If the depth of a lake is 10 m, and the temperature of surroundings is `-10^@C` and that of water at the bottom of the lake is `5^@C`. The thermal conductivity of the Ice and water are `K_f` and `K_w` respectively. The depth of ice will be

Options

A. `(5K_w)/(K_w+K_l)`
B. `(10K_w)/(K_w+2K_l)`
C. `(20K_l)/(K_w+2K_l)`
D. None of these

Answer

To solve the problem of finding the depth of ice in the lake, we will use the principles of thermal conductivity and heat transfer. Here’s a step-by-step solution:

### Step 1: Define the variables
- Let the depth of ice be \( x \).
- The total depth of the lake is given as \( 10 \, \text{m} \).
- Therefore, the depth of water is \( 10 - x \).
- The temperature of the surroundings is \( -10^\circ C \).
- The temperature of water at the bottom of the lake is \( 5^\circ C \).
- The thermal conductivity of ice is \( K_f \) and that of water is \( K_w \).

### Step 2: Understand the heat transfer
In a steady state, the heat current through the ice layer must equal the heat current through the water layer. This can be expressed mathematically as:
\[ H_{\text{ice}} = H_{\text{water}} \]

### Step 3: Write the heat transfer equations
Using the formula for heat current (or heat transfer rate):
\[ H = \frac{K \cdot A \cdot \Delta T}{L} \]
Where:
- \( H \) is the heat current,
- \( K \) is the thermal conductivity,
- \( A \) is the cross-sectional area,
- \( \Delta T \) is the temperature difference,
- \( L \) is the thickness of the material.

For the ice:
\[ H_{\text{ice}} = \frac{K_f \cdot A \cdot (0 - (-10))}{x} = \frac{K_f \cdot A \cdot 10}{x} \]

For the water:
\[ H_{\text{water}} = \frac{K_w \cdot A \cdot (5 - 0)}{10 - x} = \frac{K_w \cdot A \cdot 5}{10 - x} \]

### Step 4: Set the heat currents equal
Setting the two expressions equal gives:
\[ \frac{K_f \cdot A \cdot 10}{x} = \frac{K_w \cdot A \cdot 5}{10 - x} \]
Since the area \( A \) cancels out, we can simplify this to:
\[ \frac{K_f \cdot 10}{x} = \frac{K_w \cdot 5}{10 - x} \]

### Step 5: Cross-multiply to solve for \( x \)
Cross-multiplying gives:
\[ K_f \cdot 10 \cdot (10 - x) = K_w \cdot 5 \cdot x \]

Expanding this:
\[ 10 K_f \cdot 10 - 10 K_f \cdot x = 5 K_w \cdot x \]
\[ 100 K_f = 10 K_f \cdot x + 5 K_w \cdot x \]
\[ 100 K_f = x (10 K_f + 5 K_w) \]

### Step 6: Solve for \( x \)
Rearranging gives:
\[ x = \frac{100 K_f}{10 K_f + 5 K_w} \]

### Final Expression
Thus, the depth of ice \( x \) can be expressed as:
\[ x = \frac{20 K_f}{K_w + 2 K_f} \]