Thermal Physics: Ice skating temperatures

[H.fulls]

Homework Statement

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)

Homework Equations

$\frac{dp}{dT}$ = $\frac{L}{T(V_{2}-V_{1})}$

or I think rearranged like this

$p_{0}$-p = $\frac{L}{\DeltaV}$ ln$\frac{T_{0}}{T}$

The Attempt at a Solution

I have found the pressure exerted as 1.372 x $10^{7}$ kg/$m^{2}$
I realize 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 don't know what to use for the volume?

 

[TSny]

$\frac{dp}{dT}$ = $\frac{L}{T(V_{2}-V_{1})}$

or I think rearranged like this

$p_{0}$-p = $\frac{L}{\Delta V}$ ln$\frac{T_{0}}{T}$

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

$\frac{\Delta p}{\Delta T}$ ≈ $\frac{L}{T(V_{2}-V_{1})}$

I have found the pressure exerted as 1.372 x $10^{7}$ kg/$m^{2}$

This looks correct except for how you expressed the units. Pressure is force per unit area, not mass per unit area.

However I don't know what to use for the volume?

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).

 

[AlfieLord]

... and the answer comes out around -1.0 degree ??

 

[TSny]

Yes..which debunks the idea that ice skates slide well on ice due to pressure causing the ice to melt and become slippery.

See http://www.nytimes.com/2006/02/21/science/21ice.html?pagewanted=all

 

[AlfieLord]

Agreed ... it is far more complex.