The thickness of ice on a Jake is 6 cm and the temperature of air is `- 10^(@) C`. At what rate is the thickness of ice increasing ? Thermal conductivity of ice `= 2 "Wm"^(-1) "K"^(-1)`. Density of ice `= 920 "kg m"^(3)`, specific latent heat of ice `= 336 "KJ kg"^(-1)`

Explanation

To solve the problem of determining the rate at which the thickness of ice on a lake is increasing, we can follow these steps:

### Step 1: Understand the Problem
We have a thickness of ice on a lake of 6 cm, and the temperature of the air is -10°C. We need to find the rate at which the thickness of the ice is increasing. We are given the thermal conductivity of ice, its density, and the specific latent heat of ice.

### Step 2: Identify the Relevant Equations
The heat current (rate of heat transfer) through the ice can be expressed using Fourier's law of heat conduction:
\[ \frac{dq}{dt} = \frac{k \cdot A \cdot \Delta T}{y} \]
where:
- \( dq/dt \) = heat current (W)
- \( k \) = thermal conductivity of ice (2 W/m·K)
- \( A \) = area of the ice surface (m²)
- \( \Delta T \) = temperature difference (0 - (-10) = 10 K)
- \( y \) = thickness of the ice (converted to meters)

### Step 3: Relate Heat Current to Mass and Latent Heat
The heat transferred can also be expressed in terms of the mass of the ice layer that is forming:
\[ dq = m \cdot L \]
where:
- \( m \) = mass of the ice layer
- \( L \) = specific latent heat of ice (336 kJ/kg = 336,000 J/kg)

The mass \( m \) can be expressed as:
\[ m = \rho \cdot V = \rho \cdot A \cdot dy \]
where:
- \( \rho \) = density of ice (920 kg/m³)
- \( V \) = volume of the ice layer (A·dy)

### Step 4: Set Up the Equation
Equating the two expressions for heat transfer:
\[ \frac{k \cdot A \cdot \Delta T}{y} = \rho \cdot A \cdot dy \cdot L \]
The area \( A \) cancels out:
\[ \frac{k \cdot \Delta T}{y} = \rho \cdot dy \cdot L \]

### Step 5: Solve for \( \frac{dy}{dt} \)
Rearranging the equation gives:
\[ \frac{dy}{dt} = \frac{k \cdot \Delta T}{y \cdot \rho \cdot L} \]

### Step 6: Substitute the Values
Now we can substitute the known values:
- \( k = 2 \, \text{W/m·K} \)
- \( \Delta T = 10 \, \text{K} \)
- \( y = 0.06 \, \text{m} \) (6 cm converted to meters)
- \( \rho = 920 \, \text{kg/m}^3 \)
- \( L = 336,000 \, \text{J/kg} \)

Substituting these values into the equation:
\[ \frac{dy}{dt} = \frac{2 \cdot 10}{0.06 \cdot 920 \cdot 336000} \]

### Step 7: Calculate the Rate
Calculating the right-hand side:
\[ \frac{dy}{dt} = \frac{20}{0.06 \cdot 920 \cdot 336000} \]
Calculating the denominator:
\[ 0.06 \cdot 920 \cdot 336000 = 18432000 \]
Thus,
\[ \frac{dy}{dt} = \frac{20}{18432000} \approx 1.084 \times 10^{-6} \, \text{m/s} \]

### Final Answer
The thickness of the ice is increasing at a rate of approximately \( 1.084 \, \mu m/s \).