Question

You may have heard that ice skaters actually skate on a thin layer of water rather than solid ice (which reduces friction on the blade and allows them to skate faster). Let’s work through a calculation to determine if this statement is reasonable. Estimate the pressure exerted by a 200-lb hockey player, standing on two blades that are 0.1 mm  20 cm. Calculate the melting point of ice below the player, assuming that the density of ice under these conditions is approximately 0.915 g cm^–3 and that of liquid water is 0.998 g cm^–3. Assuming a typical temperature of ice in a skating rink is 27 °F, does a hockey player skate on ice or water?

Answer 1

The thing which you heard is true. The ice skaters actually skate on thin layer of water rather than a solid ice.

The pressure exerted by person standing is given as below.

Pressure = Force / Area

Force = m.g

m = mass & g = gravity

1 pounds = 0.4535 kgs

200 pounds = 90.7 kgs

g = 9.81 m/s2

Force = mg = 90.7 x 9.81 = 889.77 N

As per the information in the question, the hockey player is standing on two blades that are 0.1 mm & 20 cm. Normally the length of the skate blades are 50cm. So we will calculate the area of the blades in contact with the Ice. Assuming that the blades are in rectangular shape. So area will given by length & width.

Area of blade 1 (0.1mm) = 50 cm x 0.01 cm = 0.5 cm2 = 5 x 10^-5 m2

Area of blade 2 (20cm) = 50 cm x 20 cm = 1000 cm2 = 0.1 m2

Pressure 1 (0.1 mm blade) = 889.77 / (5 x 10^-5) = 177.75 x 10^5 N/m2 = 17.775 MPa

Pressure 2 (20 cm blade) = 889.77 / 0.1 = 8897.7 N/m2 = 0.008898 MPa

Thus, we can clearly see that Pressure 1 (exerted by 0.1 mm blade) is much higher compared to that of 20cm blade.

Now, to find out if the hockey player will skate on Ice or water, we will being using the Clausius Clapeyron Equation.

This equation is used to understand the discontinuous phase transition between two phases of matter for a single constituent.

dP / dT = L / (T dv)

where L = specific latent heat; in this case its latent heat of fusion = 3.34 x 10^5 J/kg

T = the temperature in K = 273 K

dv = specific volume change of the phase transition; in this case from solid to liquid = -9.05 x 10^-5 m3 / Kg

dP / dT = (3.34 x 10^5) / (273 x -9.05 x 10^-5) = -13518709 J/m3 K = -13.5 x 10^6 Pa / K = -13.5 MPa / K

So looking at this value, Pressure 1 (exerted by 0.1 mm blade) is higher than that required to melt the ice. So blade 1 (0.1mm) will be skating on water & blade 2 (20cm) will be skating on ice.

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