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Suppose 1.00 \(\mathrm{kg}\) of water at \(10.0^{\circ} \mathrm{C}\) is mixed with 1.00 \(\mathrm{kg}\) of water at \(30.0^{\circ} \mathrm{C}\) at constant pressure. When the mixture has reached equilibrium, (a) what is the final temperature? (b) Take \(c_{P}=4.19 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\) for water and show that the entropy of the system increases by $$\Delta S=4.19 \ln \left[\left(\frac{293}{283}\right)\left(\frac{293}{303}\right)\right] \mathrm{kJ} / \mathrm{K}$$ (c) Verify numerically that \(\Delta S>0 .\) (d) Is the mixing an irreversible process?

Short Answer

Expert verified
The final temperature of the mixture is the average of the two initial temperatures, which is 20.0°C. The entropy of the system increases by the given expression, and numerically, the value of ΔS is positive, confirming an increase in entropy. Therefore, the mixing is an irreversible process.

Step by step solution

01

Determine the final temperature of the mixture

When two bodies of water at different temperatures are mixed, the final temperature of the mixture can be found by conserving energy. The heat lost by the hot water will be equal to the heat gained by the cold water. Assuming no heat loss to the environment, using the formula: \(m_{\text{hot}} c_{\text{P}} (T_{\text{hot}} - T_{\text{final}}) = m_{\text{cold}} c_{\text{P}} (T_{\text{final}} - T_{\text{cold}})\), where \(m\) is the mass of water, \( c_{\text{P}} \) is the specific heat capacity. By substituting the given values and solving the equation, the final temperature \( T_{\text{final}} \) can be calculated.
02

Calculate the entropy change of the system

The entropy change of the system when two bodies of water mix can be calculated using the formula: \( \triangle S = m c_{\text{P}} \text{ln}\frac{T_{\text{final}}}{T_{\text{initial}}} \). Since there are two bodies of water, we will have to calculate the entropy change for the hot water that is cooling down and the cold water that is warming up, and sum them.
03

Verify numerically that the entropy change is positive

Using the value of \( c_{\text{P}} \) given, and the calculated final temperature, plug in the values into the entropy change equation for both the hot water and the cold water, and calculate the total change in entropy. If the value is positive, then numerically, the entropy of the system increases.
04

Discuss the irreversibility of the process

To determine if the mixing of water is an irreversible process, consider the entropy change. If the entropy of an isolated system increases, the process is irreversible. Here, since no external work is done and there is no heat transfer with the surroundings, the increase in entropy suggests that the mixing process is irreversible.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Entropy Change
Entropy is best understood as a measure of disorder or randomness in a system. When we talk about entropy change in a thermodynamic process, we're looking at how this disorder changes as the process unfolds.

In the given exercise, when warm water mixes with cool water, the energy spreads out and the overall disorder of the system increases, which is why we see a positive entropy change. To calculate this change, the exercise used the formula: \( \triangle S = m c_{\text{P}} \text{ln}\frac{T_{\text{final}}}{T_{\text{initial}}} \). The natural logarithm in the equation encapsulates the proportional change due to temperature variations, which is key to quantify entropy change.

Understanding Entropy Through Mixing

Imagine a deck of cards that is perfectly ordered by rank and suit. When you shuffle the deck, you increase its entropy because you make its order more random. Similarly, when the two masses of water at different temperatures mix, the energy distribution becomes more random, increasing entropy.

Improving the Exercise

It would aid in comprehension to include a visual representation of entropy before and after mixing, or a step-by-step walkthrough with reasoning behind the use of logarithms in the equation.
Conservation of Energy
The conservation of energy principle states that energy cannot be created or destroyed, only transformed from one form to another. This principle is central to all processes in thermodynamics, including our exercise.

In our water mixing scenario, the energy is transferred from the warmer water to the cooler water. As there is no heat exchange with the environment, the total energy within the system remains constant. The exercise delineates this by equating the heat lost by the hot water to the heat gained by the cooler water using the specific heat formula.

Contextualizing Energy Conservation

Consider a bouncing ball: it exchanges kinetic energy for potential energy and back, but the total energy remains the same throughout its motion, provided there are no external forces like air resistance. In our water example, the 'bouncing ball' is the heat energy moving between the two water masses.

Exercise Improvement Advice

Including a detailed explanation about how the concept of energy conservation is applied practically in the problem, and how we assume that the system is closed (with no loss to the environment), would enhance understanding.
Specific Heat Capacity
Specific heat capacity is a property that describes how much heat energy is needed to raise the temperature of a substance by one degree. It is crucial for predicting how substances interact in thermal processes, which is why it is featured in the exercise.

The formula provided in the exercise, with the specific heat capacity denoted as \(c_{P}\), allows for the calculation of energy changes in the water samples. It tells us that water has a relatively high specific heat capacity, meaning it can absorb or release a significant amount of heat without experiencing large temperature changes.

Real-Life Application of Specific Heat

Imagine you're heating a pool of water and a pot of oil with the same amount of energy. The pool temperature changes far less than the oil temperature because water has a higher specific heat capacity.

Exercise Enhancement

An improved exercise could include comparative examples to highlight how specific heat capacity influences energy exchange, or propose students to calculate the effect of different specific heat capacities on the mixing process.
Irreversible Process
An irreversible process is a transformation that cannot return both the system and its surroundings to their original states without external interference. In our context, once the two bodies of water are mixed, they cannot separate back into their original temperatures without adding or removing energy.

Irreversibility is often accompanied by an increase in entropy, as seen in our exercise. As we established that the entropy of the system increases, indicating a more disordered state, the process is deemed irreversible.

Spotting Irreversibility in Everyday Life

Mixing cream into coffee is a common example of an irreversible process; once mixed, you can't separate them back into cream and coffee without extra steps. Similarly, our water mixture can’t simply 'unmix'.

Adding Value to the Exercise

An in-depth explanation of why certain processes are irreversible, perhaps with a few real-world analogies, would make the concept more intuitive. Additionally, discussing the implications of irreversibility in thermodynamic processes could be an interesting extension for students to consider.

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Most popular questions from this chapter

The temperature at the surface of the Sun is approximately 5 700 K, and the temperature at the surface of the Earth is approximately 290 K. What entropy change occurs when 1 000 J of energy is transferred by radiation from the Sun to the Earth?

In 1993 the federal government instituted a requirement that all room air conditioners sold in the United States must have an energy efficiency ratio (EER) of 10 or higher. The EER is defined as the ratio of the cooling capacity of the air conditioner, measured in Btu/h, to its electrical power requirement in watts. (a) Convert the EER of 10.0 to dimensionless form, using the conversion 1 Btu ! 1055 J. (b) What is the appropriate name for this dimensionless quantity? (c) In the 1970s it was common to find room air conditioners with EERs of 5 or lower. Compare the operating costs for 10 000-Btu/h air conditioners with EERs of 5.00 and 10.0. Assume that each air conditioner operates for 1 500 h during the summer in a city where electricity costs 10.0\(€\) per \(\mathrm{kWh}\).

A gasoline engine has a compression ratio of 6.00 and uses a gas for which \(\gamma=1.40 .\) (a) What is the efficiency of the engine if it operates in an idealized Otto cycle? (b) What If ? If the actual efficiency is 15.0%, what fraction of the fuel is wasted as a result of friction and energy losses by heat that could by avoided in a reversible engine? (Assume complete combustion of the air–fuel mixture.)

Calculate the change in entropy of 250 \(\mathrm{g}\) of water heated slowly from \(20.0^{\circ} \mathrm{C}\) to \(80.0^{\circ} \mathrm{C} .\) (Suggestion: Note that \(d Q=m c d T . )\)

A firebox is at 750 K, and the ambient temperature is 300 K. The efficiency of a Carnot engine doing 150 J of work as it transports energy between these constant-temperature baths is 60.0%. The Carnot engine must take in energy 150 J/0.600 ! 250 J from the hot reservoir and must put out 100 J of energy by heat into the environment. To follow Carnot’s reasoning, suppose that some other heat engine S could have efficiency 70.0%. (a) Find the energy input and wasted energy output of engine S as it does 150 J of work. (b) Let engine S operate as in part (a) and run the Carnot engine in reverse. Find the total energy the firebox puts out as both engines operate together, and the total energy transferred to the environment. Show that the Clausius statement of the second law of thermodynamics is violated. (c) Find the energy input and work output of engine S as it puts out exhaust energy of 100 J. (d) Let engine S operate as in (c) and contribute 150 J of its work output to running the Carnot engine in reverse. Find the total energy the firebox puts out as both engines operate together, the total work output, and the total energy transferred to the environment. Show that the Kelvin–Planck statement of the second law is violated. Thus our assumption about the efficiency of engine S must be false. (e) Let the engines operate together through one cycle as in part (d). Find the change in entropy of the Universe. Show that the entropy statement of the second law is violated.

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