Tài liệu Physics exercises_solution: Chapter 17 pdf

17.46: a Let the man be designated by the subscript m and the “‘water” by w, and T is the final equilibrium temperature... 17.63: The steam both condenses and cools, and the ice melts an

17.7: 1K1C 59F, so a temperature increase of 10 K corresponds to an increase

of 18F Beaker B has the higher temperature

17.8: For (b),  10.0C.Then for (a),    5910.0C

C 5

9 F K

17.10: (In these calculations, extra figures were kept in the intermediate calculations

to arrive at the numerical results.) a)TC 400273.15127C,TF (9/5)(126.85)

F.1079.232)1055.1)(

5/9(C,1055.115.27310

)15.178)(

5/9(C,17815

.27395b)

F

260

32

7 7

F 7 7

C

F C

T T

Fahrenheit scale extends from 460F to32F and conclude that the triple point is about 492 R.

17.15: From the point-slope formula for a straight line (or linear regression, which,

while perhaps not appropriate, may be convenient for some calculators),

C,33.282Pa

1080.4Pa106.50

Pa1080.4)

C0.100()C01.0

b) Equation (17.4) was not obeyed precisely If it were, the pressure at the triple

15

37310 Pa50

17.20: α T (2.010 5(C) 1)(5.00C19.5C)2.910 4.

17.21: ( ) (  )2.310 4 m  40.12510 2m 25.0C 

0 T L L α

K 10 1 5 10 50

17.24: The temperature change is T 18.0C32.0C14.0C The volume of ethanol contracts more than the volume of the steel tank does, so the additional amount of ethanol that can be put into the tank is VsteelVethanolβsteelβethanolV0T

 

3.6105 C175105(C)1

 2.80m3 14.0C0.0280m3

17.25: The amount of mercury that overflows is the difference between the volume

change of the mercury and that of the glass;

1000cm  55.0  1.7 10  C .

cm95.8K

100

3

3 1

5 glass

2 2

17.28: (a ) No, the brass expands more than the steel.

(b) call D the inside diameter of the steel cylinder at 20CAt 150C:DST DBR

cm026.25

)C130)(

)C(102.1(1

)C130)(

)C(10(2.01cm)25(1

)cm(1

25

)cm25(cm25T

Dcm000.25

1 5

1 5 ST

BR

BR ST

BR ST

T α D

T α

D α D

D D

)m10C)(2.01110

)(

)C(10Pa)(2.010

9.0(

4

2 4 1

5 11

K10Pa)(1.210

0.2



Yα

17.33: a) (37C(20C))(0.50L)(1.310 3 kg L)(1020J kgK)38J

b) There will be 1200 breaths per hour, so the heat lost is

J

104.6J)

)C7)(

KkgJ3480)(

kg70

Q

17.35: Using Q=mgh in Eq (17.13) and solving for Τ gives

.C53.0)kg.KJ4190(

)m225)(

sm80.9

17.36: a) The work done by friction is the loss of mechanical energy,

J

1054.1

)sm50.2(2

19.36sin)m00.8)(

sm80.9()kg0.35()(

2

1

3

2 2

2 2

2 1

1060

2 2

1 2

K T

17.39: 85.0C20.0C  1.50kg910J kgK  1.80kg4190J kgK 

5.79105 J

Q c

b) An overstimate; the heat Q is in reality less than the power times the time

gQ T

m

Q c

50.0

min5.1minJ000,10

Q c T mc Q Liquid

C30kg50.0

min5.1minJ000,10

min0.1minJ000,10:

T m

Q

c

Solid

17.46: a) Let the man be designated by the subscript m and the “‘water” by w, and T is

the final equilibrium temperature

w w w m m

mC T m C T

 mmCmTTmmwCwTTw

mmCmTm TmwCwT Tw

w w m m

w w w m m m

C m C m

T C m T C m

C

 , and the mass of water is 355 kg, we get

)kgJ4190(kg)355.0()kg

J3480(kg)0.70(

)0.12(kgJ4190(kg)355.0()0.37(KkgJ3480(kg)0.70

(

C C

C C

C T

measuring a body temperature due to cooling of the mouth

17.47: The rate of heat loss is   ,or Interestingnumbers,

) ( Q t

t mC

t T

mC t

t Q

or d,005.0

day J 10

7

C) C)(0.15 kg.

J kg)(3480

taking the temperature of a sick child several minutes after the child has something to

1043.1

kgJ10334K)0.18)(

KkgJ(4190kg)350.0(

17.49: Qm(ciceTice Lf cwaterTwater LV)

Btu

34.5kcal8.69J1064.3

kgJ102256K)

kgJ)(4190C

100(

kgJ10334)CK)(10.0kg

J(2100kg)100.12(

4

3

3 3

K) K)(15.0kg

Jkg)(2100550

.0(

Q t

01.7s)

(86,400

)kgJ10334()kglb2.205lb)4000

17.52: a) (   )(25.010 3kg)(4190J kgK)(66.0K)2256103 J kg

v

L T c m

6.33104 J b)mcT (25.010 3kg)(4190J kgK)(66.0K)6.91103J

c) Steam burns are far more severe than hot-water burns

17.53: With Qm(cT L )andK (1/2)mv2,settingQK

(130J Kg K)(302.3C  24.5 10 J kg) 357m s



v

17.54: a) 101g.

)kgJ10(2.42

K)K)(1.00kg

Jkg)(34800

.70(

6 v

b) This much water has a volume of 101 cm3,about a third of a can of soda

17.55: The mass of water that the camel saves is

kg,45.3)

kgJ10(2.42

K)K)(6.0kg

Jkg)(3480400

(

6 v

which is a volume of 3.45 L

17.56: For this case, the algebra reduces to

C

1.35)

.KkgJ4190)(

kg240.0(

)KkgJ390)(

kg1000.3)(

200((

)C0.20)(

kgJ4190)(

kg240.0(

)C0.100)(

KkgJ390))(

kg1000.3)(

200((

))KkgJkg)(4190(0.170

)KkgJkg)(390500

.0((

C))(85.0KkgJkg)(470(0.250

C))(20.0KkgJkg)(4190(0.170

)KkgJkg)(390500

17.58: The heat lost by the sample is the heat gained by the calorimeter and water, and

the heat capacity of the sample is

)Ckg)(73.9(0.0850

)C))(7.1KkgJkg)(390(0.150

)KkgJkg)(4190200

.0((

Q c

=1010J kgK,

or 1000J kg to the two figures to which the temperature change is known.K

17.59: The heat lost by the original water is

J,10714.4)C0.45)(

KkgJ4190)(

kg250.0

ice ice

T c L T c

Q m

KkgJ4190()kgJ10334()C0.20(K)kgJ(2100

J)10714.4(

kgJ10(334

K)))(19.5KkgJg)(2800(6.0

)KkgJg)(22500

.16((

3 f

)CK)(750kg

Jkg)(23400

17.62: Equating the heat lost by the lead to the heat gained by the calorimeter

(including the water-ice mixtue),

.)

()200

Pb b

K)kgJkg)(390100

.0(

K)kgJkg)(4190178

.0(

kgK)Jkg)(130750

.0(

kg)J10kg)(334018

.0(

)CK)(255kg

Jkg)(130750

.0(

17.63: The steam both condenses and cools, and the ice melts and heats up along with

the original water; the mass of steam needed is

kg

0.190

)C0.72)(

kgJ4190(kgJ102256

)CK)(28.0kg

Jkg)(4190(2.85

kg)J10kg)(334450

.0(

3

3 steam

WK]

mK(222

K)100)(

m10250.1(

s)100.60(s)

600(

kg)1050.8()kgJ10334(

L dt

dm L k

17.67: (Although it may be easier for some to solve for the heat flow per unit area,

part (b), first the method presented here follows the order in the text.) a) See Example 17.13; as in that example, the area may be divided out, and solving for temperature T at the boundary,

 foamfoam foamin  woodwood woodwoodout

foam

L k L

k N

T L k T L k

m102.2

C767.5C0.19m

W010

wood foam

m100.3

C0.10C

767.5Km W080

17.68: a) From Eq.17.21,

m100.4

K140m

1040Km W040

or 200Wto two figures b) The result of part (a) is the needed power input

17.69: From Eq 17.23, the energy that flows in time t is

BtuhFft30

F34ft

17.70: a) The heat current will be the same in both metals; since the length of the

copper rod is known,

m00.1

K0.35m10400Km W0

H

b) The length of the steel rod may be found by using the above value of Hin

Eq.17.21 and solving for L or, since 2, H and A are the same for the rods,

        0.242m

K0.35Km W0.385

K0.65Km W2.50m00.1

2 2

k L L

17.71: Using dm dt (seeProblem17.66)in Eq.(17.21),

v L

H 

kA

L dt

dm L



)m150.0)(

Km(50.2

m)1085.0(s)

180(

)kg390.0()kgJ102256

)m1098.4(4

4D

2D

m1098.4A

m0.500

K300K

m

W2.50sJ150

2

2 3

2 2

2 3

π πR A

A L

T kA t Q

17.73: Ha  Hb(aaluminum,bbrass)

b b

b a

T A k H L

T A

Solving

m500.0

C)0K)(

mW0.109(m

800.0

)C0.150K)(

mW

2050

(

C)0()

C0

T T

L

T A k L

T A

k

b

b a

17.75: Repeating the calculation with Ts 273K5.0C278K givesH 167 W

17.76: The power input will be equal to H as given in Eq (17.26);net

W

1054

4

)K)290(K)3000)((

KmW1067.5)(

35.0)(

m)1050.1(

4

(

)(

P

3

4 4

4 2 8

2 2

4 s 4

17.77:

K2450K

mW1067.50.35

17.78: The radius is found from

44

2

T πσ

H π

T σ H π

mW1067.54

W107

2 4

2 8

mW1067.54

W1010

2 4

2 8

17.79: a) normal melting point of mercury: 30C0.0

normal boiling point of mercury: 357C100.0

1C;

17.81: The tube is initially at temperature T has sides of length 0, L volume ,0 V0

density ,ρ0 and coefficient of volume expansion β

a) When the temperature increase to T0 T, the volume changes by an amount ,

m ρ

.1

0 0 0

0

0

T β

ρ ρ V T βV

using the expression 1xn  1nx, where n1,ρρ01βT

b) The copper cube has sides of length

C

50.0C

0.20C0.70and

m,.0125

kg





ρ

17.82: (a) We can use differentials to find the frequency change because all length

changes are small percents Let m be the mass of the wire

m FL L

m F μ F

mL

F L

m FL v

f

L

v f

2

12

)lfundamenta(

2 and

mL F f

f

mL F mL

())(

(

2 1

2 2

1 2

)()

FL

L m F m FL v

17.83: Both the volume of the cup and the volume of the olive oil increase when the

temperature increases, but β is larger for the oil so it expands more When the oil starts

to overflow,

A A

Voil V0,oilβoilT (9.9cm)AβoilT

Vglass V0,glassβ glassT (10.0cm)AβglassT

(9.9cm)Aβ T (10.0cm)Aβ T (1.00 10 3m)A

glass oil

17.84: Volume expansion: dV  βV dT

V V

dT dV

Construct the tangent to the graph at 2Cand8C and measure the slope of this line

3 3

cm 0.10 and 1000cmSlope

3

)C(103cm

1000

C3cm

4cm24

0 and 1000cmslope

3

)C(106cm

1000

C4cm24

T2 T1T 415C

17.86: a) The change in height will be the difference between the changes in

volume of the liquid and the glass, divided by the area The liquid is free to

expand along the column, but not across the diameter of the tube, so the increase

in volume is reflected in the change in the length of the columns of liquid in the stem

A

V A

V V

 liquid glass ( liquid glass )

)m100.50

(

)m10100

2 6

3 6

17.87: To save some intermediate calculation, let the third rod be made of

fractions f1and f2 of the original rods; then f1 f2 1and f1(0.0650)

.0580.0)

0350.00580.0

1 2

and the lengths are f1(30.0cm)23.0cmand f2(30.0cm)7.00cm

17.88: a) The lost volume, 2.6 L, is the difference between the expanded volume

of the fuel and the tanks, and the maximum temperature difference is

,C78.2

)m100.106)(

)C(102.7)(C105.9(

)m106.2(

)(

3 3 1

5 1

4

3 3 0

A1 fuel

V T

or 28C to two figures; the maximum temperature was 32C b) No fuel can spill if the tanks are filled just before takeoff

17.89: a) The change in length is due to the tension and heating

 αΔT Y

A, F T

expands, and so

Pa

101.92

C)120)(

)(C102.1)(C100.2Pa)(

1020

)(

8

1 5 1

5 10

steel brass steel

17.90: In deriving Eq (17.12), it was assumed that L0; if this is not the case when there are both thermal and tensile stresses, Eq (17.12) becomes

For the situation in this problem, there are two length changes which must sum to zero,

and so Eq (17.12) may be extended to two materials a and b in the form

.0b

0b a

F T α

)C0.60(Pa))107m(0.250Pa)

1020m)350.0((

m)250.0)(

)(C104.2(m)350.0)(

)(C10(1.2

))(

)((

8

10 10

1 5 1

5 -

b 0b a

oa

0b b 0a

L α L α A

in.) 0020 0 ( 0

αR R

warmed to 87C b) the difference in the radii was initially 0.0020 in., and this must be the difference between the amounts the radii have shrunk Taking R to be the same for 0

both rings, the temperature must be lowered by an amount

αbrass αsteelR0

R T

in

0020.0

1 5 1

5

to two figures, so the final temperature would be 80C

17.92: a) The change in volume due to the temperature increase is βV and the T,change in volume due to the pressure increase is V BpEq.11.13  Setting the net change equal to zero, βVT V B p,or pBβV b) From the above,

1.61011Pa3.010 5 K 1 15.0K8.64107 Pa



17.93: As the liquid is compressed, its volume changes by an amount V pkV0.When cooled, the difference between the decrease in volume of the liquid and the

decrease in volume of the metal must be this change in volume, or α1αmV0T V

Setting the expressions for V equal and solving for T gives

K108.4K1090.3

Pa1050.8Pa10065.5

1 4 1

5

1 10 6

1 m

17.94: Equating the heat lost be the soda and mug to the heat gained by the ice and

solving for the final temperature T 

0.15KkgJ2100kg

120

0

C0.20KkgJ910kg257.0KkgJ4190kg00

910J kg K600C  54.3.2

sm77002

v T

cm

mv Q

2m100.5N520

or 180W to two figures The net torque that the rope exerts on the capstan, and hence the net torque that the capstan exerts on the rope, is the difference between the forces of the ends times the radius A larger number of turns might increase the force, but for given forces, the torque is independent of the number of turns

K)molJ470kg)(

(6.00

W)182

dt dQ dt dT

17.97: a) Replacing m with nM and nMc with nC,

4 1

4 2 3

3 3

T

T T dt nk T T

nk dQ Q

For the given temperatues,

J

6.83)K)0.10(K)0.40((

K)281(4

K)molJ1940mol)(

50.1

17.98: Setting the decrease in internal energy of the water equal to the final

gravitational potential energy, LfρwVwCwρwVwΔT mgh. Solving for h, and inserting

numbers:

km

108m1008

1

)smkg)(9.870

(

C)C)(37kg

J4190(kgJ10334)m1.8.9.1)(

mkg1000

(

)Δ(

5

2

3 3

3

w f w w

h

17.99: a) (90)(100 W)(3000s)2.7107 J

)m3200)(

mkgK)(1.20kg

J(1020

J107.2

Q cm

Q T

or 6.9C to the more apropriate two figures c) The answers to both parts (a) and (b) are multiplied by 2.8, and the temperature rises by 19.3C

17.100: See Problem 17.97 Denoting C by CabT, a and b independent of

temperature, integration gives

.)(

2)

1

2 2 1

In this form, the temperatures for the linear part may be expressed in terms of Celsius

temperatures, but the quadratic must be converted to Kelvin temperatures,

K

500and

Jmol)(29.500

.3



Q (4.1010 3 J molK2)((500K)2 (300K)2))

1.97104 J

17.101: a) To heat the ice cube to 0.0 heat must be lost by the water, which C,

means that some of the water will freeze The mass of this water is

g.4.72kg1072.4)

kgJ10334(

)CK)(10.0kg

Jkg)(2100075

.0

3 f

ice ice

K)(65.0kg

J4190(

K)K)(42.0kg

J4190(

3 v

s

w

w w

so 0.0696kgof steam supplies the same heat as 1.00kg of water Note the heat

capacity of water is used to find the heat lost by the condensed steam

17.103: a) The possible final states are steam, water and copper at 100 water, C,ice and copper at 0.0 or water and copper at an intermediate temperature C

Assume the last possibility; the final temperature would be

C1.86K)

kgJkg)(41900950

.0(

K)kgJkg)(390(0.446

K)kgJkg)(41900350

.0

(

kg)J10kg)(3340950

.0(

kg)J102256)

CK)(100kg

Jkg)((41900350

.0

This is indeed a temperature intermediate between the freezing and boiling points,

so the reasonable assumption was a valid one b) There are 0.13kgof water

17.104: a) The three possible final states are ice at a temperature below 0.0 an ice-C,water mixture at 0.0 or water at a temperature above C 0.0C To make an educated guess at the final possibility, note that (0.140kg)(2100J kgK)(15.0C) 4.41kJ are needed to heat the ice to 0.0 and C, (0.190kg)(4190J kgK)(35.0C) 27.9kJ must removed to cool the water to 0.0 so the water will not freeze Melting all of the ice C,would require an additional (0.140kg)(334103 J kg)46.8kJ, so some of the ice melts but not all; the final temperature of the system is 0.0C

Considering the other possibilities would lead to contradictions, as either water

at a temperature below freezing or ice at a temperature above freezing

b) The ice will absorb 27.9 kJ of heat energy to cool the water to 0C Then,

070.0

17.105: a) If all of the steam were to condense, the energy available to heat the water

would be (0.0400kg)(2256103 J kg)9.02104 J.If all of the water were to be heated

to 100.0 the needed heat would be C, (0.200kg)(4190J kgK)(50.0C)4.19104J.Thus, the water heats to 100.0 and some of the steam condenses; the temperature of Cthe final state is 100C

b) Because the steam has more energy to give up than it takes to raise the water temperature, we can assume that some of the steam is converted to water:

kg

019.0kgJ102256

J1019.4

Thus in the final state, there are 0.219 kg of water and 0.021 kg of steam

17.106: The mass of the steam condensed 0.525kg0.490kg0.035kg.The heat lost

by the steam as it condenses and cools is

),K0.29)(

KkgJ4190)(

kg035.0()kg035.0

and the heat gained by the original water and calorimeter is

J.108.33K)))(56.0K

kgJkg)(4190(0.340

K)kgJkg)(420150