Chapter 19: The Second Law of Thermodynamics

Directions of thermodynamic processes

Irreversible and reversible processes Thermodynamic processes that occur in nature are all irreversible

processes which proceed spontaneously in one direction but notthe other.

In a reversible process the system must be capable of being returnedto its original state with no other change in the surroundings.

A reversible process proceeds slowly through equilibrium states.

Heat engines

Heat engine Any device that transforms heat partly into work or mechanicalenergy is called a heat engine.

A quantity of matter inside the engine undergoes inflow and outflow ofheat, expansion and compression, and sometimes change of phase.

This matter inside the engine is called the working substance of theengine.

The simplest kind of engine to analyze is one in which the workingsubstance undergoes a cyclic process.

All heat engines absorb heat from a source at a relatively hightemperature, perform some mechanical work, and discard or rejectsome heat at a lower temperature.

WQWQUUU ==== 012

For a cyclic process

Heat enginesEnergy flow diagram for a heat engine

HT

0>HQ

CT

0

Heat engines (contd)

Energy flow diagram for a heat engine (contd) Net heat Q absorbed per cycle:

CHCH QQQQQ =+=Net work done W per cycle:

CHCH QQQQQW =+==Thermal efficiency :

H

C

H

C

H QQ

QQ

QW

=+== 11

Heat engines (contd)

Steam engines (external combustion)

Internal-combustion engines

Gasoline engine (a heat engine)

First a mixture of air and gasoline vapor flows into a cylinder through an open intakevalve while the piston descends, increasing the volume of the cylinder from aminimum of V to a maximum of rV (r: compression ratio 8-10).

At the end of intake stroke, the intake valve closes and the mixture is compressed,approximately adiabatically, back to volume V during the compression stroke.

Then the mixture is ignited by the spark plug, and the heated gas expands, approximately adiabatically, back to volume rV, pushing on the piston-power stroke.

Internal-combustion engines (contd)The Otto cycle

:ba

:cb

:dc

:ad

compression stroke(adiabatic compression)ignite fuel (heating at constant volume)power stroke (adiabatic expansion)Reject heat to environment(cooling at constant volume)

0)(: >= bcVH TTnCQcb

0)(: =56%

Internal-combustion engines (contd)The Otto cycle

Internal-combustion engines (contd)The Diesel cycle

:ba

:cb

:dc

:ad

compression stroke(adiabatic compression)ignite fuel (heating at constant pressure)power stroke (adiabatic expansion)Reject heat to environment(cooling at constant volume)

)(),(,0:

abVabab

abVabab

TTnCUWTTnCUQba

====

))((),(),(:

bcVpbcbcbc

bcVabbcpbc

TTCCnUQWTTnCUTTnCQcb

==

==

)(),(,0:

cdVcdcd

cdVcdcd

TTnCUWTTnCUQdc

====

)(),(,0:

daVdacdda

daVdada

TTnCWUQTTnCUWad

=+===

)()( adVbcpdacdbcabnet

TTnCTTnCWWWWW

=+++=

)5,15%56()(

1 ===

==

c

a

b

a

bc

ad

H

net

VV

VV

TTTT

QW

Internal-combustion engines (contd)The Diesel cycle

RefrigeratorsRefrigerator

A refrigerator works as a reversed heat engine.

RefrigeratorsEnergy flow of a refrigerator

HH QQWW == ,

From the first law of thermodynamics

WQQWQQ CHCH

A refrigerator takes heat from lowtemperature reservoir and gives itoff to high temperature reservoir.

==+ 0

WQQ CH +=Coefficient of performance

CH

C

CH

CC

TTT

QQQ

WQ

K

=

== ( Carnot refrigerator)

RefrigeratorsAir conditioner

A air conditioner works on exactlythe same principle as a refrigerator.

For air conditioners, the quantities ofgreatest practical importance are therate of heat removal (the heat currentH from the region being cooled) andthe power input P=W/t.

PH

PtHt

QQQ

WQ

KCH

CC ==

==

H in Btu/h, P in watts

H/P (energy efficiency rating EER)in (Btu/h)/W. Note 1 W=3.413 Btu/h.EER~7-10.

The second law of thermodynamicsThe second law of thermodynamics

It is impossible for any system to undergo a process in which it absorbsheat from a reservoir at a single temperature and converts the heatcompletely into mechanical work, with the system ending in the samestate in which it began (Kelvin-Planck statement)

The kinetic and potential (due to interactions between molecules)energies associated with random motion constitute the internal energy.

When a body sliding on a surface comes to rest as a result of friction,the organized motion of the body is converted to random motion ofmolecules in the body and in the surface.

Since we cant control the motions of individual molecules, we cantconvert this random motion COMPLTELY back to organized motion.

It is impossible for any process to have as its sole result the transfer ofheat from a cooler to a hotter body (Clausius statement).

The second law of thermodynamics (contd)

Proof : Kelvin-Planck statement = Clausius statement

(a)A workless refrigerator violates2nd law (K-P statement). If it existed,it could be used to make a 100%efficient engine, which violate K-Pstatement.

(b) An impossible 100% efficient engineviolates C statement. If it existed, itcould be used to make a worklessrefrigerator, which violates C state-ment.

K-P statement C-statement

impossible impossible

The second law of thermodynamics (contd)The second law of thermodynamics

The conversion of work to heat, as in friction or viscous fluid flow,and heat flow from hot to cold across a finite temperature gradient,are irreversible processes.

Two equivalent statements of the 2nd law state that these processescan be only partially reversed

Gases, for example, always seep spontaneously through an openingfrom a region of high pressure to a region of low pressure.

The 2nd law is an expression of the inherent one-way aspect of thisand many other irreversible processes.

In any process in which the temperature of the working substanceof the engine is intermediate between TH and TC , there must beNO heat transfer between the engine and either reservoir becausesuch heat transfer could not be reversible.

Any process in which the temperature of the working substancechanges must be adiabatic.

As heat flow through a finite temperature drop is an irreversible,during heat transfer, there must be NO finite temperature difference.

To have the maximum efficiency of a heat engine, we must avoidall irreversible processes. Answer: Carnot cycle

The Carnot cycle

The Carnot cycle

reversible expansion

reversible compression

adiabatic expansionadiabatic compression

The Carnot cycle

The Carnot cycle : ideal gas approximation 0,0: >>= WQQba H

0,0:

The Carnot cycle (contd)

The Carnot refrigerator A reversed Carnot engine = a Carnot refrigerator

H

C

H

C

HC

HC

CH

CC

TT

QQ

QQQQ

QQQ

WQ

K =

=

== ,/1

/

CH

C

TTTK

= coefficient of performance of a Carnot refrigerator

When the temperature difference is small, K is much larger thanunity, in which case a lot of heat can be pumped from the lowerto the higher temperature with only a little expenditure of work.

The Carnot cycle (contd)

The Carnot cycle and the second law

impossibleMore efficientthan a Carnotengine

A Carnot refrigeratorSince each step in the Carnotcycle is reversible, the entirecycle may be reversed. If yourun the entire backward, theengine becomes a refrigerator.

The Carnot cycle (contd)

The Carnot cycle and the second law (contd)

The refrigerator does work W=-|W|,takes in heat QC from the cold reservoir,and expels hear |QH| to the hot reservoir.

The superefficient heat engine expels heat|QC|, but to do so, it takes in a greateramount of heat QH+. Its work output isthen W+.

The net effect of the two machines togetheris to take a quantity of heat and convertit completely into work, which violates the2nd law.

No engine can be more efficient than a Carnot engine. All Carnot engines operating between the same two temperatureshave the same efficiency, irrespective of the nature of the workingsubstance.

Entropy

Entropy and disorderEntropy provides a quantitative measure of disorder.

Define the infinitesimal entropy change dS during an infinitesimalreversible process at absolute temperature T as:

TdQdS =

If a total amount of heat Q is added during a reversible isothermalprocess at absolute temperature T, the total entropy change is givenby:

TQSSS == 12

Higher temperature means greater randomness of motion Adding heat Q causes a substantial fractional increase in molecularmotion and randomness

Entropy (contd)Entropy and disorder (contd)

Generalization of definition of entropy change is to include ANYreversible process leading from one state to another, whetherit is isothermal or not:

Let us represent the process as a series of infinitesimal reversiblesteps. During a step, an infinitesimal quantity of heat dQ is addedto the system at absolute temperature T. Then the change ofentropy for the entire process is:

=2

1 TdQS entropy change in a

reversible process

The change in entropy does not depend on the path leading fromthe initial to the final state but is the same for all possible processes.

Since entropy is a function only of the current state of the system,we can compute entropy change in irreversible processes using theabove formula. ( we need to invent a path connecting the given initialand final state that consist entirely of reversible, equilibrium processesand compute the total entropy change as for the actual path).

Entropy (contd)

Entropy in cyclic processes

For any reversible process(e.g. Carnot cycle):

0= TdQ

==

=+

b

aII

a

bII

b

aI

a

bII

b

aI

TdQ

TdQ

TdQ

TdQ

TdQ 0

The entropy of a system ina given state is independentof the path taken to get there,and is thus a state variable.

Entropy (contd)Entropy in irreversible processesConsider a combined system with a Carnot engine anda cyclic system.

heat reservoir TR

RdQ

Carnotengine rev

dW

TCyclicsystem

sysdW

dQ

CsysrevR dUdWdWdQ =+ )(

CdW

RRR

R dQT

dQTT

dQT

dQ==

com

bine

d cy

clic

sys

temCarnot

cycle

CRC dUTdQTdW =

== )(0, cyclicdUdUTdQTW CCRC Q

Entropy (contd)Entropy in irreversible processesConsider a combined system with a Carnot engine anda cyclic system.

heat reservoir TR

RdQ

Carnotengine rev

dW

TCyclicsystem

sysdW

dQ

com

bine

d cy

clic

sys

tem

== )(0, cyclicdUdUTdQTW CCRC Q

= TdQTW RC

If WC > 0, a cyclic device exchangingheat with a single heat reservoir andproducing an equivalent amount of work.- Violation of K-P statement of 2nd law-

00 TdQWC

-the 2nd law-The equality is true for areversible process.Clausius inequality

Entropy (contd)Entropy in irreversible processes (contd)

Processes: aAb and aBb reversibleProcess : aCb irreversible T

S

A

B

C

A reversible heat transfer causes changesin entropy of both the system and thereservoir (at least one needed except foradiabatic process)

In an irreversible heat transfer process,there is a finite source of energy insteadof an energy reservoir, and the temperaturedifference during the heat transfer process is finite.

TdQdS

TdQSSS

subtractedT

dQT

dQTdQ

TdQ

TdQ

TdQ

TdQ

bCax

baba

bBaaAbbCax

bBa

bCax

aAbaAbCax

>=

=+=0.

00 =+= aAb bBaaAbBa TdQ

TdQ

TdQ

Entropy (contd)Entropy and the second law

When all systems taking part in a process are included, the entropyeither remains constant or increases.

ORNo process is possible in which the total entropy decreases, when allsystems taking part in the process are included.

ExercisesProblem 1

A Carnot engine takes 2000 J of heat from a reservoir at 500 K, doessome work, and discards some heat to a reservoir at 350 K.How much work does it do, how much heat is discarded, and whatis the efficiency?

SolutionThe heat discarded by the engine is:

.1400500350)2000( J

KKJ

TTQQ

H

CHC ===

From the first law, the work done by the engine is:

.600)1400(2000 JJJQQW CH =+=+=

The thermal efficiency is:

30.050035011 ===

KK

TT

H

C

ExercisesProblem 2 One kilogram of water at 0oC is heated to 100oC. Compute itschange in entropy.

SolutionIn practice, the process described would be done irreversibly, perhapsby setting a pan of water on an electric range whose cooking surface ismaintained at 100oC. But the entropy change of the water depends onlyon the initial and final states of the system, and is the same whetherthe process is reversible or not. Hence we can imagine that the temp.of the water is increased reversibly in a series of infinitesimal steps,in each of which the temp is raised by an infinitesimal amount dT.

======

2

1

./1031.1 31

22

112

T

TKJ

TTnmc

TdTmc

TdQSSS

mcdTdQ

l

Example: Refrigerator

A refrigerator pumps heat from the inside of the freezer (-5C) to the room (25C). What is the maximum coefficient of performance?

KCT

KCTo

H

oL

29827325

2682735

=+=

=+=

9.8268298

268=

=

=

KKK

TTTK

CH

Cideal

i.e. 8.9 Joules of heat would be pumped from the freezer for every Joule of work done by the compressor. (typical K = 3-5)

ExercisesProblem 3

What is the entropychange in a freeexpansion process, when the volume isdoubled.

SolutionThe work done by n moles of ideal gas in an isothermal expansionfrom V1 to V2 is: )/( 12 VVnnRTW l=

.22 nnRTVVnnRTWQ ll ===

Therefore the entropy change for n=1 is:

./76.5)2)](/(314.8)[1(2 KJnKmolJmolnnRTQS ==== ll

ExercisesProblem 4A physics student immerses one end of a copper rod in boiling water atat 100oC and the other end in an ice-water mixture at 0oC. The sides of therod are insulated. After steady-state conditions have been achieved in therod, 0.160 kg of ice melts in a certain time interval. For this time interval find(a) the entropy change of the boiling water; (b) the entropy change of theice-water mixture; (c) the entropy change of the copper rod; (d) the totalentropy change of the entire system. Solution(a) S=Q/T=-mLf/T=-(0.160 kg)(334x103 J/kg)/(373.15 K)=-143 J/K.(b) S=Q/T=mLf/T=(0.160 kg)(334x103 J/kg)/(273.15 K)= 196 J/K.(c) From the time equilibrium has been reached, there is no net heat

exchange between the rod and its surroundings, so the entropychange of the copper rod is zero.

(d) 196 J/K-143 J/K=53 J/K.

ExercisesProblem 5

Solution

An experimental power plant at the Natural Energy Lab generateselectricity from the temperature gradient of the ocean. The surface anddeep-water temperatures are 27oC and 6oC, respectively. (a) What isthe maximum theoretical efficiency of this power plant? (b) If the powerplant is to produce 210 kW of power, at what rate must heat be extractedfrom the warm water? At what rate must heat be absorbed by the coldwater? (c) The cold water that enters the plant leaves it at a temperatureof 10oC. What must be the flow rate of cold water through the system?

(a) %0.715.30015.2791 ==

KK

(b) .8.2)210)(1/1(2100.3,0.3070.0/210/ MWkWkWMWMWkWPout ====

(c) ./106/106)4)](/(4190[)/3600)(108.2(/ 556 hhkg

KKkgJhsW

TcdtQd

dtdm

l==

=

=

Chapter 19: The Second Law of ThermodynamicsExercisesExample: Refrigerator