Solucion Fisica Un_20

7/27/2019 Solucion Fisica Un_20

1/25

20.1: a) %.8.33338.0b)J.6500J4300J2200 65002200 ===+

20.2: a) %.9.28289.0b)J.2600J6400J9000 J9000J2600 ===

20.3:a) %.0.23230.0100,16

3700

== b) J.12,400J3700J100,16 =

c) .g350.0kgJ1060.4

J100,164 =

d) hp.298kW222s)60.0J)(3700( ==

20.4: a) J.1043.6 5)280.0(s)W)(1.0010180(1

3

=== PtQe

b) J.104.63s)W)(1.0010180(J1043.6 535 == PtQ

20.5: a) MW.970MW330MW1300b)%.2525.0MW1300MW330 ====e

20.6: Solving ,for(20.6)Eq. r

or)1ln(ln)1( er =

.8.13)350.0()1( 5.211

=== er

If the first equation is used (for instance, using a calculator without the yx function), notethat the symbol e is the ideal efficiency , not the base of natural logarithms.

20.7: a) C.453K726K)(9.5)15.295( 0.401 ==== ab rTT

b) Pa.1099.1Pa)(9.50)1050.8( 64 === ab rpp

20.8: a) From %.5858.0)8.8(11(20.6),Eq. 40.01 ==== re

b) %,60)6.9(1 40.0 = an increase of 2%. If more figures are kept for theefficiencies, the difference is 1.4%.

20.9: a) J.1062.1 410.2

J1040.3 4 === K

QCW

b) J.1002.5)1( 41CCH =+=+= KQWQQ


2/25

20.10: )(1

fC TcL

t

m

KtK

Q

t

WP p +

=

=

=

( ) W.128K)K)(2.5kgJ(485kg)J1060.1(s3600

kg0.8

8.2

1 5 =+

=

20.11: a) ,b)W.767s0.60

J1080.9J1044.1 45PHEER ==

or

.27.7)413.3(W767

W1633)413.3(

]s)(60J)108.9(s)(60J)1044.1([

s)(60J)108.9(45

4

==

=EER

20.12: a) )( waterwatericeiceC TcTcLmQ f ++=

( )J.1090.8

K)K)(25.0kgJ4190(K)K)(5.0kgJ2100(kgJ10334kg)80.1(5

3

=++=

b) J.1037.3 540.2

J1008.8|| 5C === K

QW

c) ==+=+= ||that(noteJ101.14J108.08J1037.3|||| H655

CH QQWQ

).)1(|| 1C KQ +

20.13: a) J.215J335J550|||| CH == QQ

b) K.378J)550JK)(335620(|)|||( HCHC === QQTT

c) %.39J)550J335(1)||||(1 HC == QQ

20.14: a) From Eq. (20.13), the rejected heat is J.103.72J)6450)(( 3K520K300 =

b) J.102.73J103.72J6450 33 = c) From either Eq. (20.4) or Eq. (20.14), e=0.423=42.3%.

20.15: a)C

Hf

C

HCH ||||

T

TmL

T

TQQ ==

J,10088.3K)(273.15K)(287.15kg)J10kg)(3340.85( 73 ==

or === ))(1(||||||||b)figures.twotoJ1009.3 HHCH7

TTQQQW C

J)1009.3( 7 J.1049.2))15.29715.273(1( 6=


3/25

20.16: a) From Eq. (20.13), J.492J)415)(( K270K320 = b) The work per cycle is

J,77J415J492 = and W,212)75.2(s1.00

J77 ==P keeping an extra figure.

c) 5.4.K)(50K)270()( CHC == TTT

20.17: For all cases, .|||||| CH QQW = a) The heat is discarded at a higher

temperature, and a refrigerator is required; == J)1000.5()1)((|||| 3CHC TTQW

J.665)1)15.263298.15(( = b) Again, the device is a refrigerator, andJ.190)1)15.263/15.273((|||| C == QW c) The device is an engine; the heat is taken

form the hot reservoir, and the work done by the engine is = J)1000.5(|| 3W J.285)1)15.263248.15(( =

20.18: For the smallest amount of electrical energy, use a Carnot cycle.

( )J102.09

K)J10kg)(334(5.00K)20(4190kg)(5.00

6

3

KkgJ

FwaterfreezeC0water toCoolin

=

+=+=+=

mLTmcQQQ

Carnot cycle:K293K268

J1009.2 out6

hot

out

cold

in Q

T

Q

T

Q=

=

room)theJ(into1028.2 6out =Q

J1009.2J1028.2 66inout == QQW

energy)alJ(electric1095.15

=W

20.19: The total work that must be done is

J104.90m)100)(smkg)(9.80500( 52tot === mgyW

J250H =Q Find CQ so can calculate workW done each cycle:

HT

T

Q

Q C

H

C =

[ ] J7.120K)(773.15K)(373.15J)250()( HHCC === QTTQ J3.129HC =+= QQW

The number of cycles required is cycles.3790J3.129

J1009.4 5tot =

=W

W


4/25

20.20: For a heat engine, ( ) ( ) J,7500600.01)J3000(1/CH === eQQ J.4500)J7500)(600.0(thenand H === eQW This does not make use of the given value

of ( ) ( )( ) K320600.01K8001engine,Carnotaforthenused,isIf. HCHH === eTTTT ,/and CHCH TTQQ = which gives the same result.

20.21: ( ) J10336.1J/kg10334kg0400.0 43fC === mLQ

( ) ( ) ( ) ( )[ ]J1089.4

J10825.1K15.273K373.15J10336.13

HC

44CCHH

H

C

H

C

=+=

+===

=

QQW

QTTQ

T

T

Q

Q

20.22:The claimed efficiency of the engine is .%58J1060.2

J1051.1

8

8

=

While the most efficient

engine that can operate between those temperatures has efficiency %.381 K400K250

Carnot ==e

The proposed engine would violate the second law of thermodynamics, and is not likely tfind a market among the prudent.

20.23: a) Combining Eq. (20.14) and Eq. (20.15),

( ) ( )( ).

1

11

1

/1

/

HC

HC

e

e

e

e

TT

TTK

=

=

=

b) As ),0(heatnoexhaustsengine)1(perfecta;0,1 C == QeKe and this isuseless as a refrigerator. As )0(uselessa;,0 = eKe engine does no work

),0( =W and a refrigerator that requires no energy input is very good indeed.

20.24:( ) ( )

( ).KJ428

k15.273

kgJ10334kg350.0a)

3

C

f

C

=

==T

mL

T

Q

.KJ392

K298.15

J1017.1b)

5

=

c) K.J36)KJ392(KJ428 =+=S (If more figures are kept in theintermediate calculations, or if K))298.151(K)15.2731(( = QS

K.J35.6used,is =S


5/25

20.25: a) Heat flows out of the C0.80 water into the ocean water and the C0.80 water cools to C0.20 (the ocean warms, very, very slightly). Heat flow for an isolatedsystem is always in this direction, from warmer objects into cooler objects, so this

process is irreversible.

b) kg100.0 of water goes form == TmcQisflowheattheandC20.0toC0.80 J102.154)60.0CK)(kgJkg)(4190(0.100 4=

This Q comes out of the 0.100 kg of water and goes into the ocean.For the 0.100 kg of water,

KJ78.02353.15)293.15ln(K)kgJkg)(4190100.0()ln( 12 === TTmcS

For the ocean the heat flow is J10154.2 4+=Q and occurs at constant T:

KJ76.85K293.15

J10154.2 4+=

==

T

QS

KJ7.7KJ85.76KJ02.78oceanwaternet +=+=+= SSS

20.26: (a) Irreversible because heat will not spontaneously flow out of 15 kg of water intoa warm room to freeze the water.

(b) roomice SSS +=

room

F

ice

F

T

mL

T

mL+=

K293kg)J10kg)(3340.15(

K273)kgJ10kg)(3340.15(

33

+=

KJ250,1+=

This result is consistent with the answer in (a) because 0>S for irreversible processes.

20.27: The final temperature will be

C,60kg)(3.00

C)kg)(80.0(2.00C)kg)(20.000.1(

=

+

and so the entropy change is

K.J4.47K353.15

K333.15lnkg)00.2(

K293.15

K333.15lnkg)(1.00K)kgJ4190( =

+


6/25

20.28: For an isothermal expansion,

K.J31.6isentropyofchangeThe.and0,0K293.15

J1850 =====T

QWQUT

20.29: The entropy change is .and, vmLQTQS == Thus,

K.J644K)216.4(

)kgJ10kg)(2.0913.0( 4=

=

=

T

mLS v

20.30: a) K.J1005.6 3K)15.373()kgJ10kg)(225600.1( 3v ====

T

mL

T

QS Note that this is the change

of entropy of the water as it changes to steam. b) The magnitude of the entropy changeis roughly five times the value found in Example 20.5. Water is less ordered (morerandom) than ice, but water is far less random than steam; a consideration of the densitychanges indicates why this should be so.

20.31: a) K.J109K)15.373(

)kgJ10kg)(2256100.18( 33v =

===

T

mL

T

QS

b) KJ8.72K)34.77(

kg)J10kg)(201100.28(:N

33

2 =

KJ2.102K)2466(

kg)J10kg)(233610(107.9:Ag

33

=

KJ6.86K)630(

kg)J10kg)(27210(200.6:Hg

33

=

c) The results are the same order or magnitude, all around KJ100 .The entropy

change is a measure of the increase in randomness when a certain number (one mole)goes from the liquid to the vapor state. The entropy per particle for any substance in a

vapor state is expected to be roughly the same, and since the randomness is much higherin the vapor state (see Exercise 20.30), the entropy change per molecule is roughly thesame for these substances.


7/25

20.32: a) The final temperature, found using the methods of Chapter 17, is

C,94.28K)kgJkg)(4190(0.800K)kgJkg)(39050.3(

)CK)(100kgJkg)(39050.3(=

+

=T

or C9.28 to three figures. b) Using the result of Example 20.10, the total change inentropy is (making the conversion to Kelvin temperature)

K.J2.49

K273.15

K302.09lnK)kgJkg)(4190800.0(

K373.15

K302.09lnK)kgJkg)(39050.3(

=

+

=S

(This result was obtained by keeping even more figures in the intermediate calculation.Rounding the Kelvin temperature to the nearest K01.0 gives the same result.

20.33: As in Example 20.8,

K.J74.6m0280.0

m0.0420lnK)molJ5mol)(8.31400.2(ln

3

3

1

2 =

=

=

V

VnRS

20.34:

a) On the average, each half of the box will contain half of each type ofmolecule, 250 of nitrogen and 50 of oxygen. b ) See Example 20.11. The total change inentropy is

ln(2))(ln(2)ln(2) 2121 kNNkNkNS +=+=

K.J105.74ln(2)K)J10381.1)(600( 2123 ==

c) See also Exercise 20.36. The probability is ( ) ( ) ( ) ,104.2212121 181600100500 == and is not likely to happen.

The numerical result for part (c) above may not be obtained directly on somestandard calculators. For such calculators, the result may be found by taking the log baseten of 0.5 and multiplying by 600, then adding 181 and then finding 10 to the power of

the sum. The result is then .104.21010 18187.0181 =


8/25

20.35: a) No; the velocity distribution is a function of the mass of the particles, thenumber of particles and the temperature, none of which change during the isothermal

expansion. b) As in Example 20.11, 231

1 wwN= (the volume has increased, and 12 ww < );

andln(3),)(3ln)ln( 12 NwwN == K.J18.3ln(3))3ln(ln(3) ==== nRknNkNS A

c) As in Example 20.8,( )

ln(3),ln12

nRVVnRS == the same as the expression used inpart (b), and K.J3.18=S

20.36: For those with a knowledge of elementary probability, all of the results for thisexercise are obtained from

( ) ,2

1

)!4(!

!4)1()(

4

== kk

ppkPknkn

k

where P(k) is the probability of obtaining kheads, 21

1and4 === ppn for a fair coin.This is of course consistent with Fig. (20.18).

a) ( ) ( ) 1614!4!0 !44!0!4 !4 2121 == for all heads or all tails. b) ( ) .21 41!3!1 !44!1!3 !4 == c) ( ) .21

834

2!2!4! = d) .122 8

341

161 =++ The number of heads must be one of 0, 1, 2, 3 or

4, and there must be unit probability of one and only one of these possibilities.

20.37: a) J300J,400H +=+= WQ J100so, HCHC ==+= QWQQQW

Since it is a Carnot cycle,H

C

H

C

T

T

Q

Q =

[ ] C73K200J)(400J)100(K)15.800()(HC =+=== HC QQTT

( )

cycles1034.3cycleJ100

J103.34

isrequiredcyclesofnumberthesoJ,100iscycleonefor

J1034.3kgJ10334kg0.10isrequiredTotalb)

46

63fC

=

==

CQ

mLQ


9/25

20.38: a) Solving Eq. (20.14) for,1

1, CHH

eTTT

= so the temperature change

( ) K.8.27600.0

1

55.0

1K15.183

1

1

1

1CHH =

=

=

eeTTT

b) Similarly, ( ) H,HHC ifand,1 TTeTT ==

( ) K.3.15600.0

050.0K15.183

1CCC =

=

=e

eeTTT

20.39: The initial volume is .m1062.8 331 11 ==

p

nRTV a) At point 1, the pressure

is given as atmospheric, and Pa,1001.1 51 =p with the volume found above, =1V

Pa1003.22and,m1062.8.m1062.8 5112

33

12

33

1

2 ===== pppVVT

T

( )( )

( )( )

%.50500.01isefficiencythermaltheK,600and

K300betweenoperatingenginecycle-CarnotaFor%.4.10104.0e)

J.227J1956J2183iscycleoneforenginetheintoflowheatthe2,-1processthe fornscalculatioin thefiguresextraKeepingd)J.227J559J786isdonenet work

Thec).J1956)K192)(KmolJ3145.8)(27)(mol350.0(

andJ1397)K192)(KmolJ3145.8)(25(J,559

)K192)(KmolJ3145.8)(mol350.0(isobaric;is1-3process

The0).(J786)K108)(KmolJ3145.8)(25)(mol350.0(

and,0adiabatic,is3-2processTheJ.1018.2K300KJ/mol3145.8

25mol350.0.0so0isochoric,is2-1Processb)

.m1041.1andPa1001.1Pa).10013.1using(

600300

J2183J227

3

3213

513

5

1

3

==

===

=

=

+===

====

====

>===

===

=====

=====

e

WUTnC

QnTnCU

TnRVpW

WTnCW

UQ

TnCQUWV

VVppp

p

V

V

V

T

T

a


10/25

20.40: (a) The temperature at point c is ,fromsinceK1000 nRTpVTc == the maximumtemperature occurs when the pressure and volume are both maximum. So

( )( )( ) ( )

mol.16.2K1000KmolJ3145.8

m0300.0Pa1000.6 35=

==c

cc

RT

Vpn

(b) Heat enters the gas along paths ab andbc, so the heat input per cycle is =HQ .

acacacUWQ += Path ab has constant volume and path bc has constant pressure, so

J1020.1)m0100.0m0300.0)(Pa1000.6()(0 4335 ==+=+= bccbcabac VVpWWWFor an ideal gas,

mol.J46.28,COFor.using,)()T( 2 ==== VaaccVacVac CRpVnTRVpVpCTnCUso

48.5))mPa)(0.01001000.2()mPa)(0.03001000.6((KmolJ3145.8

KmolJ46.28 3535 =

= acU

Then J.106.68J105.48J1020.1 444H =+=Q

(c) Heat is removed from the gas along paths cdandda, so the waste heat per cycle is.

cacacaCUWQQ +== Path cdhas constant volume and path da has constant pressure,

so

10400.0)m0300.0mPa)(0.01001000.2()(0 4335 ==+=+= daddacdca VVpWWW

From (b),

J.105.88J105.48J100.400soJ,1048.5 444C4 ==== QUU acca

(d) The work is the area enclosed by the rectangular path abcd,

J.8000J105.86J1068.6or),)(( 44CH ==+== QQWVVppW acac

(e) 0.120.J)10(6.68J)8000( 4H === QWe


11/25

20.41: a) K15.290K,15.268J,00.1 HC === TTW

For the heat pump 0Qand0 HC Q

H

C

H

CHC withthiscombining;

T

T

Q

QQQW =+= gives

J2.13290.15)268.15(1J00.1

1 HCH === TT

WQ

b) Electrical energy is converted directly into heat, so an electrical energy input of 13.2 Jwould be required.

c) From part (a), CHHC

H asdecrease1

TQTT

WQ

= decreases.

The heat pump is less efficient as the temperature difference through which the

heat has to be pumped increases. In an engine, heat flows from CH to TT and work isextracted. The engine is more efficient the larger the temperature difference throughwhich the heat flows.


12/25

20.42: (a) bcab QQQ +=in

caQQ =out

K600C327max ==== cb TTT

K200K)600(31 ==== b

b

aa

b

bb

a

aa TPPT

TVP

TVP

3

5Kmole

J

m0332.0Pa103.0

K)600)(1moles)(8.32( =

===

b

bbbbb

P

nRTVnRTVP

%6767.011)(

%2121.0J102.10

J104.4

J104.4J101.66J1010.20(b)

J101.66K)400(Kmole

J31.8

2

5moles)(2.00

J1010.2

J101.103lnK)600(Kmole

J31.8moles)(2.00

3lnln

J109.97K)400(KmoleJ31.823moles)2(

and:gasMonatomic

m0997.01

3)m0332.0(

K600K200

cannotmax

4

3

in

344outin

4out

4in

4

3

25

23

33

=====

==

==

===+=+=

=

===

=+=

=

=

=====

===

==

==

===

h

C

T

T

capca

bcab

c

bb

b

cb

bc

bbcbc

abVab

PV

a

c

bbc

c

cc

b

bb

eec

QWe

QQWWwUQ

TnCQQ

QQQ

nRTV

VnRTdV

V

nRTPdVWQ

TnCQ

RCRC

VP

PVV

T

VP

T

VP


13/25

20.43: a)

Pa1057.2m1000.5

K)K)(773molJ5mol)(8.31400.2(,

volume.minimumandpressuremaximumthehasstate(a),partFrome)

cycleeachheatofJ206wastesJ;206d)

58.8%J)500(J)294(c)

C45K318J)]500(J)206(K)[773()(

J206J500J294,

J294m)00.2)(smkg)(9.800.15(

J500b)

6

33

C

H

HCHC

H

C

H

C

HCHC

2

H

=

===

=

===

=+===

=

===+=

===

+=

V

nRTpnRTpV

a

Q

QWe

QQTT

T

T

Q

Q

QWQQQW

mgyW

Q

20.44: a) ===== kW210MW3.0MW,0.3b)%.0.71 0.070kW210

K300.15K15.279 out

e

pe

MW.2.8kW)210)(1(1 =e

c) hr.L106hrkg106K)(4K)kgJ4190(

hr)s3600(W)108.2( 556

==

== Tc

dtQd

dt

dm C

20.45: There are many equivalent ways of finding the efficiency; the method presentedhere saves some steps. The temperature at point 3 is soand,4 03 TT =

,2

192)3(

2

5)2)(2()( 00000000031313H VpVpnRTVVpTTnCWUQ V =+=+=+=

where 000 VpnRT = has been used for an ideal gas. The work done by the gas during one

cycle is the area enclosed by the blue square in Fig. (20.22), ,00VpW = and so the

efficiency is %.5.10192

H===

Q

We


14/25

20.46: a) 00.212 == pp atm,

L,00.6L.6.00L)(3/2)00.4( 2312 12 ===== VVVV

T

T 111.1)9/5(223 23 === ppp

T

Tatm,

67.1)2/3(334 43 === ppp

V

Vatm. As a check, 00.2)5/6(441 4

1 === pppT

Tatm. To

summarize,

=),( 11 Vp (2.00 atm, 4.00 L) =),( 22 Vp (2.00 atm, 6.00 L)=),( 33 Vp (1.111 atm, 6.00 L) =),( 44 Vp (1.67 atm, 4.00 L).

b) The number of moles of oxygen is ,1

11

RT

Vpn = and the heat capacities are those in

Table (19.1). The product J;4.810valuethehas11 =xVp using this and the ideal gas law,

J,14222)1J)(4.810)(508.3(1:i1

2 ==

==

T

Tx

R

CTnCQ PP

J.4052)1J)(4.810(11

21 ==

==

T

TxVpW

0.WJ,13553)2J)(4.810)(508.2(:ii1

23 ===

==

T

TTx

R

CTnCQ VV

WQV

V

T

Tx

V

VnRTW ===

=

= J,2743)2(ln6)5J)(4.810(lnln:iii

3

4

1

3

3

43

0.WJ,3396)1J)(4.810)(508.2(1:iv1

4 ===

==

T

Tx

R

CTnCQ VV

In the above, the terms are given to nearest integer number of joules to reduce roundoff

error.

c) The net work done in the cycle is J.131J274-J405 =

d) Heat is added in steps i and iv, and the added heat is J1761J339J1422 =+ and the

efficiency is 7.5%.or,075.0J1761J131 = The efficiency of a Carnot-cycle engine operating

between %.4444.01isK450andK250 450250 ==


15/25

20.47: a) ==== Pa)10363(J,106.52kJ1005kJ1657 35 VpWU

J,1039.8)m2202.0m4513.0( 433 = and so J.1036.7 5=+= WUQ b) Similarly,

VpUQ =H

)m0682.0m6Pa)(0.009410(2305kJ)1969kJ1171(

333

+= J.1033.9 5=

c) The work done during the adiabatic processes must be found indirectly (thecoolant is not ideal, and is not always a gas). For the entire cycle, ,0=U and so the net

work done by the coolant is the sum of the results of parts (a) and (b), J.1097.1 5 The

work done by the motor is the negative of this, .74.3d)J.1097.1J101.97

J1036.755

5

===

W

QcK

20.48: For a monatomic ideal gas, .Cand23

25 RRC VP ==

a) ab: The temperature changes by the same factor as the volume, and soJ.1025.2)mPa)(0.3001000.3)(5.2()( 535 ==== baa

PP VVp

R

CTnCQ

The work Vp is the same except for the factor of J.1090.0so, 525 =W

J.101.35 5== WQU

bc: The temperature now changes in proportion to the pressure change, and

J,1040.2)mPa)(0.8001000.2)(5.1()( 53523 === bbc VppQ and the work is zero

J.1040.2).0( 5=== WQUV

ca: The easiest way to do this is to find the work done first; Wwill be the negative of areain thep-Vplane bounded by the line representing the process ca and the verticals from

points a andc. The area of this trapezoid is + Pa)101.00Pa1000.3( 5521

J,1000.6)m500.0m(0.800 433 = and so the work is bemustJ.1060.0 5 U

0(sinceJ101.05 5 = U for the cycle, anticipating part (b)), and so Q must be

J.1045.0 5=+ WU b) See above; .0J,1030.0 5 === UWQ

c) The heat added, during process ab andca, is 2.25 J100.45J10 55 + J102.70 5= and the efficiency is %.1.11111.05

5

H 1070.2

1030.0 ===

QW


16/25

20.49: a) ab: For the isothermal process, ==== )ln(.0and0 1 ab VVnRTWUT

)ln(1/1 rnRT ),ln(1 rnRT= and ).ln(1 rnRTWQ == bc: For the isochoric process,).(;0and0 12 TTnCTnCUQWV VV ===== cd: As in the process ab,

).ln(and0 2 rnRTQWU === da: As in process bc, ;0and0 == WV

).( 21 TTnCQU V == b) The values ofQ for the processes are the negatives of eachother. c) The net work for one cycle is ),ln()( 12net rTTnRW = and the heat added(neglecting the heat exchanged during the isochoric expansion and compression, as

mentioned in part (b)) is ),ln(2cd rnRTQ = and the efficiency is ).(1 21 TTcdnet

Q

W = This is

the same as the efficiency of a Carnot-cycle engine operating between the twotemperatures.

20.50: The efficiency of the first engine isH

H

1 T

TTe

= and that of the second is ,C2 TTT

e =

and the overall efficiency is

.C

H

H21

==

T

TT

T

TTeee

The first term in the product is necessarily less than the original efficiency since ,CTT > and the second term is less than 1, and so the overall efficiency has been reduced.

20.51: a) The cylinder described contains a mass of air )L,dm 4( 2= and so the total

kinetic energy is .Lv)dK 228(= This mass of air will pass by the turbine in a time,vLt= and so the maximum power is

.)8(32

vdt

K

P ==

Numerically, the product .msW5.0mkg5.08( 543air =)

b) h.km50sm14m)97)(msW5.0(

)25.0(W)102.3(3/1

254

63/1

2==

=

=kd

ePv

c) Wind speeds tend to be higher in mountain passes.


17/25

20.52: a) ( ) h.L89.9gal1

L788.3

km1.609

mi1

mi25

gal1hkm105 =

b) From Eq. (20.6), %.5.57575.0)5.8(11 40.01 ==== re

c) ( )( )( ) hp.72.1W1038.5575.0kgJ1060.4Lkg740.0hrs3600

hL89.9 47 ==

d) Repeating the calculation gives hp,19W104.1 4 = about 8% of the maximumpower.


18/25

20.53: (Extra figures are given in the numerical answers for clarity.) a) The efficiency

is 611.01 40.0 == re , so the work done is J.78||andJ122 C == QeQH b) Denote the

length of the cylinder when the piston is at point a by 0L and the stroke as s. Then,

isvolumeand,100

0 sLrr

r

sL

L

==

.m100.51)m1025.41()m104.86(6.9

6.10

134233

0 ==

= sA

r

rAL

c) The calculations are presented symbolically, with numerical values substituted at the

end. At point a, the pressure is 344 m1010.5isvolumethePa,1050.8 == aa Vp as

found in part (b) and the temperature is ,isvolumethe,pointAtK.300 rVVbT aba == the

pressure after the adiabatic compression is ab rpp = and the temperature is1=

abrTT .

During the burning of the fuel, from ,to cb the volume remains constant and so

.rVVVabc

== The temperature has changed by an amount

( )( ) ( )

( )( )( ),

KmolJ5.20m1010.5Pa1050.8

J200KmolJ3145.8

344

HHH

aa

a

VaaVaaaV

TfT

TCVp

RQ

CRTVp

Q

nC

QT

=

=

===

wherefis a dimensionless constant equal to 1.871 to four figures. The temperature at c is

then ( )frTTfTT aabc

+=+= 1 . The pressure is found from the volume and

temperature, ( )frrpp ac += 1 . Similarly, the temperature at point dis found byconsidering the temperature change in going from dto a,

).)1(1(so,)1()1( HC feTTTfenC

Qe

nC

Qada

VV+=== The process from dto a is

isochoric, so .thatnotecheck,aAs).)1(1(and, cdadad rppfeppVV=+== To

summarize,

( )( ) ))1(1(11

)()( 11

1

feTVfepd

frTrVfrrpc

rTrVrpb

TVpa

TVp

aaa

aa

a

aa

a

aaa

++

++

Using numerical values (and keeping all figures in the intermediate calculations),


19/25

20.54: (a) waterandfurnaceforL

TAk

t

Q =

( )

sKJ0494.0

K313

1

K523

1K210

cm100

m1cm15

m65.0

)KmW5.79(

11

2

2

wf

wf

waterfurnace

+=

+

=

+

=

+=

+

=

TTL

TkA

T

LTkA

T

LTkA

t

S

t

S

t

S

(b) 0>S means that this process is irreversible. Heat will not flow spontaneously fromthe cool water into the hot furnace.


20/25

20.55: a) Consider an infinitesimal heat flow HdQ that occurs when the temperature of

the hot reservoir is :T

||||

)/(

HH

CC

HCC

HCC

STT

dQTQ

T

dQ

TdQ

dQTTdQ

C ==

=

=

b) The 1.00 kg of water (the high-temperature reservoir) goes from 373 K to 273 K.

J.106.2J104.19J1057.3Then

J1057.3soengine,theofoutcomes

J1057.3)KJ1308)(K273(||gives(a)partofresultThe

J/K1308)373273ln()KkgJ4190)(kg00.1()(lnS

J1019.4)K100)(KkgJ4190)(kg00.1(

455HC

5CC

5C

12

5H

=+=+=

=

==

===

===

QQW

QQ

Q

TTmc

TmcQ

h

c) 2.00 kg of water goes from 323 K to 273 K

J104.3

J1085.3||

KJ1041.1)323/273(ln)KkgJ4190)(kg00.2()(ln

J1019.4)K50)(KkgJ4190)(kg00.2(

4HC

5hCC

312

5H

=+=

==

===

===

QQW

STQ

TTmcS

TmcQ

h

d) More work can be extracted from 1.00 kg of water at 373 K than from 2.00 kg ofwater at 323 K even though the energy that comes out of the water as it cools to 273 K isthe same in both cases. The energy in the 323 K water is less available for conversioninto mechanical work.

20.56: See Figure (20.15(c)), and Example 20.8.a) For the isobaric expansion followed by the isochoric process, follow a path

from =+=== 21ln2lngettoddorddUse.to2to VppV nCnCSTnCQTnCQTTT

ln2.ln2)( nRCCnVp

=

b) For the isochoric cooling followed by the isobaric expansion, follow a path from2.lnln)(2lnlnThen.to2/to

21 nRCCnnCnCSTTT VppV ==+=


21/25

20.57: The much larger mass of water suggests that the final state of the system will bewater at a temperature between C.0.60andC0 This temperature would be

( )( )( )( ) ( )( )(

C,83.34)KkgJ4190)(kg650.0(

)kgJ10334

C0.15KkgJ2100kg0500.0

C0.45KkgJ4190kg0.600

3

=

+

=T

keeping an extra figure. The entropy change of the system is then

=318.15

307.98K)lnkgJkg)(4190(0.600S

K.J5.10

273.15

307.98lnK)kgJ4190(

K273.15

kgJ10334

258.15

273.15lnK)kgJ(2100

kg)0500.0(

3

=

+

+

+

(Some precision is lost in taking the logarithms of numbers close to unity.)

20.58: a) For constant-volume processes for an ideal gas, the result of Example 20.10may be used; the entropy changes are ).ln(and)ln( daVbcV TTnCTTnC b) The total

entropy change for one cycle is the sum of the entropy changes found in part (a); the

other processes in the cycle are adiabatic, withQ =

0 and .0=S The total is then.lnlnln

=+=

da

acV

d

aV

b

cV

TT

TTnC

T

TnC

T

TnCS

From the derivation of Eq. (20.6), ,and 11 d

ca

b TrTTrT == and so the argument of the

logarithm in the expression for the net entropy change is 1 identically, and the net entropychange is zero. c) The system is not isolated, and a zero change of entropy for anirreversible system is certainly possible.


22/25

20.59: a)

b) From Eq. (20.17), and,soand, dSTdQdST

dQ ==

== dSTdQQ

which is the area under the curve in the TSplane. c) HQ is the area under the rectangle

bounded by the horizontal part of the rectangle at HT and the verticals. || CQ is the area

bounded by the horizontal part of the rectangle atC

T and the verticals. The net work is

then |,| CH QQ the area bounded by the rectangle that represents the process. The ratioof the areas is the ratio of the lengths of the vertical sides of the respective rectangles, and

the efficiency is .H

CH

H T

TT

QWe

== d) As explained in problem 20.49, the substance that

mediates the heat exchange during the isochoric expansion and compression does notleave the system, and the diagram is the same as in part (a). As found in that problem, theideal efficiency is the same as for a Carnot-cycle engine.

20.60: a) K.J143)K15.373(kg)J10(334kg)160.0( 3f ====

T

mL

T

QS

K.J196b)K)15.273(

)kgJ10kg)(334160.0( 3f ==== T

mL

T

QS

c) From the time equilbrium has been reached, there is no heat exchange between therod and its surroundings (as much heat leaves the end of the rod in the ice as enters atthe end of the rod in the boiling water), so the entropy change of the copper rod iszero. d) K.J53KJ143KJ196 =


23/25

20.61: a) )ln( 12 TTmcS=

K.J150

K)293.15K5K)ln(338.1kgJkg)(419010250( 3

=

=

b) K.J120K15.393K)293.15KK)(338.15kgJkg)(419010250( 3

element ===

T

Tmc

S c) The sum of the result ofparts (a) and (b) is K.J30system =S d) Heating a liquid is not reversible. Whatever the

energy source for the heating element, heat is being delivered at a higher temperaturethan that of the water, and the entropy loss of the source will be less in magnitude thanthe entropy gain of the water. The net entropy change is positive.


24/25

20.62: a) As in Example 20.10, the entropy change of the first object is )ln( 111 TTcm and

that of the second is )ln(222

TTcm , and so the net entropy change is as given. Neglecting

heat transfer to the surroundings, ,0)()(,0 22211121 =+=+ TTcmTTcmQQ which is thegiven expression. b) Solving the energy-conservation relation for Tand substituting

into the expression forS

gives.1n1ln

2

1

222

1122

1

11

+

=

T

T

T

T

cm

cmcm

T

TcmS

Differentiating with respect to T and setting the derivative equal to 0 gives

+=

2

1

2

2211

222112211

)(1

)1)()((0

T

T

T

Tcmcm

Tcmcmcm

T

cm.

This may be solved for

,2211

222111cmcm

TcmTcmT +

+=

which is the same as T when substituted into the expression representing conservationof energy.

Those familiar with Lagrange multipliers can use that technique to obtain the relations

T

QS

TT

QS

T

=

=

,

and so conclude that TT = immediately; this is equivalent to treating the differentiationas a related rate problem, as

02211

=

+= dTTd

T

cm

T

cm

STd

d

and using TTcm

cm

dT

Td == gives22

11 with a great savings of algebra.

c) The final state of the system will be that for which no further entropy change ispossible. If ,TT < it is possible for the temperatures to approach each other whileincreasing the total entropy, but when ,TT = no further spontaneous heat exchange is

possible.


25/25

20.63: a) For an ideal gas, ,RCC VP += and taking air to be diatomic,

.and,57

25

27 === RCRC VP Referring to Fig. (20.6),

).()(27

27

H bbccbc VpVpTTRnQ == Similarly, ).(25

C ddaa VpVpRnQ = What needsto be done is to find the relations between the product of the pressure and the volume at

the four points.For an ideal gas, ,b

bb

c

cc

T

Vp

T

Vp = so ( ).a

c

T

T

aacc VpVp = For a compression ratio r, and

given that for the Diesel cycle the process ab is adiabatic,

.11

=

= aa

b

aaabb rVp

V

VVpVp

Similarly, .

1

=

a

cccdd

V

VVpVp Note that the last result uses the fact that process da is

isochoric, and bcad ppVV == also,; (process bc is isobaric), and so ( ).ac

T

T

bc VV = Then,

=

=

==

rT

T

V

V

VT

VT

T

T

V

V

T

T

T

T

V

V

T

T

V

V

a

c

b

a

bb

aa

a

c

b

a

b

a

a

b

a

b

b

c

a

c

1

1

Combining the above results,

2

a

caadd r

T

TVpVp

=

Subsitution of the above results into Eq. (20.4) gives

( )

=

1

21

7

51

r

r

a

c

T

T

T

Te a

c

,)167.3(

1)002.5(

4.1

11

40.0

56.0

=

r

r

where 4.1,167.3 ==a

c

T

Thave been used. Substitution of 0.21=r yields

%.8.70708.0 ==e

Solucion Fisica Un_20

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