Calculate the rate of increment of the thickness of ice layer on a lake when thickness of ice is 10 cm and the air temperature is `-5^(@) C`. If thermal conductivity of ice is 0.008 cal `"cm"^(-1) "s"^(-1)°"C"^(-1)`, density of ice is `0.91 xx 10^(3) "kg m"^(-3)` and latent heat is `79.8` cal `"gm"^(-1)`. How long will it take the layer to become `10.1` cm ?

Explanation

To solve the problem step by step, we will calculate the rate of increment of the thickness of the ice layer on a lake and determine how long it will take for the layer to increase from 10 cm to 10.1 cm.

### Step 1: Understand the given parameters
- Thickness of ice, \( h = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Air temperature, \( T_{\text{air}} = -5^\circ C \)
- Thermal conductivity of ice, \( k = 0.008 \, \text{cal} \, \text{cm}^{-1} \, \text{s}^{-1} \, \text{°C}^{-1} = 0.008 \times 10^2 \, \text{cal} \, \text{m}^{-1} \, \text{s}^{-1} \, \text{°C}^{-1} = 0.8 \, \text{cal} \, \text{m}^{-1} \, \text{s}^{-1} \, \text{°C}^{-1} \)
- Density of ice, \( \rho = 0.91 \times 10^3 \, \text{kg/m}^3 \)
- Latent heat of fusion, \( L = 79.8 \, \text{cal/g} = 79.8 \times 10^3 \, \text{cal/kg} \)
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