/*! 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 62 An isolated system of total mass... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 62

An isolated system of total mass \(m\) is formed by mixing two equal masses of the same liquid initially at the temperatures \(T_{1}\) and \(T_{2}\). Eventually, the system attains an equilibrium state. Each mass is incompressible with constant specific heat \(c\). (a) Show that the amount of entropy produced is $$ \sigma=m c \ln \left[\frac{T_{1}+T_{2}}{2\left(T_{1} T_{2}\right)^{1 / 2}}\right] $$ (b) Demonstrate that \(\sigma\) must be positive.

Short Answer

Expert verified
The total entropy produced, _sigma, is equal to \[ 蟽 = m c ln[ \frac {T_{1}T_2}{2T_1 T_2 ^ o.5} \]. Entropy cannot be negative, hence the term must be always positive.

Step by step solution

01

- Expression for Final Temperature

Before finding the entropy produced, find the final equilibrium temperature. The final equilibrium temperature for an isolated system with equal masses at different temperatures is the average of the two temperatures:\[ T_f = \frac{T_1 + T_2}{2} \]
02

- Calculate Initial and Final Entropy

Calculate the initial entropy (S_i) and final entropy (S_f). The entropy for each mass is given by the formula \( S = m c \frac{dQ}{T} \). Therefore, for ea ch mass initially:\[ \frac{dm}{T} = mc dT \]Hence, from T1:\[ S_1 = m c \ln\frac{2T_1}{T_1} , S_2=mc \ln\frac{T_1}{2T_2} \]Similarly for T2:\[S_i = m c\left(\frac{T_{1}+T_{2}}{2}\right) , S_f=mc ln \left(\frac{\frac{T_1+T_2}{2}}{T1}\right)+mc ln\left(\frac{\frac{T1+T2}{2}}{T_2} \right)\]
03

- Total Change in Entropy

The total entropy change (蟽) of the system will be the difference between the final entropy and the initial entropy:\[ 蟽 = S_f - S_i \]Therefore,\[S_i - S_f=m c ln \left[\frac {T_{1}+ T_2}{2(T1+T2)^ }\right] \]
04

- Proof that Entropy is Positive

We know that the only term which could be negative here is the term inside the logarithm, but since ln of a fraction is never negative hence the term given before must be positive.

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 91影视!

Key Concepts

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

Isolated System
An isolated system is a specific type of thermodynamic system that does not exchange energy or matter with its surroundings. In the context of the exercise, an isolated system is made up of two equal masses of the same liquid, each initially at different temperatures, but with no external influence. This means that any changes in the system are due to the interchange of energy within the system itself, and not due to any external factors. Such conditions help simplify calculations since we only need to focus on internal exchanges.
Entropy Change
Entropy is a measure of disorder or randomness in a system. In simpler terms, it's a way to quantify the amount of energy in a system that is not available to do work. When the two equal masses of liquid mix in the isolated system, they move from their initial states to a final equilibrium state, resulting in entropy production. The change in entropy \(\sigma\) can be calculated by comparing the entropy initially and the entropy when the system reaches equilibrium. This highlights the second law of thermodynamics, which states that the total entropy of an isolated system will always increase over time, leading towards a state of maximum entropy.
Equilibrium Temperature
The equilibrium temperature is the final stable temperature reached by the system after internal heat exchanges cease. For an isolated system made of two equal masses of the same liquid at different initial temperatures \(T_1\) and \(T_2\), the equilibrium temperature \(T_f\) is the average of these two temperatures:\[T_f = \frac{T_1 + T_2}{2}\]This calculation is straightforward because it leverages the symmetry of the system and the fact that there are no external energy exchanges.
Specific Heat
Specific heat \(c\) is a property of a material that indicates how much energy is required to change the temperature of a unit mass of the substance by one degree. In this exercise, specific heat is assumed to be constant. This simplifies our calculations of entropy changes because we can directly use this value to measure the heat involved in changing the temperatures of the liquid masses. Knowing the specific heat allows us to relate temperature change to energy change effectively.
Thermodynamic Equilibrium
Thermodynamic equilibrium is the state when all parts of the system are at uniform temperature, pressure, and chemical potentials, indicating no net flows of energy or matter within the system. In our scenario, once the two masses of liquid reach the equilibrium temperature \(T_f\), they are in thermodynamic equilibrium. This is when the entropy is maximized for the system, and there are no more spontaneous changes occurring because the system has reached its most disordered state. Understanding thermodynamic equilibrium helps explain why and how entropy is produced during the process.

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

Two insulated tanks are connected by a valve. One tank initially contains \(0.45 \mathrm{~kg}\) of air at \(93^{\circ} \mathrm{C}, 1\) bar, and the other contains \(0.9 \mathrm{~kg}\) of air at \(38^{\circ} \mathrm{C}, 2\) bar. The valve is opened and the two quantities of air are allowed to mix until equilibrium is attained. Employing the ideal gas model with \(c_{v}=0.7 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), determine (a) the final temperature, in \({ }^{\circ} \mathrm{C}\). (b) the final pressure, in bar. (c) the amount of entropy produced, in \(\mathrm{kJ} / \mathrm{K}\).

An electric motor operating at steady state draws a current of \(10 \mathrm{amp}\) with a voltage of \(220 \mathrm{~V}\). The output shaft rotates at 1000 RPM with a torque of \(16 \mathrm{~N} \cdot \mathrm{m}\) applied to an external load. The rate of heat transfer from the motor to its surroundings is related to the surface temperature \(T_{\mathrm{b}}\). and the ambient temperature \(T_{0}\) by \(\mathrm{hA}\left(T_{\mathrm{b}}-T_{0}\right)\), where \(\mathrm{h}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, \mathrm{A}=0.195 \mathrm{~m}^{2}\), and \(T_{0}=293 \mathrm{~K} .\) Energy transfers are considered positive in the directions indicated by the arrows on Fig. P6.51. (a) Determine the temperature \(T_{\mathrm{b}}\), in \(\mathrm{K}\). (b) For the motor as the system, determine the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\). (c) If the system boundary is located to take in enough of the nearby surroundings for heat transfer to take place at temperature \(T_{0}\), determine the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), for the enlarged system.

Air enters an insulated diffuser operating at steady state at 1 bar, \(7^{\circ} \mathrm{C}\), and \(200 \mathrm{~m} / \mathrm{s}\) and exits with a velocity of \(100 \mathrm{~m} / \mathrm{s}\). Employing the ideal gas model and ignoring potential energy, determine (a) the temperature of the air at the exit, in \({ }^{\circ} \mathrm{C}\). (b) the maximum attainable exit pressure, in bar.

The temperature of a 0.3-L can of soft drink is reduced from 25 to \(6^{\circ} \mathrm{C}\) by a refrigeration cycle. The cycle receives energy by heat transfer from the soft drink and discharges energy by heat transfer at \(25^{\circ} \mathrm{C}\) to the surroundings. There are no other heat transfers. Determine the minimum theoretical work input required, in \(\mathrm{kJ}\), assuming the soft drink is an incompressible liquid with the properties of liquid water. Ignore the aluminum can.

A pump operating at steady state receives saturated liquid water at \(50^{\circ} \mathrm{C}\) with a mass flow rate of \(20 \mathrm{~kg} / \mathrm{s}\). The pressure of the water at the pump exit is \(1 \mathrm{MPa}\). If the pump operates with negligible internal irreversibilities and negligible changes in kinetic and potential energy, determine the power required in \(\mathrm{kW}\).
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Famous Physicists

Read Explanation

Fundamentals of Physics

Read Explanation

Solid State Physics

Read Explanation

Modern Physics

Read Explanation

Scientific Method Physics

Read Explanation
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.