Ice has formed on a shallow pond, and a steady state has been reached, with the air above the ice at `-5.0^(@)C` and the bottom of the pond at `40^(@)C`. If the total depth of ice + water is `1.4 m`, (Assume that the thermal conductivities of ice and water are `0.40` and `0.12 cal//m C^(@)s`, respectively.)
The thickness of ice layer is

Options

A. 1.1 m
B. 0.4 m
C. 2.1 m
D. 3.6 m

Answer

To solve the problem of finding the thickness of the ice layer on a pond, we can use the concept of steady-state heat conduction. The heat transfer through the ice and water can be described using Fourier's law of heat conduction.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Temperature of air above the ice, \( T_a = -5.0^\circ C \)
- Temperature at the bottom of the pond, \( T_b = 40.0^\circ C \)
- Total depth of ice and water, \( L = 1.4 \, m \)
- Thermal conductivity of ice, \( k_{ice} = 0.40 \, \text{cal/m} \cdot \text{s} \cdot {}^\circ C \)
- Thermal conductivity of water, \( k_{water} = 0.12 \, \text{cal/m} \cdot \text{s} \cdot {}^\circ C \)

2. **Define Variables:**
- Let \( L_{ice} \) be the thickness of the ice layer.
- Let \( L_{water} = L - L_{ice} \) be the thickness of the water layer.

3. **Set Up the Heat Transfer Equation:**
Since the system is in steady state, the heat flow through the ice layer must equal the heat flow through the water layer. We can express this as:
\[ \frac{k_{ice} \cdot A \cdot (T_b - 0)}{L_{ice}} = \frac{k_{water} \cdot A \cdot (0 - T_a)}{L_{water}} \]
Here, \( T_x = 0^\circ C \) is the temperature at the interface between the ice and water.

4. **Simplify the Equation:**
The area \( A \) cancels out from both sides:
\[ \frac{k_{ice} \cdot (T_b - 0)}{L_{ice}} = \frac{k_{water} \cdot (0 - T_a)}{L_{water}} \]
Substituting \( L_{water} = L - L_{ice} \):
\[ \frac{k_{ice} \cdot T_b}{L_{ice}} = \frac{k_{water} \cdot (-T_a)}{L - L_{ice}} \]

5. **Substitute Known Values:**
Plugging in the values we have:
\[ \frac{0.40 \cdot 40}{L_{ice}} = \frac{0.12 \cdot 5}{1.4 - L_{ice}} \]

6. **Cross Multiply:**
Cross multiplying gives:
\[ 0.40 \cdot 40 \cdot (1.4 - L_{ice}) = 0.12 \cdot 5 \cdot L_{ice} \]

7. **Expand and Rearrange:**
Expanding both sides:
\[ 16 \cdot (1.4 - L_{ice}) = 0.6 \cdot L_{ice} \]
\[ 22.4 - 16L_{ice} = 0.6L_{ice} \]
\[ 22.4 = 16.6L_{ice} \]

8. **Solve for \( L_{ice} \):**
\[ L_{ice} = \frac{22.4}{16.6} \approx 1.352 \, m \]

9. **Calculate the Thickness of Ice Layer:**
Since the total depth is \( 1.4 \, m \):
\[ L_{water} = 1.4 - L_{ice} \approx 1.4 - 1.352 \approx 0.048 \, m \]
Thus, the thickness of the ice layer is approximately \( 1.352 \, m \).

### Final Answer:
The thickness of the ice layer is approximately **1.352 m**.