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

Explanation:

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

          $\frac{dP}{dT} = \frac{L}{T \times \Delta V}$

Also, $\frac{L}{T} = \Delta S^{fusion}_{m}$

where, $\Delta V^{fusion}_{m}$ is the difference in specific volumes.

Hence,    $\frac{dP}{dT} = \frac{\Delta S^{fusion}_{m}}{\Delta V^{fusion}_{m}}$

As, $\Delta S^{fusion}_{m} = \frac{L}{T} = \frac{6010}{273.15}$ = 22.0 J/mol K

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

where,    $d_{H_{2}O}$ = density of water

              $d_{ice}$ = density of ice

             M = molar mass of water = $18.02 \times 10^{-3} kg$

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

        $\Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}$

                       = $\frac{18.02 \times 10^{-3}}{997} - \frac{18.02 \times 10^{-3}}{920}$  

                       = $-1.51 \times 10^{-6}$        

Therefore, calculate the required pressure as follows.

              $\frac{dP}{dT} = \frac{22}{-1.51 \times 10^{-6}}$

                              = $1.45 \times 10^{7} Pa/K$

or,                           = 145 bar/K

Hence, for change of 1 degree pressure the decrease is 145 bar  and for 4.7 degree change dP = $145 \times 4.7 bar$

                              = 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.