Question

Does pressure-induced melting explain ice skating? Suppose an ice skater weighs \(m=70 \mathrm{~kg}\). An ice skate exerts a pressure \(\Delta p=m g / A\) on ice, where \(A=1.0 \mathrm{~cm}^{2}\) is the surface area of a skate and \(g\) is the gravitational acceleration constant. (a) Compute the decrease in freezing temperature of water caused by this applied pressure. (b) Is this pressure sufficient to melt ice?

Short answer

Pressure alone decreases the freezing temperature by about 0.056 K, which is insufficient to melt the ice during ice skating.

Step-by-step solution

Determine the pressure exerted

Calculate the pressure \(\Delta p\) exerted by the ice skater using the formula: \[\Delta p = \frac{m g}{A}\]Given: - mass \(m = 70 \, \mathrm{kg}\)- gravitational acceleration \(g = 9.8 \, \mathrm{m/s^2}\)- area \(A = 1.0 \, \mathrm{cm^2} = 1.0 \times 10^{-4} \, \mathrm{m^2}\)\( \Delta p = \frac{70 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2}}{1.0 \times 10^{-4} \, \mathrm{m^2}} = 6.86 \times 10^6 \, \mathrm{Pa}\)

Calculate the decrease in freezing temperature

Using the Clausius-Clapeyron relation: \[ \Delta T_f = \frac{T_f \Delta p}{L \rho} \]where:- Initial freezing point of water \(T_f = 273.15 \, \mathrm{K}\)- Latent heat of fusion for water \(L = 3.34 \times 10^5 \, \mathrm{J/kg} \)- Density of water \(\rho = 1000 \, \mathrm{kg/m^3}\)Calculate the decrease in freezing temperature:\[ \Delta T_f = \frac{273.15 \mathrm{K} \times 6.86 \times 10^6 \mathrm{Pa}}{3.34 \times 10^5 \, \mathrm{J/kg} \times 1000 \, \mathrm{kg/m^3}} \approx 0.056 \mathrm{K} \]

Evaluate pressure sufficiency to melt ice

The decrease in freezing temperature computed is approximately \(0.056 \mathrm{K}\). Since typical ice skating conditions are below \(273.15 \mathrm{K}\), this slight decrease is not sufficient to melt ice. Therefore, the pressure alone does not cause the ice to melt.

Key concepts

pressure on ice

An ice skater exerts pressure on the ice due to their weight being concentrated on the small surface area of the skate blades. This pressure is calculated using the formula: \[ \Delta p = \frac{m g}{A} \] where \(\frac{m g}{A}\) represents the force applied by the skater divided by the area of the skate blade contacting the ice. In the provided exercise, the weight of the skater is 70 kg, and the surface area of the skate blade is 1.0 cm² (or 1.0 × 10^{-4} m²). Using the acceleration due to gravity (9.8 m/s²), the exerted pressure is determined to be \(6.86 \times 10^6 \mathrm{Pa}\).
This pressure is significant and impacts the ice beneath the skates.

Clausius-Clapeyron relation

The Clausius-Clapeyron relation describes the change in the melting (or boiling) temperature of a substance as a function of pressure. The equation is given by: \[ \Delta T_f = \frac{T_f \Delta p}{L \rho} \] Here, \( \Delta T_f \) is the change in freezing temperature, \( T_f \) is the initial freezing point, \( \Delta p \) is the pressure change, \( L \) is the latent heat of fusion, and \( \rho \) is the density of water.
In the exercise, these values are: \( T_f = 273.15 \mathrm{K}, \Delta p = 6.86 \times 10^6 \mathrm{Pa}, L = 3.34 \times 10^5 \mathrm{J/kg}, \rho = 1000 \mathrm{kg/m^3}\).
Plugging these into the formula calculates the decrease in freezing temperature: \[ \Delta T_f = \frac{273.15 \mathrm{K} \times 6.86 \times 10^6 \mathrm{Pa}}{3.34 \times 10^5 \mathrm{J/kg} \times 1000 \mathrm{kg/m^3}} \approx 0.056 \mathrm{K} \] This shows that the freezing point of water only decreases by a small amount under pressure from the skater.

phase transitions

Phase transitions refer to the changes from one state of matter to another, such as from solid to liquid. When pressure is applied to ice, the Clausius-Clapeyron relation helps understand how the freezing point changes. In this scenario, as the ice skater applies pressure, the freezing point of the ice decreases slightly.
However, the small decrease of approximately 0.056 K is not enough to melt the ice completely. Ice melting requires a significant decrease in the freezing point, especially since the conditions are already below 0°C (273.15 K).
Thus, although pressure can induce a phase transition from ice to water, the pressure exerted by an ice skater alone is insufficient to cause noticeable melting. This means that other factors, like friction and heat from the environment, also play crucial roles.
Understanding pressure-induced melting can be expanded to various applications, such as in glaciers and high-pressure physics.