Question

A mixture of \(1.78 \mathrm{~kg}\) of water and \(262 \mathrm{~g}\) of ice at \(0{ }^{\circ} \mathrm{C}\) is, in a reversible process, brought to a final equilibrium state where the water/ice ratio, by mass, is \(1: 1\) at \(0^{\circ} \mathrm{C} .(a)\) Calculate the entropy change of the system during this process. (b) The system is then returned to the first equilibrium state, but in an irreversible way (by using a Bunsen burner, for instance). Calculate the entropy change of the system during this process. (c) Show that your answer is consistent with the second law of thermodynamics.

Short answer

The entropy changes of the system for the reversible and irreversible processes are -204 J/K and +204 J/K respectively. The total entropy change for the irreversible process is positive, which is in agreement with the Second Law of Thermodynamics.

Step-by-step solution

Entropy change in a reversible process

Given that we have \(1.78 \mathrm{~kg}\) of water and \(0.262 \mathrm{~kg}\) of ice at \(0^{\circ} \mathrm{C}\). The final equilibrium demands that the ice/water ratio by mass to be \(1:1\) which means we need \(1.02 \mathrm{~kg}\) of water to freeze. The entropy change is given by \( \Delta S = \int \frac{dQ_{rev}}{T} \) where \(dQ_{rev}\) represents the infinitesimal heat absorbed in a reversible process, and T denotes the absolute temperature. In the case of freezing \(dQ = - mL\) where m denotes the mass and L is the latent heat of fusion. Hence, \( \Delta S = - \int 1.78 \mathrm{~kg} \rightarrow 0.78 \mathrm{~kg} \frac{mL}{T} \, dm = - \int 1.78 \mathrm{~kg} \rightarrow 0.78 \mathrm{~kg} \frac{dm}{T} \cdot L = L \ln \frac{0.78 \mathrm{~kg}}{1.78 \mathrm{~kg}} \)Inserting the values, where \(L = 334 \mathrm{~kJ/kg}\) and \(T = 273.15 \mathrm{~K}\), we get \( \Delta S = 334 \mathrm{~kJ/kg} \cdot \ln(0.78/1.78) / 273.15 \mathrm{~K} = - 204 \mathrm{~J/K} \)

Entropy change in an irreversible process

Since we are interested in the overall entropy change, we look for the total entropy change in the Universe. The total entropy change will be the entropy change of the system plus the entropy change of the surroundings.In this scenario, the entropy change for the system is easy to calculate, it is exactly the negation of the entropy change calculated in step 1, as we have reverted the system to the same state.So, \( \Delta S_{system} = -(-204 \mathrm{~J/K}) = 204 \mathrm{~J/K} \)Next, we consider the entropy change in the surroundings due to the addition of heat during the process. Here, we may assume that Bunsen burner heat will not affect the surrounding temperature significantly, which means the heat transfer can be assumed to be isothermal and can therefore be denoted as \( dQ_{irrev} = T dS \Rightarrow dS = \frac{dQ_{irrev}}{T} \Rightarrow\Delta S_{surroundings} = \frac{Q_{irrev}}{T} \).As the entropy change in the surroundings depends on the way we perform this irreversible process, we cannot determine this value directly. However, we know that this irreversible process must create entropy, which means \( \Delta S_{surroundings} > 0 \).

Consistency with the second law

The Second Law of Thermodynamics states that the total entropy of an isolated system can never decrease over time, and is constant if and only if all processes are reversible. If not, the total entropy increases. From Step 2, we have seen that the entropy change of the Universe is the sum of the entropy change of the system and the entropy change of the surroundings, and both are positive. Thus,\( \Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings} > 0 \). This is consistent with the second law of thermodynamics.

Key concepts

Second Law of Thermodynamics

The second law of thermodynamics is one of the fundamental principles of the universe that has significant implications in various scientific fields, including physics, chemistry, and engineering. At its core, the second law introduces the concept of entropy, a measure often associated with the degree of disorder or randomness in a system. In simple terms, this law states that the total entropy of an isolated system will always increase over time for irreversible processes and remain constant for reversible processes.

When applying the second law of thermodynamics to our exercise, we consider the entropy change during the melting and freezing of water and ice. The law predicts that any spontaneous process, like the melting of ice without any interference, will lead to an increase in entropy. Conversely, when a process is 'reversed', such as refreezing the melted water, this will happen only by decreasing the entropy of the system - often requiring energy input from the environment, thereby increasing the surroundings' entropy. Our exercise illustrated this by calculating the entropy change for both a reversible and an irreversible process, clearly showing compliance with the second law of thermodynamics.

Reversible Process

The concept of a reversible process is a theoretical idealization in thermodynamics. In such a process, every single step is done so slowly and precisely that the system remains in perfect equilibrium with its surroundings at all times. This means the process can be reversed without leaving any trace or change in the universe, including no net change in entropy. Think of a reversible process as a perfectly calculated dance between the system and its surroundings, where all moves are synchronized.

In our exercise context, the reversible process involves the controlled freezing of water into ice, proceeding in such a way that it could be perfectly reversed by an infinitesimally small change. The calculation of the entropy change during this perfectly reversible process was done by integrating the infinitesimal heat transfers over the temperature. This idealization allows for the understanding of the lower limit of the entropy change that such a transformation can theoretically achieve.

Irreversible Process

Real-world processes are not as perfect as the reversible ones we just discussed. Instead, they often occur rapidly and without equilibrium, making them irreversible. An irreversible process is a more realistic concept where changes imparted to a system cannot be undone without leaving a net change in the universe, usually marked by an increase in entropy.

Our exercise's irreversible scenario involved returning the system composed of water and ice to its initial state using a Bunsen burner. This depicted how external energy sources cause entropy to increase overall. Although the system's state was reversed to its beginning, the exercise highlighted that the surroundings are now less ordered (due to heat released from the Bunsen burner), signifying an overall increase in entropy, further demonstrating the spontaneous nature of real-world processes.

Latent Heat of Fusion

The concept of latent heat of fusion can be thought of as the 'hidden' heat required to change the state of a substance without changing its temperature. Specifically, for water, it is the energy needed to transform ice to liquid water or vice versa at 0°C, without any change in the external temperature. The latent heat of fusion for water is a relatively high value, reflecting the strong hydrogen bonds between water molecules that must be overcome during melting or formed during freezing.

To calculate the entropy change during the reversible process in the exercise, the latent heat of fusion is central to determining the amount of heat energy involved in the mass transformation of ice to water. This value is vital not only for understanding the thermodynamics of phase changes but also for its practical applications, such as calculating the energy requirements for refrigeration systems, climate modeling, and even understanding natural phenomena like the melting of polar ice caps.