chapter 05 entropy (pp 100-118)

CH 05 ENTROPY 100

CHAPTER 05 ENTROPY

5-1 INTRODUCTION

Example 5-1

Calculate the change in entropy when kJ25 of energy is

transferred reversibly and isothermally as heat to a large

block of ice at (a) C00 and (b) C

0100 .

Solution (a) The change in entropy is defined as

T

QS

∆=∆

KJS /922730

1025 3

=+

×=∆

(b) The change in entropy is defined as

T

QS

∆=∆

KJS /67273100

1025 3

=+

×=∆

Example 5-2

An ideal gas undergoes a reversible isothermal expansion at

C0132 . The entropy of the gas increases by KJ /2.46 . How

much heat is absorbed? P.U. B.Sc. 2007

Solution The change in entropy is defined as

T

QS

∆=∆

)273132)(2.46()( +=∆=∆ TSQ

KkJKJQ /71.18/10871.1 4 =×=∆

CH 05 ENTROPY 101

Example 5-3

The ends of a well insulated metal rod are made in contact

with two reservoirs at C060 and C

015− . During certain time

J200 of heat are transferred from hot reservoir to cold

reservoir. What are the entropy changes at the two

reservoirs and net change in entropy? K.U. B.Sc. 2000

Solution The entropy change in the hot reservoir is given by

kgJT

QS

H

H

H /601.0)27360(

200−=

+−=

∆=∆

The entropy change in the cold reservoir is given by

kgJT

QS

L

L

L /775.0)27315(

200=

+−=

∆=∆

The net change in the entropy of the system is

LH SSS ∆+∆=∆

kgJS /174.0775.0601.0 =+−=∆

Example 5-4

A kg1500 car traveling at hkm /100 crashes into a

concrete wall. If the temperature of the air is C020 ,

calculate the entropy change in the universe.

Solution The entropy change is defined as

T

vm

T

QS

)2/( 2

=∆

=∆

Now kgm 1500=

smsmhkmv /9

250/

3600

)1000)(100(/100 ===

KKCT 293)27320(200 =+==

Therefore

KJS /1975)293(2

)9/250)(1500( 2

==∆

CH 05 ENTROPY 102

Example 5-5

g250 of ice melts (reversible) to water, the temperature

remaining at C00 . Calculate the entropy change for ice if its

heat of fusion is kgkJ /334 . P.U. B.Sc. 2004, 2008

Solution For an isothermal and reversible process, the change in entropy

is given by

T

Lm

T

QS

f=

∆=∆

KJS /306273

)10334)(10250( 33

=××

=∆−

Example 5-6

A lump of ice whose mass is g235 melts (reversibly) to

water, the temperature remaining at C00 throughout the

process. What is the entropy change for the ice? The heat of

fusion of ice is kgkJ /334 . K.U. B.Sc. 2004

Solution The change in entropy is given by

T

Lm

T

QS

f=

∆=∆

Now kgkggm 13 1035.210235235 −− ×=×==

kgJKJKkJL f /1034.3/10334/334 53 ×=×==

KCT 27300 ==

Hence

KJS /288273

)1034.3)(1035.2( 51

=××

=∆−

Example 5-7

Calculate the change of entropy when g6.54 of ice at C00

is converted into water at C00 at normal pressure. The

latent heat of fusion of ice is 80 calories per gram.

B.U. B.Sc. 1997A

CH 05 ENTROPY 103

Solution The change in entropy at constant temperature is given by

T

Lm

T

QS

f=

∆=∆

KcaloriesS /273

)80)(6.54(=∆

KJKJS /976.66/273

)186.4)(80)(6.54(==∆

Jcalorie 186.41 =Θ

Example 5-8

Calculate the change of entropy when g3.27 of ice at C00

is converted into water at C00 at normal pressure. The

latent heat of fusion of ice is 80 calories per gram.

K.U. B.Sc. 1999

Solution The change in entropy at constant temperature is given by

T

Lm

T

QS

f=

∆=∆

Now kgkggm 23 1073.2103.273.27 −− ×=×==

gramcaloriesL f /80=

kgJkgJL f /103488.3/10

)186.4)(80( 5

3×==

−

KCT 27300 ==

Hence

kgJS /488.33273

)103488.3)(1073.2( 52

=××

=∆−

Example 5-9

An ice tray contains g500 of liquid water at C00 .

Calculate the change in entropy of water as it freezes slowly

and completely at C00 .

CH 05 ENTROPY 104

Solution For a freezing process

T

Lm

T

QS

f−=

∆=∆

KJS /612273

)1034.3)(10500( 53

−=××

−=∆−

Example 5-10

Calculate the change in entropy when kg10 of water at

C0100 is converted into steam at same temperature. The

latent heat of vapourization for water is kgJ /10256.2 6× .

Solution Now

T

Lm

T

QS v=

∆=∆

KJS /1005.6273100

)10256.2)(10( 46

×=+

×=∆

Example 5-11

Calculate the change of entropy when 46.7 grams of water

at C0100 are converted into steam at C

0100 at normal

pressure. The latent heat of vapourization for water at

C0100 is 537 calories/gram. B.U. B.Sc. (Hons.) 1984A


T

Lm

T

QS v=

∆=∆

74.10273100

)537)(46.7(=

+=∆S calories / K

KJKJS /96.44/)186.4)(74.10( ==∆

Jcalorie 186.41 =Θ

CH 05 ENTROPY 105

Example 5-12

The heat of vapourization of a compound at C027 is found

to be 1288.29 −

molkJ . Calculate the mole entropy of

vapourization at C027 . K.U. B.Sc. 1999


T

Lm

T

QS v=

∆=∆

KJS /6.97)27327(

)10288.29)(1( 3

=+

×=∆

Example 5-13

The melting point of copper is K2840 . What will be the

change in entropy when kg20 of copper melts? The heat of

vapourization of copper is 161073.4 −× kgJ .


T

Lm

T

QS v=

∆=∆

KJS /10331.32840

)1073.4)(20( 46

×=×

=∆

Example 5-14

Calculate the change in entropy when g25 of steam at

C0100 is converted into water at the same temperature. The

latent heat of vapourization for water is kgJ /10256.2 6× .

Solution

T

Lm

T

QS v−=

∆=∆

163

151)273100(

)10256.2)(1025( −−

−=+

××−=∆ KJS

The negative sign indicates that the entropy of the steam

decreases.

CH 05 ENTROPY 106

Example 5-15

What is the entropy change of kg500.0 of steam at

atm00.1 pressure and C0100 when it condenses to

kg500.0 of water at C0100 ?

Solution As steam condenses to water, therefore entropy change will be

negative and given by

T

Lm

T

QS v−=

∆−=∆

136

10024.3)273100(

)10256.2)(500.0( −×−=+

×−=∆ KJS

Example 5-16

For the Carnot cycle shown in Figure, calculate (a) the heat

that enters and (b) the work done on the system.

Solution (a) Heat enters the system along the top path and given by

STQ Hin ∆=

JQin 200)10.060.0)(400( =−=

(b) Heat leaves the system along the bottom path and given

by

CH 05 ENTROPY 107

STQ Lout ∆=

JQout 125)60.010.0)(250( −=−=

For a cyclic path

0=+WQ

JQQQW outin 75)125200(}{ −=−−=−−=−=

CH 05 ENTROPY 108

5-2 ENTROPY CHANGE FOR AN IDEAL GAS 5-2(A) ISOTHERMAL PROCESS

Example 5-17

An ideal gas undergoes an isothermal expansion at C077

increasing its volume from 3.1 to L4.3 . The entropy change

of the gas is KJ /24 . How many moles of gas are present?

Solution The heat added to the system, Q∆ , will be equal to the work

done by the gas in the form of isothermal expansion and is

given by

=∆

i

f

V

VnTRnQ λ

=

∆=∆

i

f

V

VnRn

T

QS λ

)3.1/4.3()314.8(24 nn λ=

moln

n 003.3)3.1/4.3()314.8(

24==

λ

Example 5-18

One mole of an ideal gas expands isothermally to ten times

its initial volume. Calculate the entropy change.


=

∆=∆

i

f

V

VnRn

T

QS λ

KJnS /1.19)10()314.8)(1( ==∆ λ

Example 5-19

Two moles of an ideal gas undergoes a reversible isothermal

expansion from 3028.0 m to

3042.0 m at a temperature

of C025 . What is the change in entropy of the gas?

CH 05 ENTROPY 109


=

∆=∆

i

f

V

VnRn

T

QS λ

KJnS /742.6028.0

042.0)314.8)(2( =

=∆ λ

Example 5-20

Five moles of an ideal gas are allowed to expand

isothermally at C027 from a pressure of atm20 to atm4 .

Calculate the entropy change.

Solution The entropy change at constant temperature and in terms of

initial and final pressures is given by

=∆

f

i

p

pnRnS λ

KJnS /9.60)4/20()314.8)(5( ==∆ λ

Example 5-21

A sample of nitrogen gas of mass g25 at K250 and

atm2 expands isothermally until its pressure is atm3.4 .

Calculate the change in entropy of the gas. The molar mass

of nitrogen is molg /28 .

Solution The desired change in entropy is given by

=∆

f

i

p

pnRnS λ

KJnS /5.163.4

1.21)314.8(

28

25=

=∆ λ

CH 05 ENTROPY 110

5-2 ENTROPY CHANGE FOR AN IDEAL GAS 5-2(B) ISOBARIC PROCESS

Example 5-22

One kilogram of water C00 is heated to C

0100 . Find the

change in entropy. Specific heat of water is 114200 −− KkgJ .

B.U. B.Sc. 1986A


=∆

i

f

T

TnCmS λ

Now kgm 1=

114200 −−= KkgJC

KCTi 27300 ==

KKCT f 373)273100(1000 =+==

Hence

kgJnS /1311)273/373()4200)(1( ==∆ λ

Example 5-23

Calculate the entropy increase when kg5 of water from

C020 to C

080 . Given that specific heat may be assumed to

have the constant value of 1131091.4 −−× KkgJ .


=∆

i

f

T

TnCmS λ

kgJnS /10902.327320

27380)1019.4)(5( 33 ×=

+

+×=∆ λ

CH 05 ENTROPY 111

Example 5-24

Calculate the change in entropy when a body of mass mg5

is heated from C0100 to C

0300 . Given specific heat of the

body 111.0 −−= Kgcal . K.U. B.Sc. 2006


=∆

i

f

T

TnCmS λ

Now kggmgm 63 1051055 −− ×=×==

11

3

11

10

)186.4)(1.0(1.0 −−

−

−− == KkgJKgcalC

11210186.4 −−×= KkgJC

KKCTi 373)273100(1000 =+==

KKCT f 573)273300(3000 =+==

)373/573()10186.4)(105( 23 nS λ××=∆ −

kgJS /899.0=∆

CH 05 ENTROPY 112

5-2 ENTROPY CHANGE FOR AN IDEAL GAS 5-2(C) ISOCHORIC PROCESS

Example 5-25

Two moles of a monoatomic ideal gas is warmed slowly

from K300 to K400 at constant volume. Determine the

entropy change of the gas.

Solution The entropy change at constant volume is given by

=∆

i

f

vT

TnCnS λ

=∆

i

f

T

TnRnS λ

2

3 Θ For monoatomic gas RCv

2

3=

KJnS /175.7)300/400()314.8)(2(2

3==∆ λ

Example 5-26

An aluminum block of kg10 mass is heated from K300 to

K800 . Calculate the change in entropy if 11900 −−= KkgJCv for aluminum.

Solution The desired change in entropy is given by

=∆

i

f

vT

TnCmS λ

KJnS /10827.8)300/800()900)(10( 3×==∆ λ

CH 05 ENTROPY 113

5-3 MISCELLANEOUS

Example 5-27

gms100 of ice at C00 is heated to C

0100 . Calculate the

change in entropy if the specific heat of water is

)/(4200 0CkgJ • . B.U. B.Sc. (Hons.) 1987A

Solution

The increase in entropy during melting of g100 of ice at C00

into water at C00 is given by

11

11

T

Lm

T

QS

f=

∆=∆

KJS /3.122273

)1034.3)(100.0( 5

1 =×

=∆

When g100 of water is heated from C00 to C

0100 , the

increase in entropy is given by

=∆

1

22

T

TnCmS λ

KJnS /1.131273

273100)4200)(100.0(2 =

+=∆ λ

The net change in entropy is given by

21 SSS ∆+∆=∆

KJS /4.2531.1313.122 =+=∆

Example 5-28

Calculate the change of entropy when kg10 of ice at C00 is

changed into water at C02 .

(Heat of fusion of ice KJ /1034.3 5×= and

Specific heat of water )/(4200 KkgJ •= )

K.U. B.Sc. 2001, 2003

CH 05 ENTROPY 114

Solution

When kg10 of ice at C00 is melted into water at C

00 , the

increase in entropy is given by

11

11

T

Lm

T

QS

f=

∆=∆

KJS /10223.1273

)1034.3)(10( 45

1 ×=×

=∆

When kg10 of water is heated from C00 to C

02 , the increase

in entropy is given by

=∆

1

22

T

TnCmS λ

KJnS /10066.3273

2732)4200)(10( 2

2 ×=

+=∆ λ

The desired entropy change is

21 SSS ∆+∆=∆

KJS /10254.110066.310223.1 424 ×=×+×=∆

Example 5-29

g200 of aluminum at C0107 is mixed with g50 of water at

C017 . Calculate the equilibrium temperature and change of

entropy of the system. (Specific heats of aluminum and

water are 11910 −− KkgJ and 114200 −− KkgJ )

K.U. B.Sc. 2002

Solution

Now kggm 2.02001 ==

KKCT 380)273107(1070

1 =+==

11

1 910 −−= KkgJC

kggm 05.0502 ==

KKCT 290)27317(170

2 =+==

11

2 4200 −−= KkgJC

Let T be the final temperature of the mixture, then

CH 05 ENTROPY 115

Heat lost by aluminum = Heat gained by water

)()( 222111 TTCmTTCm −=−

)290)(4200)(05.0()380)(910)(2.0( −=− TT

6090021018269160 −=− TT

TT 1822106090069160 +=+

T392130060 =

CKT059332

392

130060===

The change of entropy when the temperature g50 of water

rises from 2T to T is given by

=∆

2

221T

TnCmS λ

KJnS /4.28)290/332()4200)(05.0(1 ==∆ λ

The positive sign indicates gain in entropy. Similarly the change

of entropy in g200 of aluminum, when its temperature falls

from 1T to T , is given by

=∆

1

112T

TnCmS λ

KJnS /6.24)380/332()910)(2.0(2 −==∆ λ

The negative sign indicates the loss in entropy. Hence the

change of entropy of the system is

21 SSS ∆+∆=∆

KJS /8.36.244.28 =−=∆

Example 5-30

At very low temperatures, the molar specific heat of many

solids is (approximately) proportional to 3T , that is, 3

TACv = where '' A depends on particular substance. For

aluminum we have )/(1015.3 44 KmolJA •×= − . Find the

entropy change of mol8.4 of aluminum when its

temperature is raised from K0.5 to K10 .

CH 05 ENTROPY 116


∫∫ ==∆2

1

2

1

T

T

v

T

TT

dTCm

T

dQS

∫∫ ==∆2

1

2

1

23 T

T

T

T

dTTAmT

dTTAmS

[ ] [ ]3

1

3

2

3

3

1

3

1 2

1TTAmTAmS

T

T −==∆

[ ] KJS /441.0)5()10()1015.3)(8.4(3

1 334 =−×=∆ −

CH 05 ENTROPY 117

ADDITIONAL PROBLEMS (1) Calculate the change of entropy when 3.37 gms of

water at C0100 are converted into steam at same

temperature at normal pressure. The latent heat of

steam of steam at C0100 is 537 calories per gm.

B.U. B.Sc. 2009A

(2) g100 of ice at C00 is heated to C

0100 . Find the


)/(4200 KkgJ • . B.U. B.Sc. (Hons.) 1987A

(3) Calculate the change in entropy when 3.37 grams of

water at C0100 are converted into steam at C

0100 .

The latent heat of steam at C0100 is 537 calories per

gram. B.U. B.Sc. (Hons.) 1988A

(4) Twenty-five kilograms of water at C015 is heated to

C0100 . Calculate the change in entropy if the

specific heat of water is )/(4200 KkgJ • .

B.U. B.Sc. 1999A

(5) Calculate the change of entropy when 3.27 grams of

ice at C00 are converted into water at C

00 , at

normal pressure. The latent heat of fusion of ice is

80 calories per gram. K.U. B.Sc. 1999

(6) kg10 of ice at C00 is heated to C

0100 . Find the


)/(4200 KkgJ • . B.U. B.Sc. 2001A

(7) Calculate the change in entropy when g50 of ice at

C00 changes to water at C

02 . (Heat of fusion of ice

kgJ /1034.3 5×= and specific heat of

water )/(4200 KkgJ •= . K.U. B.Sc. 2005

(8) An ideal gas is confined to a cylinder by a piston. The

piston is pushed down slowly so that the gas

temperature remains at C020 . During the

compression 730 joules of work is done on the gas.

Find the entropy change of the gas. K.U. B.Sc. 2007

CH 05 ENTROPY 118

(9) Calculate the change of entropy when 3.27 gms of

water at C00 are converted into steam at same

temperature at normal pressure. The latent heat of

steam of steam at C0100 is 537 calories per gm.

B.U. B.Sc. 2009S

(10) Calculate the change of entropy when 250 gm of ice

is melted at C00 . Given that

51034.3 × joules of energy is

required to melt 1 kg of ice at C00 .

B.U. B.Sc.(Hons.) 1988A

Answers

(1) KJorKcalories /49.15/7.3 (2) KJ /4.253

(3) Kcalories /7.53 or KJ /79.224 (4) 1410715.2 −× KJ

(5) Kcalories /8 or KJ /5.33 (6) KJ /10311.1 4×

(7) KJ /52.631 (8) KJ /49.2 (9) 8 calories/K

(10) J306

chapter 05 entropy (pp 100-118)

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