It has been suggested that the surface melting of ice plays a role in enabling speed skaters to achieve peak performance. Carry out the following calculation to test this hypothesis. At 1 atm pressure, ice melts at 273.15 K (delta)Hfusion= 6010J/mol, the density of ice is 920 kg/m3, and the density of liquid water is 997 kg/m3.

Part A:
What pressure is required to lower the melting temperature by 4.7 degrees Celsius?
Part B:
Assume that the width of the skate in contact with the ice has been reduced by sharpening to 0.019 cm, and that the length of the contact area is 18 cm. If a skater of mass 79kg is balanced on one skate, what pressure is exerted at the interface of the skate and the ice?
Part C:
What is the melting point of ice under this pressure?

a) Pf = 689.4 bar

b) P = 226.6 bar

c) T = 269.99 K

Explanation:

a)

The molar volume of ice is equal to:

Vi = m/p = (18.02x10^-3 kg H2O/1 mol H2O)*(1 m^3/920 kg) = 1.96x10^-5 m^3 mol^-1

The molar volume of liquid water is equal to:

Vl = (18.02x10^-3 kg/1 mol)*(1 m^3/997 kg) = 1.8x10^-5 m^3 mol^-1

The change in volume is equal to:

ΔVchange = Vi-Vl = 1.96x10^-5 - 1.8x10^-5 = 1.5x10^-6 m^3 mol^-1

using the Clapeyron equation:

Pf = Pi + ((ΔHf*ΔT)/(ΔVf*Ti)) = 1.013x10^5 Pa + ((6010 J mol^-1 * 4.7 K)/(1.5x10^-6 * 273.15 K)) = 6.89x10^7 Pa = 689.4 bar

b)

For the pressure we will use the equation:

P = (m*g)/A, where m is the mass, g is the acceleration of gravity and A is the area. Replacing values:

P = (79 kg * 9.81 m s^-2)/(1.9x10^-4 m * 0.18 m) = 2.26x10^7 Pa = 226.6 bar

c)

From Clapeyron´s expression we need to clear ΔT:

ΔT = ((Pf-Pi)*ΔV*Ti)/ΔHf = ((2.26x10^7 - 6.89x10^7)*1.5x10^-6*273.15)/6010 = -3.16 K

you can evaluate the new melting point of ice:

T = Ti + ΔT = 273.15 K - 3.16 K = 269.99 K

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