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 `4.0^(@)C`. If the total depth of ice `+` water is 2.0m, how thick is the ice ? (Assume that the thermal conductivity of ice and water are 0.40 and 0.12 `cal//m.""^(@)C.s`, respectively. )

Answer

To solve the problem of determining the thickness of ice on a pond given the temperatures and thermal conductivities, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Temperature above ice, \( T_1 = -5.0^\circ C \)
- Temperature at the bottom of the pond, \( T_2 = 4.0^\circ C \)
- Total depth of ice and water, \( D = 2.0 \, m \)
- Thermal conductivity of ice, \( k_{ice} = 0.40 \, \text{kcal/m} \cdot ^\circ C \cdot s \)
- Thermal conductivity of water, \( k_{water} = 0.12 \, \text{kcal/m} \cdot ^\circ C \cdot s \)

2. **Define Variables:**
- Let \( x \) be the thickness of the ice.
- Therefore, the thickness of the water layer will be \( D - x = 2.0 - x \).

3. **Set Up Heat Transfer Equations:**
At steady state, the heat transfer through the ice layer must equal the heat transfer through the water layer. The heat transfer rate \( Q \) can be expressed as:
\[ Q = \frac{k \cdot A \cdot (T_{hot} - T_{cold})}{d} \]
where \( A \) is the area, \( T_{hot} \) and \( T_{cold} \) are the temperatures at the two sides, and \( d \) is the thickness.

For the ice:
\[ Q_{ice} = \frac{k_{ice} \cdot A \cdot (T_2 - T_{ice}}{x} \]
For the water:
\[ Q_{water} = \frac{k_{water} \cdot A \cdot (T_{ice} - T_1)}{2 - x} \]

4. **Assume Temperature at the Ice-Water Interface:**
At steady state, we can assume the temperature at the interface between ice and water is \( T_{ice} = 0^\circ C \).

5. **Substitute Values into the Equations:**
Substitute \( T_{ice} = 0^\circ C \) into the equations:
\[ Q_{ice} = \frac{k_{ice} \cdot A \cdot (4 - 0)}{x} = \frac{0.40 \cdot A \cdot 4}{x} \]
\[ Q_{water} = \frac{k_{water} \cdot A \cdot (0 - (-5))}{2 - x} = \frac{0.12 \cdot A \cdot 5}{2 - x} \]

6. **Set the Heat Transfer Rates Equal:**
Since \( Q_{ice} = Q_{water} \):
\[ \frac{0.40 \cdot A \cdot 4}{x} = \frac{0.12 \cdot A \cdot 5}{2 - x} \]

7. **Cancel Area \( A \) from Both Sides:**
\[ \frac{0.40 \cdot 4}{x} = \frac{0.12 \cdot 5}{2 - x} \]

8. **Cross Multiply to Solve for \( x \):**
\[ 0.40 \cdot 4 \cdot (2 - x) = 0.12 \cdot 5 \cdot x \]
\[ 1.6(2 - x) = 0.6x \]
\[ 3.2 - 1.6x = 0.6x \]
\[ 3.2 = 2.2x \]
\[ x = \frac{3.2}{2.2} \approx 1.45 \, m \]

9. **Conclusion:**
The thickness of the ice is approximately \( 1.45 \, m \).