In an `20` m deep lake, the bottom is at a constant temperature of ` 4^(@)C`. The air temperature is constant at `-10 ^(@)C`. The thermal conductivity of ice in `4` times that water. Neglecting the expansion of water on freezing, the maximum thickness of ice will be

Options

A. ` (22)/(11)` m
B. `(200)/(11)` m
C. `20 m`
D. `10` m

Explanation

To solve the problem of determining the maximum thickness of ice that can form on a lake with a given temperature profile, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- The lake is 20 m deep.
- The bottom temperature is constant at 4°C.
- The air temperature is -10°C.
- The thermal conductivity of ice is 4 times that of water.

2. **Defining Variables**:
- Let \( x \) be the thickness of the ice.
- The thickness of the water below the ice is \( 20 - x \).

3. **Heat Flow Concept**:
- At steady state, the heat flow through the ice must equal the heat flow through the water.
- The heat flow rate can be expressed using Fourier's law of heat conduction:
\[ \frac{dQ}{dt} = \frac{kA \Delta T}{\Delta x} \]
- Here, \( k \) is the thermal conductivity, \( A \) is the cross-sectional area, \( \Delta T \) is the temperature difference, and \( \Delta x \) is the thickness.

4. **Setting Up the Heat Flow Equations**:
- For the ice:
\[ \frac{dQ}{dt} = 4kA \frac{(0 - (-10))}{x} = 4kA \frac{10}{x} \]
- For the water:
\[ \frac{dQ}{dt} = kA \frac{(4 - 0)}{(20 - x)} = kA \frac{4}{20 - x} \]

5. **Equating the Heat Flows**:
- Since the heat flow through the ice equals the heat flow through the water at steady state:
\[ 4kA \frac{10}{x} = kA \frac{4}{20 - x} \]

6. **Canceling Common Terms**:
- We can cancel \( kA \) from both sides:
\[ 4 \cdot 10 \cdot (20 - x) = 4x \]

7. **Simplifying the Equation**:
- This simplifies to:
\[ 40(20 - x) = 4x \]
- Expanding gives:
\[ 800 - 40x = 4x \]
- Rearranging results in:
\[ 800 = 44x \]

8. **Solving for \( x \)**:
- Dividing both sides by 44:
\[ x = \frac{800}{44} = \frac{200}{11} \approx 18.18 \text{ m} \]

9. **Conclusion**:
- The maximum thickness of ice that can form on the lake is approximately \( \frac{200}{11} \) m.