Problem 32 - An Introduction to Thermal Physics

Question

The density of ice is \(917 \mathrm{kg} / \mathrm{m}^{3}.\) (a) Use the Clausius-Clapeyron relation to explain why the slope of the phase boundary between water and ice is negative. (b) How much pressure would you have to put on an ice cube to make it melt \(a t-1^{\circ} \mathrm{C} ?\) (c) Approximately how deep under a glacier would you have to be before the weight of the ice above gives the pressure you found in part (b)? (Note that the pressure can be greater at some locations, as where the glacier flows over a protruding rock.) (d) Make a rough estimate of the pressure under the blade of an ice skate, and calculate the melting temperature of ice at this pressure. Some authors have claimed that skaters glide with very little friction because the increased pressure under the blade melts the ice to create a thin layer of water. What do you think of this explanation?

Step-by-Step Solution

Answer

Negative slope due to decreased volume on melting. Calculate \(dP\) to find needed pressure.

Step 1: Understand Clausius-Clapeyron Relation

The Clausius-Clapeyron relation describes how the pressure of a phase boundary depends on temperature. For water and ice, it can be expressed as \(\frac{dP}{dT} = \frac{L}{T \Delta V}\), where \(L\) is the latent heat of fusion, \(T\) is the temperature, and \(\Delta V\) is the change in volume upon melting. Since ice melts into a denser liquid, \(\Delta V\) is negative, leading to a negative slope.

Step 2: Calculate Pressure Needed to Melt Ice at -1°C

Use the modified ice melting point with the relation \(dT = -1^{\circ}C\) and the known Clausius-Clapeyron relation \(\frac{dP}{dT} = \frac{L}{T \Delta V}\). Given \(L = 334,000 \text{ J/kg}\), \(T = 273 \text{ K}\), and \(\Delta V = \frac{1}{\rho_{water}} - \frac{1}{\rho_{ice}}\), calculate \(dP\) to find the pressure for \(T = -1^{\circ}C\).

Key Concepts

Phase Boundary

The phase boundary refers to the line or surface in a phase diagram where distinct phases, such as solid and liquid, coexist in equilibrium. For water and ice, this boundary is intriguing because it behaves uniquely under varying pressure and temperature conditions.
For instance, according to the Clausius-Clapeyron relation, the slope of the water-ice phase boundary is negative. This indicates that as pressure increases, the temperature at which ice melts decreases, unlike many other substances. Why? Because the melting of ice into liquid water results in a decrease in volume. Therefore, when the pressure increases, it encourages the melting process at lower temperatures rather than higher ones.
This negative slope is not just a mathematical outcome. It has physical implications that affect natural processes like glacier movement and gas-liquid systems in various environments.

Latent Heat

Latent heat is the amount of energy required for a substance to change its phase, such as from a solid to a liquid, without changing its temperature. For example, the latent heat of fusion (or equivalent) of ice turning into water is significant. This energy is required to break the molecular bonds that hold the ice crystal structure together.
In the context of the Clausius-Clapeyron relation, latent heat ( \( L \) ) contributes to how the phase boundary is determined. The relation \( \frac{dP}{dT}= \frac{L}{T \Delta V} \) incorporates \( L \) to highlight how the pressure change \( dP \) is influenced significantly by this heat when transitioning between phases at equilibrium.
Molecules need enough energy to overcome the attractions holding them in place, allowing transitions like melting or vaporization, critical in thermodynamic systems.

Density of Ice

Density is a measure of mass per unit volume. Ice has a density of \( 917 \text{ kg/m}^3 \) , which is less than that of liquid water, which stands at around \( 1000 \text{ kg/m}^3 \). This peculiar property explains why ice floats in water, a fundamental fact that affects everything from climate patterns to natural habitats.
In terms of the Clausius-Clapeyron relationship, the density difference between ice and water affects the \( \Delta V \) in the equation \( \frac{1}{\rho_{water}} - \frac{1}{\rho_{ice}} \). Since \( \rho_{ice} \) is less than \( \rho_{water} \), \( \Delta V \) is negative, contributing to the negative slope of the phase boundary described earlier.
This concept illustrates the molecular arrangements in phases and their impact on equilibrium processes.

Pressure and Temperature Relationship

Pressure and temperature are interrelated with phase changes, especially in the realm of thermodynamics. The Clausius-Clapeyron equation helps describe this relationship, where a change in one will affect the other during a phase transition.
For ice and water, an increase in pressure can decrease the melting temperature. This happens because increased pressure "squeezes" the ice, promoting it to turn into the denser liquid state, so less heat (or higher temperature) is needed for melting.
When considering everyday occurrences like glaciers or ice skating, this relationship explains phenomena such as pressure-induced melting. Under high-pressure conditions, like under a glacier's weight or an ice skate's blade, ice melting happens quicker, illustrating the practical implications of this thermodynamic principle.