# Homework Help: Thermal Physics: Ice skating temperatures

1. Nov 23, 2012

### H.fulls

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

or I think rearranged like this

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

3. The attempt at a solution
I have found the pressure exerted as 1.372 x $10^{7}$ kg/$m^{2}$
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. Nov 23, 2012

### TSny

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

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

3. Nov 25, 2012

### AlfieLord

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

4. Nov 25, 2012

### TSny

5. Nov 25, 2012

### AlfieLord

Agreed ... it is far more complex.