/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 A "cold" object, \(T_{1}=300 \ma... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A "cold" object, \(T_{1}=300 \mathrm{~K}\), is briefly put in contact with a "hot" object, \(T_{2} \simeq 400 \mathrm{~K}\), and \(60 \mathrm{~J}\) of heat flows from the hot object to the cold one. The objects are then separated, their temperatures having changed negligibly due to their large sizes. (a) What are the changes in entropy of each object and the system as a whole? (b) Knowing only that these objects are in contact and at the given temperatures, what is the ratio of the probabilities of their being found in the second (final) state W that of their being found in the first (initial) state? What does chis result suggest?

Short Answer

Expert verified
The change in entropy of the hot and cold objects are -0.15 J/K and +0.20 J/K respectively, making total change in entropy of the system to be +0.05 J/K. The final state of the objects being in contact is massively more probable than the initial state.

Step by step solution

01

Calculation of change in entropy for each object

The formula for change in entropy \(\Delta S\) when heat flows at a constant temperature is given by \(\Delta S = Q / T\), where Q is the heat transferred and T is the temperature. For the hot object, the heat Q is flowing out, so it is -60 J. The initial temperature \(T_{2}\) is 400 K. Hence, the change in entropy for the hot object \(\Delta S_{2} = -60 / 400 = -0.15 \ J/K\). For the cold object, the heat Q is flowing in, so it is +60 J. The initial temperature \(T_{1}\) is 300 K. Hence, the change in entropy for the cold object \(\Delta S_{1} = +60 / 300 = +0.20 \ J/K\). Remember, the entropy change is positive for heat gain and negative for heat loss at constant temperature.
02

Calculation of change in entropy for the whole system

The total entropy change of the system is the sum of the entropy changes of the hot and cold objects. Thus, \(\Delta S_{total} = \Delta S_{1} + \Delta S_{2} = +0.20 - 0.15 = +0.05 \ J/K\). As expected, the entropy of a system increases with time, due to the second law of thermodynamics that predict the arrow of time.
03

Calculation of the ratio of probabilities of the states of the objects

According to the principle of equal a priori probabilities, the probability of a certain state is proportional to the number of ways, W, that this state can be realized. According to Boltzmann's entropy formula, \(S = k \ln(W)\), where k is Boltzmann's constant, and W the number of microstates. The ratio of probabilities follows from this by exponentiating: \(W_{final} / W_{initial} = e^{(\Delta S / k)}\), with \(\Delta S = S_{final} - S_{initial} = +0.05\ J/K\), and k is the Boltzmann constant \(k = 1.4 \times 10^{-23}\ J/K\). Hence, \(W_{final} / W_{initial} = e^{(+0.05 / 1.4 \times 10^{-23})}\). The resulting ratio is a huge number, indicating that the final state is overwhelmingly more probable than the initial state, as expected due to the increase in entropy.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding the Second Law of Thermodynamics
The Second Law of Thermodynamics is a fundamental principle which states that the entropy of an isolated system always increases over time, or remains constant in ideal cases of reversible processes. In simple terms, entropy can be thought of as a measure of disorder or randomness: the greater the disorder, the higher the entropy.

The implication of this law is profound: it introduces the concept of a 'direction' in time, often referred to as the 'arrow of time'. In the context of heat exchange, like in our exercise, this law implies that heat will naturally flow from a hotter object to a cooler one, increasing the overall entropy of the system.

In practical terms, this manifests as an inability to fully convert heat energy into work without some loss of energy to the surroundings, hence why perfectly efficient engines are impossible according to the second law. Its importance cannot be overstated as it governs the principles of heat transfer, energy dissipation, and has implications for the ultimate fate of the universe.
The Basics of Heat Transfer
Heat transfer is a fundamental concept in thermodynamics that describes the physical act of thermal energy being exchanged between different bodies or a body and its environment. In the exercise problem, heat is transferred from the hot object at 400 K to the cold one at 300 K. This conforms with our expected behavior from the second law of thermodynamics.

There are three modes of heat transfer: conduction, which occurs through solid materials; convection, which occurs through fluids (liquids or gases) often involving fluid motion; and radiation, which involves emitting or absorbing thermal radiation. In our scenario, it's likely that conduction would be the mode of heat transfer given that the objects are in direct contact.

The current problem simplifies the situation by assuming that the heat transfer does not alter the temperatures of the objects significantly due to their large size. This means that the calculation of entropy change can be done using the heat transfer and the initial temperatures, without needing to consider the complexities of changing temperature during the process.
Boltzmann's Constant and its Role in Entropy
Boltzmann's constant, denoted as \(k\), plays a crucial role in statistical mechanics and thermodynamics, relating the micro-level dynamics to macro-level observations. It is a bridge between the molecular world and the macroscopic properties we can measure, such as temperature and entropy.

With a value of approximately \(1.38 \times 10^{-23} \ J/K\), this constant is a scaling factor that translates temperature into kinetic energy and provides a measure of the thermal energy per particle per degree of freedom. In the context of entropy, using Boltzmann's relation \(S = k \ln(W)\), Boltzmann's constant provides a way to quantify the multiplicities of microstates (W) that correspond to a macrostate characterized by a certain entropy.

In our exercise, we use Boltzmann's constant when calculating the ratio of probabilities of different states based on the change in entropy. This is a statistical method that underscores the probabilistic nature of thermodynamics, particularly in systems with a large number of particles.
Thermodynamic Probability and Microstates
Thermodynamic probability often referred to as the number of accessible microstates \(W\), is a concept that's interwoven with the fabric of statistical mechanics. It represents the number of different ways in which a system can be arranged at the microscopic level, while still appearing the same at the macroscopic level.

Understanding thermodynamic probability is key to grasping why certain events occur with a higher likelihood than others. The exercise shows a scenario where the system evolves from an initial state to a final state, with the final state having a greater number of accessible microstates — this indicates a higher thermodynamic probability, and therefore, a higher entropy.

Using Boltzmann's formula, where entropy is proportional to the logarithm of the number of microstates, we relate entropy change to thermodynamic probability. This allows us to calculate the extremely large probability ratio of finding the system in the final state compared to the initial state. This ratio portrays the quintessential tendency of isolated systems to evolve towards configurations with the greatest number of microstates, thus respecting the second law of thermodynamics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You have six shelves, one above the other and all above the floor, and six volumes of an encyciopedia, A. B. C. D. \(E\), and \(F\). (a) L.ist all the ways you can arrange the volumes with five on the floor and one on the sixth/top shelf. One way might be \(\mid \mathrm{ABCDE},-,-,-,-,-, \mathrm{F}\\}\) (b) List all the ways you can arrange them with four on the floor and two on the third shelf. (c) Show thal there are many more ways, relative to parts \((\mathrm{a})\) and \((\mathrm{b})\), to strange the six volumes with two on the floor and two eachon the first and second shelves. (There are several ways to answer this, but even listing them all won't take forever it's fewer than \(100 .)\) (d) Suddenly, a fantastic change! All six volumes are volume \(X\) - \(\mathrm{it}\) 's impossible to tell them apar. For each of the three distributions described in parts (a), (b), and (c). how many different (distinguishable) ways are there now? (e) If the energy you expend to lift a volume from the floor is proportional to a shelf's height, how do the total energies of distributions (a), (b), and (c) compare? (I) Use these ideas to atgue that the relative probabili. ties of occupying the lowestenergy states should be higher for hosons than for classically distinguishable particles. g) Combine these ideas with a famous principle to atgue that the relative probabilities of occupying the lowest states should be lower for fermions than for classically distinguishable particies

A scientifically untrained but curious friend asks, "When I walk into a room. is there a chance that all the air will be on the other side?" How do you answer this question?

When would a density of states be needed: in a sum over states? in a sum over energies? in an integral over energies? in an integral over states?

We claim that the famous exponential decrease of probability with energy is natural, the vastly most probable and disordered state given the constraints on total energy and number of particles. It should be a state of maximum entropy! The proof involves mathematical techniques beyond the scope of the text, but finding support is good exercise and not difficult. Consider a system of 11 oscillators sharing a total energy of just \(5 h \omega_{0}\). In the symbols of Section \(9.3, N=11\) and \(M=5\). (a) Using equation \((9-9),\) calculate the probabilities of \(n\) being \(0.1,2 .\) and \(3 .\) (b) How many particles. \(N_{n}\), would be expected in each level? Round each other nearest integer. (Happily. the number is still 11 . and the energy still \(\operatorname{she}_{0}\) ) What you have is a distribution of the energy that is as close to expectations is possible, given that numbers at each level in a real case are integers. (c) Entropy is related to the number of microscopic ways the macrostate can be obtained. and the number of ways of permuting particle labels with \(N_{0}\). \(N_{1}, N_{2},\) an \(d N_{1}\) fixed and totaling 11 is \(11 ! /\left(N_{0} ! N_{1} !\right.\) \(\left.N_{2} ! N_{3}^{\prime}\right) .\) (See Appendix J for the proof.) Calculate the number of ways for your distribution. (d) Calculate the number of ways if there were 6 particles in \(n=0.5\) in \(n=1\), and none higher. Note that this also has the same total energy. (e) Find at least one other distribution in which the 11 oscillators share the same energy, and calculate the number of ways. (f) What do your findings suggest?

There are more permutations of particle labels when of particles have energy 0 and two have energy 1 than when three particles have energy 0 and one has energy 2\. (The total energies are the same.) From this observation alone argue that the Boltzmann distribution should be lower than the Bose-Einstein at the lowest energy level.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.