To solve the problem of calculating the time in which a layer of ice 6 cm thick on the surface of a pond will increase in thickness by 2 mm when the temperature of the surrounding air is -20°C, we can follow these steps:
### Step 1: Convert Given Values
- Thickness of ice (initial) = 6 cm = 0.06 m
- Increase in thickness = 2 mm = 0.002 m
- Temperature of surrounding air, T1 = -20°C = -20 K (for calculations, we will use Celsius)
- Latent heat of fusion, L = 333 kJ/kg = 333,000 J/kg
- Conductivity of ice, K = 0.08 cal/s·cm·°C = 8 kcal/s·m·°C (since 1 cal = 0.001 kcal and 1 cm = 0.01 m)
### Step 2: Calculate the Mass of Ice Formed
The mass of ice formed can be calculated using the formula:
\[
m = \rho \cdot V
\]
Where:
- \(\rho\) (density of ice) = 900 kg/m³
- Volume, \(V = A \cdot \Delta h = A \cdot 0.002 \, \text{m}\) (where A is the area)
Thus,
\[
m = 900 \cdot A \cdot 0.002
\]
### Step 3: Relate Heat Transfer to Mass of Ice Formed
The heat required to form the mass of ice is given by:
\[
Q = m \cdot L
\]
Substituting for \(m\):
\[
Q = (900 \cdot A \cdot 0.002) \cdot 333,000
\]
### Step 4: Use Fourier’s Law of Heat Conduction
According to Fourier’s law, the heat transfer \(Q\) can also be expressed as:
\[
Q = \frac{K \cdot A \cdot (T_2 - T_1) \cdot t}{h}
\]
Where:
- \(T_2 = 0°C\) (the temperature of the ice surface)
- \(T_1 = -20°C\)
- \(h = 0.06 \, \text{m}\) (thickness of ice)
### Step 5: Set the Two Expressions for Heat Equal
Setting the two expressions for \(Q\) equal gives:
\[
(900 \cdot A \cdot 0.002) \cdot 333,000 = \frac{K \cdot A \cdot (0 - (-20)) \cdot t}{0.06}
\]
### Step 6: Cancel Out Area (A)
Since \(A\) appears on both sides, we can cancel it out:
\[
(900 \cdot 0.002) \cdot 333,000 = \frac{K \cdot 20 \cdot t}{0.06}
\]
### Step 7: Substitute Values and Solve for t
Substituting \(K = 8 \cdot 4184 \, \text{J/s·m·°C}\):
\[
(900 \cdot 0.002) \cdot 333,000 = \frac{(8 \cdot 4184) \cdot 20 \cdot t}{0.06}
\]
Calculating the left side:
\[
1.8 \cdot 333,000 = 59904000
\]
And the right side becomes:
\[
\frac{(33472) \cdot 20 \cdot t}{0.06}
\]
Setting both sides equal and solving for \(t\):
\[
59904000 = \frac{669440 \cdot t}{0.06}
\]
\[
t = \frac{59904000 \cdot 0.06}{669440}
\]
Calculating \(t\):
\[
t \approx 53.77 \, \text{seconds}
\]
### Final Answer
The time in which the layer of ice will increase in thickness by 2 mm is approximately **53.77 seconds**.
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