Question

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, ΔHfus=6010J⋅mol−1, the density of ice is 920. kg⋅m−3, and the density of liquid water is 997 kg⋅m−3.

Part A: What pressure is required to lower the melting temperature by 2.72 ∘C? Answer is 397 bar

Part B: Assume that the width of the skate in contact with the ice has been reduced by sharpening to 17.3×10−3cm, and that the length of the contact area is 17.6 cm. If a skater of mass 74 kg 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?Express your answer in degrees Celsius to two significant figures.

Answer 1

calculate the molar volume of ice

V_ice = mass / density of ice

use the theoretical molar mass of water

Vice = 18.02 x 10^-3 / 920 (kg/m3)

V ice = 1.96 x 10^-5 m³ / mol

now calculate the real molar volume of the liquid water

V liquid = 18.02 x 10^-3 / 997 (kg/m3) = 1.81 x 10^-5 m³ / mol

Now calculate the change in volume

Delta V = V ice - V liquid =1.96 x 10^-5 m³ / mol - 1.81 x 10^-5 m³ / mol = 1.51 x10^-6 m³ / mol

Now apply the clausius clapeyron equation

dP / dT = Delta H_fus / Delta V_m

for our case

Pf - Pi = (Delta H_fus / Delta V_m ) * (Delta T / Ti)

Pf - Pi = (6010 /1.51 x10^-6) * (2 / 273.15) = 2.91 x 10⁷ J / m³

Pf = 101 325 Pa + 3.96 x 10⁷ J / m³

Pf = 3.96 x 10⁷ Pa

Pf = 396.6 Bar

B)

Apply the next equation

P = mg / A

where m is mass, g is gravity and A is area

mass is 74 Kg

width of the skate is 17.3 x 10^-5 m and the lenght is 0.176 m

Area = 17.3 x 10^-5 m * 0.176 m = 3.04 x 10^-5 m²

P = 74 * 9.8 / 3.04 x 10^-5 = 2.38 x 10⁷ Pascals

P = 2.38 Bar

c) Take the clapeyrus equation and rearrange it to get:

Delta T = (Pf - Pi) Delta V * T / Delta H fus

Delta T = (2.38 x 10⁷ Pa - 3.96 x 10⁷ Pa) * 1.51 x10^-6 m³ / mol * 273.15 / 6010

Delta T = -1.09 K

T = Ti + Delta T

T = 273.15 - 1.09 = 272.06 K

Temperature is -1.09 C or 272.06 K