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, ΔHfusion= 6010 Jmol−1, the density of ice is 920 kgm−3, and the density of liquid water is 997 kgm−3.

A) What pressure is required to lower the melting temperature by 5.0 ∘C?

B) Assume that the width of the skate in contact with the ice has been reduced by sharpening to 19×10−3cm, and that the length of the contact area is 18 cm. If a skater of mass 75 kg is balanced on one skate, what pressure is exerted at the interface of the skate and the ice?

C) What is the melting point of ice under this pressure?

Answer 1

A) Using clapyron equation

=  

where, = Enthalpy of fusion = 6010 Jmol^-1

= change in temperature = T_initial - T_final = 5⁰c = 5K

= change in pressure

V_l = volume of liquid water per mole = 18*10^-3/997 m³/mol

V_s = volume of ice per mole = 18*10^-3/920  m³/mol

now,

=

or, = *

or, = - 5* 6010/273.15*(1.80 - 1.956)*10^-5   

or, = - 5*6010*10⁵/42.614

or, = - 705.210 *10⁵ pa = - 694.104 atm [ 1 atm = 1.013 *10⁵ pa]

now, = P_initial - p_final = - 694.104 atm

or, P_final = p_initial + 694.104 = 1+ 694.104 = 695.104 atm

Hence at 695.104 atm the melting point will be lower by 5⁰c

B)

contact area of the skate (A) = 19*10^-3*18 = 0.342 cm² = 3.42*10^-5 m²

now pressure exerted = Force/Area = mg/ A [ g is gravitational constant = 9.8 m/s²]

= 75*9.8/3.42*10^-5 = 214.91*10⁵ pa = 212 atm

c) now from part a

= *  

or, = * [6010/273.15* 0.156*10^-5]

or, =   * [273.15*0.156*10^-5/6010] = * 7.09*10^-8

now , = (1.013 - 214.91) *10⁵ pa

then , = - 213.897 *10⁵* 7.09*10^-8 = -1.51 K

hence , melting point at that pressure = 273.15 - 1.51 = 271.64 K.