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Thermal Physics: Ice skating temperatures

  1. Nov 23, 2012 #1
    1. The problem statement, all variables and given/known data
    It is said that good ice skating only occurs when the ice below the skates melts. Using the Clausius-Clapeyron equation, estimate the coldest temperature at which good ice skating can occur. (Water expands 9% on freezing, Latent heat of ice melting is 334 kJ/Kg, the contact area is 1mm by 5cm and the skater weighs 70kg, water has a molar mass of 18g)


    2. Relevant equations
    [itex]\frac{dp}{dT}[/itex] = [itex]\frac{L}{T(V_{2}-V_{1})}[/itex]

    or I think rearranged like this

    [itex]p_{0}[/itex]-p = [itex]\frac{L}{\DeltaV}[/itex] ln[itex]\frac{T_{0}}{T}[/itex]


    3. The attempt at a solution
    I have found the pressure exerted as 1.372 x [itex]10^{7}[/itex] kg/[itex]m^{2}[/itex]
    I realise that we want the ice to be melting.. so 273k at this pressure, so I need to find the temperature at normal room pressure of 101 KPa.
    However I dont know what to use for the volume?
     
  2. jcsd
  3. Nov 23, 2012 #2

    TSny

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    Often, you can get an accurate enough answer without integrating by treating the right hand side of the Claussius-Clapeyron equation as constant over the temperature changes involved. Then you can just write it as

    [itex]\frac{\Delta p}{\Delta T}[/itex] ≈ [itex]\frac{L}{T(V_{2}-V_{1})}[/itex]
    This looks correct except for how you expressed the units. Pressure is force per unit area, not mass per unit area.
    The volumes ##V_1## and ##V_2## are "specific" volumes (i.e., volumes per kg of material). Note that water is odd in that the ##V_1## (for ice) is greater than ##V_2## (for liquid water).
     
  4. Nov 25, 2012 #3
    ... and the answer comes out around -1.0 degree ??
     
  5. Nov 25, 2012 #4

    TSny

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  6. Nov 25, 2012 #5
    Agreed ... it is far more complex.
     
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