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 find the maximum thickness of ice that can form on a lake with a given depth and temperature conditions, we can use the concept of heat conduction. The heat flow through the ice and the water must be equal at thermal equilibrium.

### Step-by-Step Solution:

1. **Identify the Known Values:**
- Depth of the lake (h) = 20 m
- Temperature at the bottom of the lake (T_bottom) = 4°C
- Air temperature (T_air) = -10°C
- Thermal conductivity of ice (k_ice) = 4 × k_water

2. **Set Up the Heat Flow Equations:**
The rate of heat flow through the ice (Q_ice) and the water (Q_water) can be expressed using Fourier's law of heat conduction:
\[ Q = \frac{k \cdot A \cdot (T_1 - T_2)}{d} \]
where \( k \) is the thermal conductivity, \( A \) is the area, \( T_1 \) and \( T_2 \) are the temperatures, and \( d \) is the thickness.

3. **Heat Flow Through Ice:**
For the ice layer of thickness \( x \):
\[ Q_{ice} = \frac{k_{ice} \cdot A \cdot (T_{air} - T_{bottom})}{x} \]
Substituting the values:
\[ Q_{ice} = \frac{4k_{water} \cdot A \cdot (-10 - 4)}{x} = \frac{4k_{water} \cdot A \cdot (-14)}{x} \]

4. **Heat Flow Through Water:**
For the water layer below the ice (thickness \( 20 - x \)):
\[ Q_{water} = \frac{k_{water} \cdot A \cdot (T_{bottom} - 0)}{20 - x} \]
Substituting the values:
\[ Q_{water} = \frac{k_{water} \cdot A \cdot (4 - 0)}{20 - x} = \frac{4k_{water} \cdot A}{20 - x} \]

5. **Set the Heat Flows Equal:**
Since the heat flow through the ice must equal the heat flow through the water at equilibrium:
\[ \frac{4k_{water} \cdot A \cdot (-14)}{x} = \frac{4k_{water} \cdot A}{20 - x} \]

6. **Cancel Common Terms:**
We can cancel \( 4k_{water} \cdot A \) from both sides:
\[ \frac{-14}{x} = \frac{1}{20 - x} \]

7. **Cross-Multiply to Solve for x:**
\[ -14(20 - x) = x \]
\[ -280 + 14x = x \]
\[ 14x - x = 280 \]
\[ 13x = 280 \]
\[ x = \frac{280}{13} \approx 21.54 \text{ m} \]

8. **Conclusion:**
The maximum thickness of ice that can form on the lake is approximately \( 21.54 \) m.