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