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The thickness of ice in a lake at a certain moment is 0.03m. At what rate the thickness of the ice increasing and how long will it take for the thickness of ice layer to be doubled, if the temperature of air is -20°C. Given coefficient of thermal conductivity of ice = 0.168 Wm-1K-1 degree. Latent heat of fusion of ice = 3.35 × 10 Jkg-1 density of ice= 9.2 × 102 Kg/m3

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When the layer of ice covers the lake, the lower part of the ice is in contact with ice is at 0°c and outer surface with air at – 20°c

∴ θ1 = 0°c and θ2 = – 20°c

Let the thickness of ice increase by small amount ∆x by an area A in a time ∆t.

Then, mass of ice formed = A × ∆x * ρ

Quantity of heat, Q = A × ∆x × ρ × L …..(1)

But, Q = \(\frac{kA(θ_1-θ_2)Δt}{d}\) ….(2)

from equation (1) & equation (2)

A × ∆x × r × L = \(\frac{kA(θ_1-θ_2)Δt}{d}\)

= 3.634 × 10-7ms-1

The time taken for the ice layer to double is

= 60.57 × 10-7sec

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