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 the density of ice is 920 kg m-3 , and the density of liquid water is 997 kg m-3. What pressure is required to lower the melting temperature by 4 C?

Answer 1

Step-by-step explanation:

Relation between pressure, latent heat of fusion, and change in volume is as follows.

`(dP)/(dT) = (L)/(T * \Delta V)`

Also,
`(L)/(T) = \Delta S^(fusion)_(m)`

where,
`\Delta V^(fusion)_(m)` is the difference in specific volumes.

Hence,
`(dP)/(dT) = (\Delta S^(fusion)_(m))/(\Delta V^(fusion)_(m))`

As,
`\Delta S^(fusion)_(m) = (L)/(T) = (6010)/(273.15)` = 22.0 J/mol K

And,
`\Delta V^(fusion)_(m) = \frac{M}{d_{H_(2)O}} - (M)/(d_(ice))` ...... (1)

where,
`d_{H_(2)O}` = density of water

`d_(ice)` = density of ice

M = molar mass of water =
`18.02 * 10^(-3) kg`

Therefore, using formula in equation (1) we will calculate the volume of fusion as follows.

`\Delta V^(fusion)_(m) = \frac{M}{d_{H_(2)O}} - (M)/(d_(ice))`

=
`(18.02 * 10^(-3))/(997) - (18.02 * 10^(-3))/(920)`

=
`-1.51 * 10^(-6)`

Therefore, calculate the required pressure as follows.

`(dP)/(dT) = (22)/(-1.51 * 10^(-6))`

=
`1.45 * 10^(7) Pa/K`

or, = 145 bar/K

Hence, for change of 1 degree pressure the decrease is 145 bar and for 4.7 degree change dP =
`145 * 4.7 bar`

= 681.5 bar

Thus, we can conclude that pressure should be increased by 681.5 bar to cause 4.7 degree change in melting point.