A \) is the area, \( \Delta T = T_w - T_a = 0 - (-10) = 10^\circ C \).

3. **Express the mass of ice formed in terms of thickness:**
The mass of the ice formed can be expressed as:
\[ dm = \rho \cdot A \cdot dx \]
where \( dx \) is the increase in thickness of the ice.

4. **Relate the heat transfer to the mass of ice formed:**
The heat required to form the mass \( dm \) of ice is given by:
\[ dQ = dm \cdot L = \rho \cdot A \cdot dx \cdot L \]

5. **Set the equations equal:**
Setting the two expressions for \( dQ \) equal, we have:
\[ k \cdot A \cdot \frac{\Delta T}{x} = \rho \cdot A \cdot dx \cdot L \]
Cancel \( A \) from both sides:
\[ k \cdot \frac{\Delta T}{x} = \rho \cdot dx \cdot L \]

6. **Rearranging to find the rate of increase of thickness:**
\[ \frac{dx}{dt} = \frac{k \cdot \Delta T}{\rho \cdot L \cdot x} \]

7. **Substituting the values:**
\[ \frac{dx}{dt} = \frac{1.7 \cdot 10}{1000 \cdot 3.36 \times 10^5 \cdot 0.1} \]
\[ \frac{dx}{dt} = \frac{17}{33600000} \approx 5.06 \times 10^{-7} \, \text{m/s} \]

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

1. **Set up the differential equation:**
From the previous part, we have:
\[ dt = \frac{\rho \cdot L}{k \cdot \Delta T} \cdot \frac{x}{dx} \]

2. **Integrate to find the total time:**
Integrate from \( 0 \) to \( T \) and from \( 0 \) to \( 0.1 \, \text{m} \):
\[ T = \int_0^{0.1} \frac{\rho \cdot L}{k \cdot \Delta T} \cdot \frac{x}{dx} \]
\[ T = \frac{\rho \cdot L}{k \cdot \Delta T} \cdot \int_0^{0.1} x \, dx = \frac{\rho \cdot L}{k \cdot \Delta T} \cdot \left[ \frac{x^2}{2} \right]_0^{0.1} \]
\[ T = \frac{\rho \cdot L}{k \cdot \Delta T} \cdot \frac{(0.1)^2}{2} \]

3. **Substituting the values:**
\[ T = \frac{1000 \cdot 3.36 \times 10^5}{1.7 \cdot 10} \cdot \frac{0.01}{2} \]
\[ T = \frac{336000000}{17} \cdot 0.005 \approx 9870.59 \, \text{s} \]

4. **Convert seconds to hours:**
\[ T \approx \frac{9870.59}{3600} \approx 2.74 \, \text{hours} \]

### Final Answers:
(a) The rate of increase of thickness of the ice is approximately \( 5.06 \times 10^{-7} \, \text{m/s} \).

(b) The total time taken in forming 10 cm of ice is approximately \( 2.74 \, \text{hours} \).