Question

Assuming the blade of an ice skate has a contact area of 1.20 cm2 with the ice, calculate the melting point depression of ice as a result of the additional pressure on the ice of an 90.0 kg skater. Remember the skater has two legs! The molar volumes for ice and water liquid around 0oC are 0.0196 L/mol and 0.0180 L/mol, respectively. The standard molar heat of fusion of ice is 6.01 kJ/mol. Use g = 9.807 m.s-2 for the acceleration of gravity.

Answer 1

m = 90 kg skater...

Calculate Pressure

P = F/A = W/A = m*g/A

A = 1.2 cm2 = 0.00012 m2

P = (90)(9.807)/(0.00012) = 7355250 Pa

but Pressure is divided in 2 legs so

P =7355250/2 = 3677625 Pa

then..

apply Clasius Clapeyron equation

dV = molar Volume differences = 0.0196 -0.0180 = 0.0016 L/mol = 0.0016 *10^-3 m3/mol

dP/dT = dHfus / ( T*dV)

when solving

P2-P1 = dH/dV*(lnT2-lnT1)

3677625 - 101325 = 6010/(0.0016 *10^-6) * (ln(T2) - ln(273.15))

lnT2 = (3677625 - 101325 /6010 * (0.0016 *10^-6) + ln(273.15)

T2 = 271.3 k

T2 in C = 271.3-273.15 = -1.85°C