How Long Does It Take for Ice Thickness to Double in Cold Conditions?

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SUMMARY

The discussion focuses on calculating the time required for the thickness of ice on a pond to double from 4.50 cm to 9.00 cm under constant air temperature conditions of -11.0°C. The heat conduction equation, dQ/dt = kA delta T/x, is utilized alongside the relationship dQ = LpA dx to derive the time. After integrating the equations and substituting the latent heat of fusion (L = 3.36 x 10^5 J/m³), the calculated time is 42.5 seconds, although some participants express skepticism about the validity of this result.

PREREQUISITES
  • Understanding of the heat conduction equation and its components
  • Knowledge of latent heat of fusion and its application in phase changes
  • Basic calculus for integration and solving differential equations
  • Familiarity with physical properties of ice, including density and thermal conductivity
NEXT STEPS
  • Research the application of the heat conduction equation in different materials
  • Learn about the effects of temperature gradients on ice formation
  • Explore the concept of latent heat in various phase transitions
  • Investigate real-world examples of ice thickness measurement and prediction
USEFUL FOR

Students studying physics, particularly those interested in thermodynamics and heat transfer, as well as researchers and professionals involved in environmental science and ice dynamics.

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A pond of water at 0°C is covered with a layer of ice 4.50 cm thick. If the air temperature stays constant at -11.0°C, how much time does it take for the thickness of the ice to increase to 9.00 cm?

Hint: To solve this problem, use the heat conduction equation,

dQ/dt = kA delta T/x

and note that the incremental energy dQ extracted from the water through the thickness x is the amount required to freeze a thickness dx of ice. That is, dQ = LpA dx, where p is the density of the ice, A is the area, and L is the latent heat of fusion. (The specific gravity and thermal conductivity for ice are, respectively, 0.917 is 2.0 W/m/°C.)






I don't have much of an idea on how to attempt this question, all I've got so far is.

dQ = LpA dx
so LpA dx/dt = kA delta T/x
x/dt = L delta T/ L p dx

I guess that's useful as it gets rid of surface area in the equation ( which isn't given), but I am not sure where to go from there. Also, delta T would be zero, and so the entire equation would equal zero, which doesn't make much sense to me.

By the way I am 16 and so presume that I am very ignorant.
 
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why should dT be 0? Outside temp is -11 while water is at 0.

dQ=KA.(T1-T2)/x .dt where T1-T2=11
dQ=mL=dx.A.P.L
so we have from above eqns.
dx.A.P.L=KA.11/x .dt
xdx.P.L=K.11 dt
integrate LHS from 0.045 to 0.09 and RHS from 0 to t, where t is the required time.
P.L.(x^2)/2=22t
substituing the values (L=3.36 x 10^5)& solving i get t=42.5 sec

IMO this is too small a value, anyway, do tell the answer :)
 

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