chapter 8 dinh luat thu hai cua nhiet dong hoc

The Second Law of Thermodynamics8.1 Heat Engines and the Second Law of thermodynamics8.2 Heat Pumps and Refrigerators8.3 Reversible and Irreversible Processes8.4 The Carnot Engine8.5 Ent

TẬP ĐOÀN DẦU KHÍ VIỆT NAM

TRƯỜNG ĐẠI HỌC DẦU KHÍ VIỆT NAM

General Physics I

Lecturer : Assoc.Prof Pham Hong QuangEmail: quangph@pvu.edu.vn

Chapter 8 The Second Law of Thermodynamics

8.1 Heat Engines and the Second Law of thermodynamics

8.2 Heat Pumps and Refrigerators

8.3 Reversible and Irreversible Processes8.4 The Carnot Engine

8.5 Entropy

8.6 Entropy on a Microscopic Scale

Learning outcome

The students should be able to:

•Identify the second law of thermodynamics: If a process occurs in a closed system, the entropy of the system increases for

irreversible processes and remains constant for reversible processes; it never decreases.

•Identify that entropy is a state function (the value for a particular state of the system does not depend on how that state is

•Calculate the change in entropy for a process by integrating the

inverse of the temperature (in kelvins) with respect to the heat Q

transferred during the process.

Learning outcome

•Identify that a heat engine is a device that extracts energy from its environment in the form of heat and does useful work and that in an ideal heat engine, all processes are reversible, with no

wasteful energy transfers.

•Sketch a p-V diagram for the cycle of a Carnot engine,

•indicating the direction of cycling, the nature of the processes involved, the work done during each process (including algebraic sign), the net work done in the cycle, and the heat transferred during each process (including algebraic sign).

•Calculate the efficiency of a Carnot engine in terms of the heat

transfers and also in terms of the temperatures of the reservoirs.•Identify that there are no perfect engines in which the energy

transferred as heat Q from a high temperature reservoir goes entirely into the work Wdone by the engine.

Learning outcome

•Identify that a refrigerator is a device that uses work to transfer energy from a low-temperature reservoir to a high-temperature reservoir, and that an ideal refrigerator is one that does this with reversible processes and no wasteful losses.

•Identify that there is no ideal refrigerator in which all of the energy extracted from the low-temperature reservoir is

transferred to the high-temperature reservoir.

•Apply the relationship between the coefficient of performance K

and the heat exchanges with the reservoirs and the temperatures

of the reservoirs.

•Identify that the efficiency of a real engine is less than that of the ideal Carnot engine.

Learning outcome

•Explain what is meant by the configurations of a systemof molecules.

•Calculate the multiplicity of a given configuration.

•Identify that all microstates are equally probable but the

configurations with more microstates are more probable than the other configurations.

•Apply Boltzmann’s entropy equation to calculate the entropy associated with a multiplicity.

8.1 Heat Engines and the Second Law of Thermodynamics

First Law of Thermodynamics – Review

The first law is a statement of Conservation of Energy.The first law states that a change in internal energy in a system can occur as a result of energy transfer by heat, by work, or by both.

The first law makes no distinction between processes that occur spontaneously and those that do not.

Only certain types of conversion and transfer processes actually take place in nature.

energy-8.1 Heat Engines and the Second Law of Thermodynamics

Establishes which processes do and which do not occurSome processes can occur only in one direction according to the first law.

This directionality is governed by the second law.These types of processes are irreversible.

An irreversible process is one that occurs naturally in one direction only.

No irreversible process has been observed to run backwards.

The Second Law of Thermodynamics

8.1 Heat Engines and the Second Law of Thermodynamics

Lord Kelvin 1824 – 1907British physicist and

8.1 Heat Engines and the Second Law of Thermodynamics

A heat engine is a device that

takes in energy by heat and, operating in a cyclic process,

expels a fraction of that energy by means of work.

A heat engine carries some working substance through a cyclical process.

The working substance absorbs energy by heat from a high

temperature energy reservoir (Qh).

Work is done by the engine (Weng).Energy is expelled as heat to a

lower temperature reservoir (Q ).

Heat Engine

8.1 Heat Engines and the Second Law of Thermodynamics

Since it is a cyclical process, ΔEint = 0

Its initial and final internal energies are the same.

Therefore, Weng = Qnet = |Qh| - |Qc|

The net work done by a heat engine equals the net energy transferred to it.

8.1 Heat Engines and the Second Law of Thermodynamics

Thermal efficiency is defined as the ratio of the net work done by the engine during one cycle to the energy input at the higher temperature.

We can think of the efficiency as the ratio of what you gain to what you give.

In practice, all heat engines expel only a fraction of the input energy by mechanical work.

Therefore, their efficiency is always less than 100%.

8.1 Heat Engines and the Second Law of Thermodynamics

Second Law: Kelvin-Planck Form

It is impossible to construct a heat engine that,

operating in a cycle, produces no effect other than the input of energy by heat from a reservoir and the performance of an equal amount of work.

Weng can never be equal to |Qh|Means that Qc cannot equal 0

Some energy |Qc| must be expelled to the environment

Means that e cannot equal 100%

8.1 Heat Engines and the Second Law of Thermodynamics

No energy is expelled to the cold reservoir.It takes in some amount of energy and does an equal amount of work.

8.2 Heat Pumps and Refrigerators

Heat Pumps and Refrigerators

Heat engines can run in reverse.This is not a natural direction of energy transfer.

Must put some energy into a device to do this

Devices that do this are called heat pumps or refrigerators

A refrigerator is a common type of heat pump.

An air conditioner is another example of a heat pump.

8.2 Heat Pumps and Refrigerators

Second Law - Clausius statement for refrigerator

It is not possible for heat to flow from a colder

body to a warmer body without any work having been done to accomplish this flow

Energy will not flow

spontaneously from a low temperature object to a higher temperature

object

8.2 Heat Pumps and Refrigerators

The effectiveness of a heat pump is described by a number

called the coefficient of performance (COP).

Similar to thermal efficiency for a heat engine

It is the ratio of what you gain (energy transferred to or from a reservoir) to what you give (work input).

In cooling mode, you “gain” energy removed from a cold

temperature reservoir.

A good refrigerator should have a high COP.Typical values are 5 or 6

Coefficient of Performance

COP energy transferred at low tempQc

work done on the pumpW

8.2 Heat Pumps and Refrigerators

COP, Heating Mode

In heating mode, the COP is the ratio of the heat

transferred in to the work required.

Qh is typically higher than W

Values of COP are generally about 4For outside temperature about 25° F

COP = energy transferred at high tempQh

work done by heat pump  W

8.3 Reversible and Irreversible Processes

Reversible and Irreversible Processes

A reversible process is one in which both the system and its environment can be returned to exactly the states they were in before the process occurred

A reversible process is one in which every point along some path is an equilibrium state.

An irreversible process does not meet these requirements.All natural processes are known to be irreversible.

Reversible processes are an idealization, but some real processes are good approximations.

8.3 Reversible and Irreversible Processes

A real process that is a good approximation of a reversible one will occur very slowly.

The system is always very nearly in an equilibrium state.

A general characteristic of a reversible process is that there are no dissipative effects that convert mechanical energy to internal energy present.

No friction or turbulence, for example

8.4 The Carnot Engine

Carnot Engine

A theoretical engine developed by Sadi CarnotA heat engine operating in an ideal, reversible cycle (now called a Carnot cycle) between two reservoirs is the most efficient engine possible

This sets an upper limit on the efficiencies of all other engines.

8.4 The Carnot Engine

Carnot’s Theorem

No real heat engine operating between two

energy reservoirs can be more efficient than a Carnot engine operating between the same

two reservoirs.

All real engines are less efficient than a Carnot engine because they do not operate through a reversible cycle.

The efficiency of a real engine is further reduced by friction, energy losses through conduction, etc.

8.4 The Carnot Engine

Carnot Cycle

8.4 The Carnot Engine

The gas does work WAB

in raising the piston.

8.4 The Carnot Engine

No energy enters or leaves the system by heat.

The temperature falls

from Th to Tc.

The gas does work WBC.

8.4 The Carnot Engine

The gas is placed in

thermal contact with the cold temperature

8.4 The Carnot Engine

D → A is an adiabatic

The base is replaced by a thermally nonconducting wall.

So no heat is

exchanged with the surroundings.

The temperature of the

gas increases from Tc to

The work done on the gas

is WDA.

8.4 The Carnot Engine

The work done by the engine is

shown by the area enclosed by the

curve, Weng.

The net work is equal to |Qh| – |Qc|.

DEint = 0 for the entire cycle

Carnot showed that the efficiency of the engine depends on the

temperatures of the reservoirs.

Temperatures must be in KelvinsAll Carnot engines operating

between the same two temperatures will have the same efficiency.

8.4 The Carnot Engine

8.4 The Carnot Engine

Efficiency is 0 if Th = Tc

Efficiency is 100% only if Tc = 0 KSuch reservoirs are not availableEfficiency is always less than 100%

The efficiency increases as Tc is lowered and as Th is raised.

In most practical cases, Tc is near room temperature, 300 K

So generally Th is raised to increase efficiency.

Notes About Carnot Efficiency

8.4 The Carnot Engine

Carnot Cycle in Reverse

Theoretically, a Carnot-cycle heat engine can run in reverse.

This would constitute the most effective heat pump available.

This would determine the maximum possible COPs for a given combination of hot and

cold reservoirs.

8.4 The Carnot Engine

Carnot Heat Pump COPs

8.5 Entropy

Entropy and Heat

The original formulation of entropy dealt with the transfer of energy by heat in a reversible process.

Let dQr be the amount of energy transferred by heat when a system follows a reversible path.

The change in entropy, dS is

The change in entropy depends only on the starting point and the endpoints and is independent of the path followed.The entropy change for an irreversible process can be

determined by calculating the change in entropy for a

reversible process that connects the same initial and final points.



8.5 Entropy

ΔS for a Reversible Cycle

ΔS = 0 for any reversible cycle

In general,

This integral symbol

indicates the integral is over a closed path.



8.5 Entropy

To calculate the change in entropy in a real system,

remember that entropy depends only on the state of the system.

We find a reversible process which has the same initial and final equilibrium states and calculate the change in entropy for this process

Do not use Q, the actual energy transfer in the process.Distinguish this from Qr , the amount of energy that

would have been transferred by heat along a reversible path.

Q is the correct value to use for DS.

Entropy Changes in Irreversible Processes

8.5 Entropy

8.5 Entropy

This process is an irreversible process We replace this process by two reversible processes in which the cold

reservoir absorbs energy Q isothermally (its entropy

changes by Q/Tc. ) and at the same time, the hot reservoir

loses Q isothermally (its entropy changes by -Q/Th

Since Th > Tc , the increase in entropy in the cold reservoir is greater than the decrease in entropy in the hot reservoir.

Therefore, ∆SU > 0

For the system and the Universe

∆S in Thermal Conduction

Q = 0 but we need to find Qr

Choose an isothermal, reversible expansion in which the gas pushes slowly against the piston while

energy enters from a reservoir to keep T constant.

ΔS in a Free Expansion

ΔS in Free Expansion, cont

8.5 Entropy

The entropy of the system and

environment (Universe) increases in all real processes.

This is another statement of the second law of thermodynamics.

It is equivalent to the Kelvin-Planck and Clausius statements.

Entropy and the Second Law

Then: ∆Ssys + ∆Senvi < 0

This result against the Second Law of Thermodynamics

It is impossible to construct an



8.5 Entropy

Because the heat energy (system) undergoes an ideal cycle, ∆Ssys must equal 0.

Let us consider the change of entropy of environment:

Then: ∆Ssys + ∆Senvi < 0

This result against the Second Law of Thermodynamics

It is impossible to construct an ideal heat pump.

8.6 Entropy on a Microscopic Scale

Entropy on a Microscopic Scale

We can treat entropy from a microscopic viewpoint through statistical analysis of molecular motions.

A microstate is a particular configuration of the individual

constituents of the system.

A macrostate is a description of the conditions from a

macroscopic point of view.

A connection between entropy and the number of microstates (Ω) for a given macrostate is ) for a given macrostate is

S = kB ln Ω) for a given macrostate is

The more microstates that correspond to a given

macrostate, the greater the entropy of that macrostate.This shows that entropy is a measure of disorder.

8.6 Entropy on a Microscopic Scale

allfor smicrostateof

number total

macrostate the

toingcorresponds

number macrostate

aofy

Example 1

8.6 Entropy on a Microscopic Scale

For a system composed of two identical dice, let the macrostate be defined as the sum of the numbers showing on the top faces

What is the maximum entropy of this system in units of Boltzmann’s constant?

Example 2

8.6 Entropy on a Microscopic Scale

SumPossible microstates2(1,1)

9(3,6); (4,5); (5,4) (6,3)10(4,6); (5,5); (6,4)

11(5,6); (6,5)12(6,6)

8.6 Entropy on a Microscopic Scale

The maximum entropy corresponds to a sum of 7 on the dice For this

macrostate, Ω) for a given macrostate is = 6 with an entropy of

k

8.6 Entropy on a Microscopic Scale

Entropy can be thought of as the increase in disorder in the universe In this diagram, the end state is less ordered than the initial state – the separation between low and high

temperature areas has been lost

8.6 Entropy on a Microscopic Scale

Application of Entropy- Adiabatic Demagnetization

Key words of the chapter

Heat Engines; Heat Pumps; Refrigerators; Thermal

Efficiency; Coefficient of Performance; Reversible

Processes; Irreversible Processes; Carnot Engine; Carnot Cycle; Entropy; Number of microstates.

•The spontaneous flow of heat between objects in thermal equilibrium is always from the hotter one to the colder one •A heat engine converts heat into work

•Efficiency of a heat engine:

•A reversible engine has the maximum possible efficiency:

•Refrigerators, air conditioners, and heat pumps use work to transfer heat from a cold region to a hot region.

•Coefficient of performance of a refrigerator: COP=QC/W •Coefficient of performance for a heat pump: COP=QH/W