Question

A 0.313-m-thick sheet of ice covers a lake. The air temperature at the ice surface is -15.9 °C. In 4.70 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 0.313 m. Calculate the number of millimeters by which the ice thickens. Do not enter unit.

Answer 1

here,

The liquid water directly beneath the ice sheet is equilibrium with the solid and has therefore a temperature 0degree 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* deltaT/delta x

k thermal conductivity, for ice k = 2.33 Wm⁻¹K⁻¹
A surface area of the lake
delta T temperature difference between the two sides of the ice sheet
delta 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:
delta Q/delta t = k*A*deltaT/deltax
  
delta Q = k *deltaA*deltaT*deltat/ deltax

On the other hand the loss of heat causes some water to freeze.
This is amount is given by
delta Q = m_ice ∙ delta 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 = p_ice*V_ice = p_ice*A*s
p_ice = 916.7kgm⁻³
s is the amount by which the ice sheet thickens.

Hence:
delta Q = p_ice*A*s*deltahf

Equate the two expression for delta Q and solve for s
k*A*deltaT*deltat/deltax = p_ice*A*s*deltahf

k*deltaT*deltat/deltax = p_ice*s*deltahf

s = k*deltaT*deltat/(deltax*p_ice*deltahf)

s = 2.33 * 15.9 * 4.7* 60 / (0.313* 916.7 *333550)

s = 1.09 * 10^-4 m

s = 0.109 mm

the number of milimeters are 0.109 mm