Question

A 0.294-m-thick sheet of ice covers a lake. The air temperature at the ice surface is -17.7 °C. In 3.00 minutes, the ice thickens by a small amount. Assume that no heat flows from the ground below into the water and that the added ice is very thin compared to 0.294 m. Calculate the number of millimeters by which the ice thickens. Do not enter unit.

Answer 1

The liquid water directly beneath the ice sheet is equilibrium with the solid and has therefore a temperature 0°C (=temperature of fusion).

Since the water is hotter than the air above the ice sheet, it looses some heat due to thermal conduction though the ice. This heat flow rate can be found from Fourier's law:
dQ/dt = k∙A∙∆T/∆x

k thermal conductivity, for ice k = 2.33 Wm⁻¹K⁻¹
A surface area of the lake
∆T temperature difference between the two sides of the ice sheet
∆x thickness of the ice sheet

Assuming constant thickness of the ice layer, the amount of heat flowing through the ice can be written as:
∆Q/∆t = k∙A∙∆T∙/∆x
=>
∆Q = k∙A∙∆T∙∆t/∆x

On the other hand the loss of heat causes some water to freeze.
This is amount is given by
∆Q = m_ice ∙ ∆hf
m_ice mass of ice formed
∆hf is the enthalpy of fusion, for ice ∆hf = 333.55kJkg⁻¹

Since the mass equals density times volume:
m_ice = ρ_ice∙V_ice = ρ_ice∙A∙s
ρ_ice = 916.7kgm⁻³
s is the amount by which the ice sheet thickens.

Hence:
∆Q = ρ_ice∙A∙s∙∆hf


Equate the two expression for ∆Q and solve for s
k∙A∙∆T∙∆t/∆x = ρ_ice∙A∙s∙∆hf

k∙∆T∙∆t/∆x = ρ_ice∙s∙∆hf

s = k∙∆T∙∆t/(∆x∙ρ_ice∙∆hf)
= 2.33Js⁻¹m⁻¹K⁻¹ *17.7K *3*60s / (0.294m *916.7kgm⁻³ *333550Jkg⁻¹)
= 8.25×10⁻⁵m = 0.082 mm