Question

>> Thermodynamics <<

QUESTION )) Quantitatively explain why ice skates slide on ice for a 80 kg skater wearing 30 cm x 0.1 cm blades (consider one foot on the ice). Develop an equation for the change of the melting temperature of water ice as a function of pressure (Clapeyron or ClausiusClapyeron?). The heat of fusion for the ice-water phase transition is 335 kj/kg at 0°C and 1 bar. The density of water is 1 g/cm3 at these conditions and that of ice is 0.915 g/cm3 . Can it get too cold to ice skate? Does the prediction above seem to capture the whole story of ice skating? Would it be possible to skate on materials like solid CO2?

Answer 1

80 kg skater wearing 30 cm x 0.1 cm blades

means the pressure is 80 Kg/3 cm² = 26.67 Kg/cm² = 26.15 bar

SO the pressure applied on the ice is 26.15 bar so the melting point of ice decreases and it becomes water so it lets the person slide.

The clausius clapeyron equation is

ln p2-ln p1 = delta H /R(1/T1-1/T2)

335 kJ/kg is 335 kJ/1000/18 = 6030 J/mol

ln 26.15 - ln 1 = 6030/8.314 (1/273 - 1/T2)

3.26 - 0 = 725 (0.00366 - 1/T2)

0.00449 = 0.00366 - 1/T2

1/T2 = -8.36 x 10^-4

T2 = -1195 K SO the melting temperature is very low

So it can never be too cold to ice skate