A lake is covered with ice 2 cm thick.the temperature of ambient air is `-15^(@)C` find the rate of thickening of ice. For ice `k=4xx10^(-4)k-cal-m^(-1)s^(_1)(.^(@)C)^(-1)`. Density `=0.9xx10^(3)kg//m^(3)` and latent heat `L=80` kilo cal/kg

Options

A. 1.5cm/h
B. 0.75 cm/h
C. 3 cm/h
D. 2 cm/h

Explanation

To solve the problem of finding the rate of thickening of ice on a lake, we will follow these steps:

### Step 1: Understand the parameters given
- Thickness of ice (x) = 2 cm = 0.02 m
- Ambient temperature (T_a) = -15°C
- Temperature at the bottom of the ice (T_b) = 0°C (temperature of water)
- Thermal conductivity of ice (k) = \(4 \times 10^{-4} \, \text{k-cal/m/s/°C}\)
- Density of ice (ρ) = \(0.9 \times 10^{3} \, \text{kg/m}^3\)
- Latent heat of fusion (L) = 80 k-cal/kg

### Step 2: Calculate the temperature difference (ΔT)
\[ \Delta T = T_b - T_a = 0 - (-15) = 15 \, °C \]

### Step 3: Convert thermal conductivity to SI units
Since \(1 \, \text{k-cal} = 4184 \, \text{J}\), we convert k to J:
\[ k = 4 \times 10^{-4} \, \text{k-cal/m/s/°C} \times 4184 \, \text{J/k-cal} = 1.672 \times 10^{-3} \, \text{J/m/s/°C} \]

### Step 4: Use the heat flow equation
The rate of heat flow (dQ/dt) through the ice can be expressed as:
\[ \frac{dQ}{dt} = \frac{k \cdot A \cdot \Delta T}{x} \]
Where:
- A = area (which will cancel out later)

### Step 5: Relate heat flow to the rate of thickening
The mass of ice formed per unit time is given by:
\[ \frac{dm}{dt} = \rho \cdot A \cdot \frac{dx}{dt} \]
Where:
- \(dx/dt\) is the rate of thickening of the ice.

### Step 6: Relate heat flow to mass change
The heat required to freeze this mass is:
\[ \frac{dQ}{dt} = \frac{dm}{dt} \cdot L \]
Equating the two expressions for heat flow:
\[ \frac{k \cdot A \cdot \Delta T}{x} = \rho \cdot A \cdot \frac{dx}{dt} \cdot L \]

### Step 7: Cancel area (A) and rearrange for dx/dt
\[ \frac{dx}{dt} = \frac{k \cdot \Delta T}{\rho \cdot L \cdot x} \]

### Step 8: Substitute the values
Substituting the known values:
\[ \frac{dx}{dt} = \frac{(1.672 \times 10^{-3}) \cdot 15}{(0.9 \times 10^{3}) \cdot (80 \times 4184) \cdot (0.02)} \]

### Step 9: Calculate the rate of thickening
Calculating the denominator:
\[ \rho \cdot L = 0.9 \times 10^{3} \cdot 80 \cdot 4184 = 29952000 \, \text{J/m}^3 \]
Now substituting back:
\[ \frac{dx}{dt} = \frac{(1.672 \times 10^{-3}) \cdot 15}{29952000 \cdot 0.02} \]
Calculating:
\[ \frac{dx}{dt} = \frac{0.02508}{598080} \approx 4.19 \times 10^{-8} \, \text{m/s} \]

### Step 10: Convert to cm/hour
To convert to cm/hour:
\[ \frac{dx}{dt} \approx 4.19 \times 10^{-8} \, \text{m/s} \times 3600 \, \text{s/hour} \approx 0.000150 \, \text{m/hour} \approx 0.015 \, \text{cm/hour} \]

### Final Answer
The rate of thickening of ice is approximately \(0.015 \, \text{cm/hour}\).

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