Problem 27 - University Physics with Modern Physics

Question

20.27 A 15.0-kg block of ice at \(0.0^{\circ} \mathrm{C}\) melts to liquid water at \(0.0^{\circ} \mathrm{C}\) inside a large room that has a temperature of \(20.0^{\circ} \mathrm{C}\) . Treat the ice and the room as an isolated system, and assume that the room is large enough for its temperature change to be ignored. (a) Is the melting of the ice reversible or irreversible? Explain, using simple physical reasoning without resorting to any equations. (b) Calculate the net entropy change of the system during this process. Explain whether or not this result is consistent with your answer to part (a).

Step-by-Step Solution

Answer

(a) Irreversible; (b) Net entropy change is +1261.95 J/K, confirming irreversibility.

Step 1: Understanding Reversibility

(a) For the process to be reversible, the transfer of heat between the ice and the room must occur at temperatures that are infinitesimally close to each other. Since the ice melts at \(0^{\circ} \mathrm{C}\) and the room is at a much higher temperature of \(20^{\circ} \mathrm{C}\), there is a significant temperature difference. Therefore, the melting process occurs rapidly, making it irreversible.

Step 2: Calculating Entropy Change of the Ice

The entropy change for the ice can be calculated using the formula \(\Delta S = \frac{Q}{T}\), where \(Q\) is the heat absorbed and \(T\) is the temperature. For melting, \(Q = mL_f\) where \(m = 15.0\, \text{kg}\) and \(L_f = 334\, \text{J/g}\). Convert to Joules and use \(T = 273\, \text{K}\): \[Q = 15,000\, \text{g} \times 334\, \text{J/g} = 5,010,000\, \text{J}\]Calculate \(\Delta S_{\text{ice}} = \frac{5,010,000}{273} = 18,357.51\, \text{J/K}\).

Step 3: Calculating Entropy Change of the Room

The entropy change for the room is negative, as it loses the same amount of energy the ice gains. Use the formula \(\Delta S = \frac{Q}{T}\), where \(T=293\, \text{K}\) for the room:\[\Delta S_{\text{room}} = -\frac{5,010,000}{293} = -17,095.56\, \text{J/K}\].

Step 4: Finding Net Entropy Change

The net entropy change of the system is the sum of the entropy changes of the ice and the room:\[\Delta S_{\text{net}} = 18,357.51 + (-17,095.56) = 1,261.95\, \text{J/K}\].

Step 5: Consistency with Irreversibility

(b) The net entropy change is positive, \(\Delta S_{\text{net}} = 1,261.95\, \text{J/K}\). This positive change is consistent with the second law of thermodynamics, confirming that the process is irreversible, as entropy has increased.

Key Concepts

Reversible and Irreversible Processes

In thermodynamics, a process can be classified as either reversible or irreversible based on how the change occurs.
For a process to be reversible, it must happen infinitely slowly, allowing the system to be in equilibrium at each stage. There should be no loss of energy, such as heat, due to friction or other factors during the process.
In practical terms, a reversible process is an idealization, often not achievable in real scenarios because any finite temperature difference causes irreversible heat transfer.

• A classic example of a reversible process is the slow compression or expansion of a gas within a piston that allows the gas to maintain thermal equilibrium with its surroundings.
• An irreversible process, like the melting of ice in a warm room, occurs naturally without the opportunity to return to the original state without additional changes in external conditions.

In the given exercise, the rapid melting of the ice block due to a significant temperature difference from the surrounding room temperature makes it an irreversible process. The heat transfer is not slow enough for the process to be reversible, resulting in a net entropy change, consistent with real-world occurrences.

Second Law of Thermodynamics

The second law of thermodynamics is fundamental to understanding entropy and the behavior of physical systems.
It states that the total entropy of an isolated system can never decrease over time. In essence, natural processes tend to evolve towards a state of maximum entropy or disorder.
This law describes why certain processes are irreversible and dictates the direction of time in thermodynamic processes—from low entropy to high entropy.

• Mathematically, for any irreversible process, the entropy change delta S is always greater than zero for a closed system.
• In reversible processes, the change in entropy, delta S, would be zero, denoting no net change in disorder.

In the context of the ice melting in a warm room, the second law is affirmed by the positive net change in entropy calculated in the exercise. This indicates the irreversibility of the process, as entropy increases systematically over time.

Entropy Calculation

Understanding the calculation of entropy change is key to mastering thermodynamic processes. Entropy, denoted as S, quantifies the amount of energy in a system unavailable to do work and is generally associated with disorder.
When calculating the change in entropy, deltaS, consider both the system gaining energy and the system losing energy.
The formula used for entropy change is given by: \[ \Delta S = \frac{Q}{T} \]where Q is the amount of heat transferred and T is the temperature at which the process takes place in Kelvin.

• In the exercise, the melting ice (system gaining energy) results in positive entropy change since energy is absorbed to convert solid ice to liquid water.
• The room (system losing energy) yields a negative entropy change, as it transfers heat to the ice.

To find the net entropy change, sum the entropy changes of the ice and the room, as shown in the exercise. The result reveals a positive net entropy, aligning with the fundamental principles of the second law, signifying an overall increase in disorder.