To solve the problem, we need to determine the time required for a layer of ice on a pond to increase in thickness from 10 cm to 15 cm when the surrounding temperature is -10°C. We will use the formula derived from the concept of thermal conductivity.
### Step-by-Step Solution:
1. **Identify the Given Values:**
- Initial thickness of ice, \( x_1 = 10 \, \text{cm} \)
- Final thickness of ice, \( x_2 = 15 \, \text{cm} \)
- Change in thickness, \( \Delta x = x_2 - x_1 = 5 \, \text{cm} \)
- Coefficient of thermal conductivity of ice, \( K = 0.005 \, \frac{\text{Cal}}{\text{cm} \cdot \text{s} \cdot \degree C} \)
- Density of ice, \( \rho = 0.9 \, \text{g/cm}^3 \)
- Latent heat of fusion of ice, \( L = 80 \, \frac{\text{Cal}}{\text{g}} \)
- Temperature difference, \( \theta = 10 \, \degree C \) (since the surrounding temperature is -10°C and the ice is at 0°C)
2. **Use the Formula for Time:**
The time \( T \) required for the thickness of ice to increase can be calculated using the formula:
\[
T = \frac{\rho L}{2K \theta} (x_2^2 - x_1^2)
\]
3. **Substitute the Values:**
- Substitute \( \rho = 0.9 \, \text{g/cm}^3 \)
- Substitute \( L = 80 \, \frac{\text{Cal}}{\text{g}} \)
- Substitute \( K = 0.005 \, \frac{\text{Cal}}{\text{cm} \cdot \text{s} \cdot \degree C} \)
- Substitute \( \theta = 10 \, \degree C \)
- Substitute \( x_1 = 10 \, \text{cm} \) and \( x_2 = 15 \, \text{cm} \)
Now, calculate \( x_2^2 - x_1^2 \):
\[
x_2^2 - x_1^2 = 15^2 - 10^2 = 225 - 100 = 125
\]
Now substitute into the formula:
\[
T = \frac{0.9 \times 80}{2 \times 0.005 \times 10} \times 125
\]
4. **Calculate the Time:**
- Calculate the numerator:
\[
0.9 \times 80 = 72
\]
- Calculate the denominator:
\[
2 \times 0.005 \times 10 = 0.1
\]
- Now substitute these values into the equation:
\[
T = \frac{72}{0.1} \times 125 = 720 \times 125 = 90000 \, \text{s}
\]
5. **Convert Seconds to Hours:**
To convert seconds into hours:
\[
T = \frac{90000}{3600} \approx 25 \, \text{hours}
\]
### Final Answer:
The time in which the layer of ice will increase by 5 cm is approximately **25 hours**.
