/*! 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 63 20.63. An object of mass \(m_{1}... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

20.63. An object of mass \(m_{1}\) , specific heat capacity \(c_{1}\) , and temperature \(T_{1}\) is placed in contact with a second object of mass \(m_{2},\) specific heat capacity \(c_{2},\) and temperature \(T_{2}>T_{1} .\) As a result, the temperature of the first object increases to \(T\) and the temperature of the second object decreases to \(T^{\prime} .\) (a) Show that the entropy increase of the system is $$ \Delta S=m_{1} c_{1} \ln \frac{T}{T_{1}}+m_{2} c_{2} \ln \frac{T^{\prime}}{T_{2}} $$ and show that energy conservation requires that $$ m_{1} c_{1}\left(T-T_{1}\right)=m_{2} c_{2}\left(T_{2}-T^{\prime}\right) $$ (b) Show that the entropy change \(\Delta S\) , considered as a function of \(T,\) is a maximum if \(T=T,\) which is just the condition of thermodynamic equilibrium. (c) Discuss the result of part (b) in terms of the idea of entropy as a measure of disorder.

Short Answer

Expert verified
The entropy increase is maximized when the system reaches thermal equilibrium, maximizing disorder.

Step by step solution

01

Calculate Entropy Change for Each Object

For an object with mass \( m \), specific heat capacity \( c \), and temperature changing from \( T_i \) to \( T_f \), the change in entropy \( \Delta S \) is given by the formula \( \Delta S = m c \ln \left( \frac{T_f}{T_i} \right) \). Using this, we calculate the entropy change for the first object as \( \Delta S_1 = m_1 c_1 \ln \left( \frac{T}{T_1} \right) \) and for the second object as \( \Delta S_2 = m_2 c_2 \ln \left( \frac{T'}{T_2} \right) \). The total entropy change for the system is then \( \Delta S = \Delta S_1 + \Delta S_2 \).
02

Apply Conservation of Energy

According to the conservation of energy, the heat lost by the second object (with mass \( m_2 \)) must equal the heat gained by the first object (with mass \( m_1 \)). This gives us the equation:\[ m_1 c_1 (T - T_1) = m_2 c_2 (T_2 - T') \] This equation ensures that the system is in thermal equilibrium, where energy is neither created nor destroyed but transferred from the hotter object to the cooler one.
03

Analyze Entropy Change as a Function of Final Temperature

The entropy change \( \Delta S(T) \) can be considered as a function of \( T \) and a maximum is found by taking the derivative with respect to \( T \) and setting it to zero. Thus, \( \frac{d}{dT} \left( m_1 c_1 \ln \frac{T}{T_1} + m_2 c_2 \ln \frac{T'}{T_2} \right) = 0 \). Using the relation \( m_1 c_1 (T - T_1) = m_2 c_2 (T_2 - T') \), solve for \( T \) and check that no further increase in entropy occurs beyond this \( T \).
04

Interpret Entropy and Thermodynamic Equilibrium

The result shows that \( \Delta S(T) \) is maximized when the final temperatures are such that the entropy of the system is greatest.This condition is met when the temperatures \( T \) and \( T' \) reach a point of equilibrium where they approach one another. Therefore, the maximum entropy of the system describes a state of maximum disorder or randomness, reaffirming the concept of entropy as a measure of disorder, which increases until equilibrium is achieved.

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.

Specific Heat Capacity
Specific heat capacity is a property of matter that tells us how much heat is needed to change the temperature of a substance. It is usually given in units of joules per kilogram per degree Celsius (J/kg°C). This value varies depending on the material, indicating how much energy it can absorb before its temperature changes.
When two objects with different specific heat capacities are placed in contact, the amount of energy each can absorb or release will influence how their temperatures change. For instance:
  • A substance with a high specific heat capacity can absorb more heat without changing temperature dramatically.
  • A substance with a low specific heat capacity will change temperature more easily with the same amount of heat.
Understanding specific heat capacity helps us predict how objects will respond to heat transfer, which is fundamental in solving problems related to entropy change.
Thermodynamic Equilibrium
Thermodynamic equilibrium is a state where no net heat transfer occurs between the objects in contact. Each object in a system has the same temperature. It represents a balance where all parts of the system are in a stable condition.
Achieving thermodynamic equilibrium means:
  • The temperatures of interacting objects equalize.
  • There's no more heat movement from one object to another.
  • The overall energy is conserved within the system as a whole.
When two objects reach thermodynamic equilibrium, their temperature changes no longer affect the entropy of the system, as energy transitions have ceased. In the context of the exercise, reaching equilibrium corresponds to a state where the system entropy is maximized, reflecting maximum disorder.
Conservation of Energy
The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In thermodynamics, this principle ensures that any energy lost by one object must be gained by another.
For a system of two objects:
  • The heat lost by the warmer object must equal the heat gained by the cooler object.
  • This ensures that the total energy within the system remains constant.
The equation given in the exercise \[ m_1 c_1 (T - T_1) = m_2 c_2 (T_2 - T') \]demonstrates this principle. It states that the heat exchange between the two objects aligns with the conservation of energy, ultimately leading the system to thermodynamic equilibrium.
Heat Transfer
Heat transfer is the movement of thermal energy from one object or substance to another. It occurs when there is a temperature difference, moving from the hotter object to the cooler one. This process continues until thermal equilibrium is achieved.
The modes of heat transfer include:
  • Conduction: Direct heat transfer through a substance without any movement of the substance itself.
  • Convection: Heat transfer through a fluid (liquid or gas) caused by the moving fluid itself.
  • Radiation: Transfer of heat through electromagnetic waves, such as the heat from the sun.
In the context of the exercise, heat transfer drives the change in temperatures of the two objects, impacting the overall entropy of the system. By understanding how heat transfer works, we can better grasp why entropy changes as energy moves within the system.
Disorder in Thermodynamics
In thermodynamics, disorder refers to the distribution and arrangement of particles within a system. Increasing entropy reflects increasing disorder. This tendency towards disorder is a universal law, known as the Second Law of Thermodynamics.
Entropy measures a system's level of disorder or randomness. When two objects at different temperatures are placed together:
  • Heat flows naturally from the hot object to the cooler one.
  • Entropy increases until maximum disorder (thermal equilibrium) is achieved.
  • This aligns with the natural tendency for systems to evolve towards disorder unless energy input drives order.
As entropy reaches its peak when a system is at equilibrium, the energy distribution becomes more random. This concept of disorder and randomness is crucial for understanding why systems behave the way they do in 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

An aircraft engine takes in 9000 \(\mathrm{J}\) of heat and discards 6400 \(\mathrm{J}\) each cycle. (a) What is the mechanical work output of the engine during one cycle? (b) What is the thermal efficiency of the engine?

20.18. A Carnot device extracts 5.00 \(\mathrm{kJ}\) of heat from a body at \(-10.0^{\circ} \mathrm{C} .\) How much work is done if the device exhausts heat into the environment at \((a) 25.0^{\circ} \mathrm{C} ;(\mathrm{b}) 0.0^{\circ} \mathrm{C} ;(\mathrm{c})-25.0^{\circ} \mathrm{C} ;\) In each case, is the device acting as an engine or as a refrigerator?

20.5. A certain nuclear-power plant has a mechanical-power output (used to drive an electric generator) of 330 \(\mathrm{MW}\) . Its rate of heat input from the nuclear reactor is 1300 \(\mathrm{MW}\) . (a) What is the thermal efficiency of the system? (b) At what rate is heat discarded by the system?

20.29. You make tea with 0.250 \(\mathrm{kg}\) of \(85.0^{\circ} \mathrm{C}\) water and let it cool to room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\) before drinking it. (a) Calculate the entropy change of the water while it cools. (b) The cooling process is essentially isothernal for the air in your kitchen. Calculate the change in entropy of the air while the tea cools, assuming that all the heat lost by the water goes into the air. What is the total entropy change of the system tea \(+\) air?

20.45. An experimental power plant at the Natural Energy Laboratory of Hawaii generates electricity from the temperature gradient of the ocean. The surface and deep-water temperatures are \(27^{\circ} \mathrm{C}\) and \(6^{\circ} \mathrm{C}\) , respectively. (a) What is the maximum theoretical effciency of this power plant? (b) If the power plant is to produce 210 \(\mathrm{kW}\) of power, at what rate must heat be extracted from the warm water? At what rate must heat be absorbed by the cold water? Assume the maximum theoretical efficiency. (c) The cold water that enters the plant leaves it at a temperature of \(10^{\circ} \mathrm{C}\) . What must be the flow rate of cold water through the system? Give your answer in \(\mathrm{kg} / \mathrm{h}\) and \(\mathrm{L} / \mathrm{h}\) .

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.