Question

A \(74.6-\mathrm{g}\) ice cube floats in the Arctic Sea. The pressure and temperature of the system and surroundings are at 1 atm and \(0^{\circ} \mathrm{C},\) respectively. Calculate \(\Delta S_{\mathrm{sys}}, \Delta S_{\mathrm{surr}}\) and \(\Delta S_{\text {univ }}\) for the melting of the ice cube. What can you conclude about the nature of the process from the value of \(\Delta S_{\text {univ }}\) ? (The molar heat of fusion of water is \(6.01 \mathrm{~kJ} / \mathrm{mol} .\) )

Short answer

ΔS_univ = 0 J/K, indicating a reversible process.

Step-by-step solution

Calculate the number of moles of ice

To find the moles of ice, divide the mass of the ice cube by the molar mass of water. The molar mass of water is approximately \(18.02 \text{ g/mol}\). \[ n = \frac{74.6 \text{ g}}{18.02 \text{ g/mol}} \approx 4.14 \text{ mol} \]

Determine the entropy change for the system (ΔS_sys)

Use the formula for entropy change during a phase transition: \[ \Delta S_\mathrm{sys} = \frac{n \times \Delta H_\mathrm{fusion}}{T} \] where \( \Delta H_\mathrm{fusion} = 6.01 \text{ kJ/mol} = 6010 \text{ J/mol} \) and \(T = 273 \text{ K} (0^{\circ} \text{C})\). \[ \Delta S_\mathrm{sys} = \frac{4.14 \text{ mol} \times 6010 \text{ J/mol}}{273 \text{ K}} \approx 91.0 \text{ J/K} \]

Calculate the entropy change for the surroundings (ΔS_surr)

The entropy change for the surroundings is given by:\[ \Delta S_\mathrm{surr} = - \frac{q_\mathrm{sys}}{T} \] Since the process is at constant pressure, \(q_\mathrm{sys} = n \times \Delta H_\mathrm{fusion} = 4.14 \text{ mol} \times 6010 \text{ J/mol}\).\[ \Delta S_\mathrm{surr} = -\frac{4.14 \text{ mol} \times 6010 \text{ J/mol}}{273 \text{ K}} = -91.0 \text{ J/K} \]

Calculate the total entropy change for the universe (ΔS_univ)

The total change in entropy of the universe is given by:\[ \Delta S_\mathrm{univ} = \Delta S_\mathrm{sys} + \Delta S_\mathrm{surr} \]Substitute the calculated values:\[ \Delta S_\mathrm{univ} = 91.0 \text{ J/K} + (-91.0 \text{ J/K}) = 0 \text{ J/K} \]

Interpret the total change in entropy (ΔS_univ)

A total entropy change \( \Delta S_\mathrm{univ} = 0 \text{ J/K} \) indicates a reversible process. The system and the surroundings are in equilibrium, showing that the melting of the ice cube is a reversible process when it occurs at \(0^{\circ} \text{C}\) and 1 atm.

Key concepts

Entropy Change for System

When an ice cube melts at 0°C and 1 atm of pressure, an entropy change occurs within the system — the ice turning into water. Entropy (ΔS_{ ext{sys}}), in simple terms, is a measure of disorder or randomness. During a phase transition like melting, molecules transition from a more ordered solid state to a less ordered liquid state. This change in molecular arrangement results in an increase in entropy.

For the given problem, we calculate the entropy change using the formula:

• \[\Delta S_\mathrm{sys} = \frac{n \times \Delta H_\mathrm{fusion}}{T}\],

where:

• \( n = \text{number of moles} \)
• \( \Delta H_\mathrm{fusion} = \text{molar heat of fusion} \)
• \( T = \text{temperature in Kelvin} \)

In this scenario, the increase in entropy reflects the transformation from structured ice to more chaotic liquid water.

Entropy Change for Surroundings

In a phase transition, while the system experiences a change, the surroundings also respond. When the ice melts, it absorbs heat from its surroundings, causing the entropy of the surroundings (ΔS_{ ext{sur}}) to decrease. This is because the heat lost by the surroundings decreases the environmental randomness.

The entropy change of the surroundings is calculated by:

• \[\Delta S_\mathrm{surr} = - \frac{q_\mathrm{sys}}{T}\]

Here, the minus sign indicates the surroundings lose energy. Since the system absorbs energy equal to \(q_\mathrm{sys}\), it leads to a decrease in their entropy. Therefore, while the ice cube melts and system entropy increases, the entropy of surrounding elements decreases.

Phase Transition

Phase transitions, such as melting and freezing, are vital processes that can shift a substance from one physical state to another. In the case of an ice cube turning into water, it undergoes a melting phase transition. This process occurs when the temperature and pressure conditions are right, allowing molecules in a solid state to gain enough energy to overcome their ordered structure and move into a liquid state.

During melting, or fusion, the supplied heat energy helps break the rigid bonds in ice, aiding in its conversion to liquid water. These transitions are typically evaluated at constant temperature, ensuring that energy primarily contributes to changing the molecular structure, not increasing kinetic energy.

The characteristic feature of phase transitions is the entropy change due to the molecular structural modification. These changes, despite having specific heat needs, maintain overall energy and entropy balance when considered with the surroundings.

Reversible Process

A reversible process is essentially an idealization in thermodynamics. It's a process that can occur without net change in either the system or surroundings. In our example, the melting of an ice cube at 0°C in 1 atm pressure is considered reversible because it happens at equilibrium conditions. This means both forward and reverse processes (melting and freezing) could occur, with no noticeable effect on the total entropy when combined.

To determine if a process is reversible, we assess the total entropy change of the universe (ΔS_{ ext{univ}}). If ΔS_{ ext{univ}} = 0, the process is perfectly reversible. For example:

• \[\Delta S_\mathrm{univ} = \Delta S_\mathrm{sys} + \Delta S_\mathrm{surr} \]

In this balanced state, neither system nor surroundings gain entropy, enabling the process to reverse seamlessly. Such balanced state scenarios are theoretical benchmarks since actual processes often involve some irreversibility. Nonetheless, they are useful for calculating and understanding entropy.