Calculate Thickness of Ice Sheet on Lake: -17.8°C in 3.80 Minutes

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SUMMARY

The discussion focuses on calculating the thickness increase of a 0.315-meter ice sheet on a lake at an air temperature of -17.8 °C over a period of 3.80 minutes. Using the formula h = (k * ΔT * t) / (ρ * Lf * L), where k is the thermal conductivity of ice (2.2 J/s·m·°C), the density of ice (917 kg/m³), and the heat of fusion (Lf = 334 kJ/kg), the calculated thickness increase is approximately 0.0923 mm. This calculation assumes no heat flow from the ground and that the added ice is negligible compared to the existing thickness.

PREREQUISITES
  • Understanding of heat transfer principles, specifically conduction.
  • Familiarity with thermal conductivity and its units.
  • Knowledge of the heat of fusion for water.
  • Basic algebra for manipulating equations and units.
NEXT STEPS
  • Explore advanced heat transfer equations in different materials.
  • Learn about the effects of temperature gradients on ice formation.
  • Research the impact of environmental factors on ice thickness in lakes.
  • Study the properties of ice, including density variations with temperature.
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Students in physics or engineering, environmental scientists studying lake ecosystems, and anyone interested in the thermal properties of ice and its formation processes.

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Homework Statement


A 0.315-m-thick sheet of ice covers a lake. The air temperature at the ice surface is -17.8 °C. In 3.80 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 0.315 m. Calculate the number of millimeters by which the ice thickens.

Homework Equations


Q = k A ΔT / d
dQ/dt = k∙A∙∆T/∆x
The heat of fusion of water is about 334 kJ / kg
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

The Attempt at a Solution


I've tried a bunch of different equations, but I'm missing something. Is that equation the one to use?
 
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Never mind, I figured it out:
Density of ice=917 kg/m^3
Lf=33.5E4 J/kg
Thermal conductivity ice=2.2 J/smdegC

h=k(deltaT)t/pLfL
h=(2.2)(17.8)(228 secs)/(917)(33.5E4)(0.315)
h=9.2268E-5 m
Multiply by 1000 to get mm
0.0923 mm
 

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