RReview

for F

inal E

xam

1

Problem 1 U

nderstanding of motion of objects in 2D

(Chap. 3 + 2) Problem

2 Understanding N

ewton's laws (Chap.4 + 5 + 6) Problem

3 Understanding a collision of 2 bodies and a m

otion of 2 bodies after the collision (Chap. 8 + others such as 7)

Problem 4 U

nderstanding a rotational motion of solid objects (Chap.

9) Problem

5 Understanding wave m

otion (Chap. 12 along with 11) Problem

6 Understanding pV = nRT (Chap.15)

�B

ON

US A

SSIGN

ME

NT: H

ere (Deadlne: 12:30 pm

, Dec. 3 (Tue))

�FIN

AL E

XA

M: D

ec 6 (Fri) (3 pm - 5 pm

) at MPH

Y 203

[Fall 2013 Final Exam Schedule] [Form

ular Sheet for Exam4 and

Final] �

Old PH

YS 218 E

xam Solutions: 1. 2. 3. Final.

�T

EN

TATIV

E PL

AN

for Final Exam as of D

ec 1 (6 PM):

2

1

Kinem

atics (2D)

I

III

UUn

dersta

ndin

g Pro

jectile Motio

ns

II

3

PPro

jectile Motio

n

4

–The red ball is dropped at the sam

e time that the

yellow ball is fired horizontally. The strobe marks equal tim

e intervals.

–Projectile m

otion as horizontal m

otion with constant velocity (a

x = 0) and vertical m

otion with constant acceleration (a

y = ��g).

PPro

jectile Motio

n

5

–The red ball is dropped at the sam

e time that the

yellow ball is fired horizontally. The strobe marks equal tim

e intervals.

–Projectile m

otion as horizontal m

otion with constant velocity (a

x = 0) and vertical m

otion with constant acceleration (a

y = ��g).

Kinem

atics (2D)

FFu

rther L

ook

at P

rojectile M

otio

n

(2) ax = 0

ay = �g = �9.80 m

/s 2

(1) Choose an origin &

an x-y coordinate system

vx = constant

(3) vy = 0

(4) y = 0

6

Projectile motion

as horizontal motion with

constant velocity (a

x = 0) and vertical m

otion with constant acceleration (a

y = �g).

Kinem

atics (2D)

[Quick Q

uiz 2] Is

this a

motion

with a

constant acceleration

or with a varying acceleration?

?final �

v �

7

TTh

e Sam

e Pro

blem

?

Kinem

atics (2D)

Kinem

atics (2D)

y = 0

Vy = 0

y = �1.00m(1) C

hoose an origin &

an x-y coordinate system

(2) ax = 0

ay = �g = �9.80 m

/s 2

(3) Use kinem

atic eqs.in x and y separately.

Vx = constant

8

SSam

e Pro

blem

s?

9

R

R

y = 0 m

y = ? m

10

Kinem

atics (2D)

t = 7.6 s

Exam

ple 2: A projectileis launched from

ground level to the top of a cliff w

hich is R = 195 m aw

ay and H = 155 m

high. The projectile lands on top of

the cliff T = 7.60 safter it is fired. U

se 2sin ��cos� = sin2���if necessary. The

acceleration due to gravity is g = 9.80 m/s 2 pointing dow

n. Ignore air friction. a. Find the initial velocity of the projectile (m

agnitude v0 and direction �).

b. Find a formula of tan� in term

s of g, R, H and T.

11

Aprojectil

launchedfrom

groundlevel

tole

isla

too

thet

Th

e Sam

e Pro

blem

?

TTh

e Pain

tball G

un

�

See Example 3.3

�

Full consideration given to the m

otion after the dye-filled ball is fired.

12

Question: H

ow fast m

ust the m

otorcycle leave the cliff-top?

[Quick Q

uiz 1] Is this a m

otion with a constant acceleration or with a varying acceleration?

13

Th

e Sam

e Pro

blem

?

FFirin

g at a

More C

om

plex T

arget

�A m

oving target presents a real-life scenario. �

It is possible to solve a falling body as the target. This problem

is a “classic” on standardized exam

s.

14

http://ww

w.youtube.com

/watch?v=cxvsH

NR

XL

jw

Kinem

atics (2D)

A boy on a sm

all hill aims his w

ater-balloon slingshot horizontally, straight at second boy hanging from

a tree branch a distance d aw

ay. At the instant

the water balloon is released, the second

boy lets go and falls from the tree, hoping

to avoid being hit. Show that he m

ade the w

rong move.

15

A boy on a sm

all hill aims his w

ater-balloon

slingshot upw

ard, directly

at second boy hanging from

a tree branch. At

the instant the water balloon is released,

the second boy lets go and falls from the

tree, hoping to avoid being hit. Show that

he made the w

rong move.

Kinem

atics (2D)

Kinem

atics (2D)

Same

Concept

200 m, given �

x?

d, given � H

? H ?

?

16

117

http://link.brightcove.com/services/player/bcpid36804639001?bckey=A

Q~~,A

AAACI

JPQzk~,qiwYyU

rE_-dz5lglGrCClkfJDM

1jW3zH

&bclid=0&bctid=109459228001

Can you explain? MM

agic? o

r Ph

ysics?

Giancoli’s Textbook 3rd Ed.

We see “A

n apple in motion (x direction)

tends to stay in motion (x direction).”

�

Motion with constant velocity in x

direction.

18

2

New

ton’s Laws of M

otion

Kinem

atics (r, v, a) ��

�

�

Stru

cture o

f New

ton

ian

Mech

an

ics

Inertial Reference Fram

e (N

ewton’s 1

st Law

)

Action-R

eaction (N

ewton’s 3

rd Law

) M

ass (m

)

The N

ature of Force T

he Nature of O

bject T

he Nature of M

otion

F = m a

(New

ton’s 2nd L

aw)

�

�

Kinem

atics (r, v, a) �

�

�

K

inematics

(r, v, a) �

�

�

K

inematics

(r, v, a) �

�

�

19

New

ton’s Laws of M

otion

The force on a hokey puck

causes the acceleration If the net force on a hokey puck is zero (equilibrium

), the acceleration is zero.

0

0�

�

����

��a

F

220

FForce: A

cceleratio

n/E

qu

ilibriu

m

Acceleration ��

Kinetic Equations (see Chap. 2 &3)

New

ton’s Laws of M

otion

[A]

221

QQu

ick Q

uiz

(b)

A

hockey puck

is sliding

at a

constant velocity

across a

flat horizontal ice surface. W

hich is the correct free-body diagram

?

New

ton’s Laws of M

otion

Draw a D

iagram �

FBD �

Newton’s Laws

DD

i

22

New

ton’s Laws of M

otion

Draw a D

iagram �

FBD �

Newton’s Laws 2

3

224

WWeigh

t on

Pla

net

�M

ass is a measure of “how m

uch material do I

have?” �

Weight is “how hard do I push down on the floor?”

�If you were offered to get 9.8N

of gold on earth and 9.8N

of gold on moon, which offer do you take?

New

ton’s Laws of M

otion

a �F

N

Ff

Fg

m Fa

f ���

FN

Ff

Fg

Ignore the truck and two people!

Ff is the force on the box by the truck’s bed.

Quick Q

uiz Solution: The friction force appears as it keeps from

sliding back on the truck’s bed. T

hus, the direction of the (static) friction is pointing to the right. If you isolate the box, and draw

the free-body diagram for the box, you

find that it is consistent with N

ewtons’ 2

nd law:

25

New

ton’s Laws of M

otion

Exam

ple 6: Suppose that you are standing on a train

accelerating at

0.20g. W

hat m

inimum

coefficient of static friction m

ust exist between

your feet and the floor if you are not to slide?

a = 0.20 g

FG

FN

Ff

where F

f = �s F

N = �s (m

g) T

hus, �s (m

g) = m (0.20g)

So, �s = 0.20

Ff

Which one is correct?

New

ton’s 2nd law

: F = m

a � F

f = m (0.20g)

x

26

5(e) : In a “Rotor-ride” at a carnival,

People pay money to be rotated in a

Vertical cylindrically walled “room

.” W

hich diagram correctly show

s the Forces acting on each rider? E

xplain E

xplicitly (in words) the reasoning w

hy you choose the diagram

by labeling each arrow

(in wards: e.g., dow

nward

arrow, horizontal-left arrow

etc.).

ab

cd

eTaken from

Fig. 5-38 (Giancoli)

27

New

ton’s Laws of M

otion

Exam

ple 2: A block (mass m

1 ) is placed on a smooth

horizontal surface, connected by a thin cord that passes over a pulley to a second block (m

2 ), which hangs vertically. D

raw the

free-body diagram for each of m

1 and m2 . E

xpress the acceleration of the block in term

s of m1 , m

2 , and g.

FN

Fg1

FT

FT

Fg2

FT = m

1 a F

g2 – FT = m

2 a

m2 g – m

1 a = m2 a

m2 g = m

1 a + m2 a

a = m2 g

/ (m1 + m

2 ) a

a

x 28

New

ton’s Laws of M

otion

229

Circular M

otion

The quantity v2/R is not a force -

it doesn’t belongin the free-body

(force) diagram

Which one is correct?

F �

R vm

2

F �

F �F �

INC

OR

RE

CT

C

OR

RE

CT

F.B

.D. o

f Un

iform

Circu

lar M

otio

n

30

Thinking …

Circular M

otion

Swinging a ball on the end of string …

[Q

uick Quiz] W

hich one is correct?

(a) (b)

331

�M

odel airplane on a string

332

EExa

mple P

roblem

1

Rota

tion

� C

enter-seek

ing F

orce �

FT

r̂r v

m

am

F

2

T rad�

��

����

One revolution every 4.00 seconds

yyTr

time ce

tandis

v

2 �

��

Circular M

otion

EExa

mple P

roblem

1

A 1000-kg car rounds a curve on a flat road of radius 50.0 m

at a m

aximum

speed of 14.0 m/s w

ithout skidding. a)

Draw

the free-body diagram for the car.

b)Find the m

agnitude and direction of the friction force. c)

Find the coefficient of the friction force. d)

Is the result in c) independent of the mass of the car?

333

r̂r v

m

am

F

2

f rad�

��

����

mg

F�

N

�s m

in = Ff /F

N

Circular M

otion

A 1000-kg car rounds a curve on a banked road of radius 50.0

m at a m

aximum

speed of 14.0 m/s w

ithout skidding. The

banking angle is 22o.

a)D

raw the free-body diagram

for the car. b)

Find the magnitude and direction of the friction force.

c)Find the coefficient of the friction force.

d)Is the result in part c) independent of the m

ass of the car?

334

EExa

mple P

roblem

2

No Skidding on Banked Curve

r̂r v

m

am

2

rad�

��

mg

F

N

�s = F

f /FN

UUn

dersta

ndin

g Satellite M

otio

n

335

r̂r v

m

am

2

rad�

��

r̂r mM

GF

2�

��

•G

= 6.67 x 10�11 N

m2/kg

2

•M

E = 5.98 x 1024 kg

•R

E = 6.38 x 103 km

•

MS = 1.99 x 10

30 kg •

Mm

oon = 7.35 x 1022 kg

•R

moon = 1.74 x 10

3 km

A tetherball problem

– Example 6.2 and Figure 6.5

•Refer to the w

orked example on page 165.

336

37

3

Work and Energy

A 50.0-kg crate is pulled 40.0 m

by a constant force exerted (F

P = 100 N and � = 37.0

o) by a person. Assum

e a coefficient of friction force �

k = 0.110. Determ

ine the w

ork done byeach force acting on the crate and its net

work. Find the final velocity of the crate if d = 40 m

and v

i = 0 m/s. WW

ork

En

ergy Th

eorem

38

Wnet =

Wi

= 1302 [J] (> 0)

Wnet =

Kf – K

i

= (1/2) m v

f 2 � 0

Energy Method

Equations of Motion

39

Energy Conservation

Find h Using Energy Conservation

=

ISEE

ISEE

40

EEEEEnnnnnnnnnnnneeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeerrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooonnnnnnnnnnnnnnnnnnservation

41

42

443

Mom

entum C

onservation nnnnn

(A) Momentum Conservation

(B) E

nergy Conservation

1

2

Same C

oncept with B

allistic Pendulum

8-85

8-44, 8-83 44

Mom

entum C

onservation

(A) Momentum Conservation

(B) E

nergy Conservation

1

2

Same C

oncept with B

allistic Pendulum

A

B

C

Skeet+Pellet

45

New

ton’s Laws of M

otion

46

Similar Problem

47

48

4

Rotational M

otion

NNew

ton

’s Law

s for R

ota

tion

1st part

[s –2] 3

rd part [N

m]

2nd part

[kg m2]

Krot = (1/2) I �

2 (�K

= (1/2) m v

2)

49

Chap. 10

Angular M

omentum

Conservation

Rotational M

otion

RRota

tion

al E

nergy a

nd In

ertia m1

m2

m3

50

Rotational M

otion

RRota

tion

al In

ertia

51

MMech

an

ical E

nergy C

on

servatio

n

K = K

m + KM

Krot = (1/2) I �

2 (�K

= (1/2) m v

2)

Rotational M

otion

Solid disk (M, R

0 )

m

53

0

Mech

an

ical E

nergy C

on

servatio

n

RRace o

f the o

bjects o

n a

ram

p

�This is a classic m

ultiple-choice question from

MCA

T-style standardized tests. �

Refer to Figure 9.23.

2cm

2

2 12 1

�

IM

MgH

��

v

Icm = Large �

v = Small

55

Can you explain why?

Rotational M

otion

Analyze

the rolling

sphere in

terms

of forces

and torques: find the m

agnitudes of the velocity v and ….

56

Can

you

find v?

Rotational M

otion

57

Ktrans +

Krot =

Krolling

am

F�

�

�net

�

��

I�

net

58

Rotational M

otion

559

An

gula

r Mom

entu

m C

on

servatio

n

P10-30 (MP)

EExa

mple 2

60

�i �i ��

��f �f

P10-33 P10-34 (M

P)

61

A spinning figure

skater pulls his arms

in as he rotates on the ice. A

s he pulls his arm

s in, what

happens to his angular m

omentum

L and kinetic energy K

?

A. L and K

both increase.

B. L stays the sam

e; K increases.

C. L increases; K

stays the same.

D. L and K

both stay the same.

Q10.11

A spinning figure

skater pulls his arms

in as he rotates on the ice. A

s he pulls his arm

s in, what

happens to his angular m

omentum

L and kinetic energy K

?

A. L and K

both increase.

B. L stays the sam

e; K increases.

C. L increases; K

stays the same.

D. L and K

both stay the same.

A10.11

The four forces shown all have the

same m

agnitude: F1 = F

2 = F3 = F

4 .

Which force produces the greatest

torque about the point O (m

arked by the blue dot)?

A. F

1

B. F

2

C. F

3

D. F

4

E. not enough inform

ation given to decide

Q10.1

F1 F

2

F3

F4

O

The four forces shown all have the

same m

agnitude: F1 = F

2 = F3 = F

4 .

Which force produces the greatest

torque about the point O (m

arked by the blue dot)?

A. F

1

B. F

2

C. F

3

D. F

4

E. not enough inform

ation given to decide

A10.1

F1 F

2

F3

F4

O

Which of the four forces show

n here produces a torque about O

that is directed out of the plane of the draw

ing?

A. F

1

B. F

2

C. F

3

D. F

4

E. m

ore than one of these

Q10.2

F1 F

2

F3

F4

O

Which of the four forces show

n here produces a torque about O

that is directed out of the plane of the draw

ing?

A. F

1

B. F

2

C. F

3

D. F

4

E. m

ore than one of these

A10.2

F1 F

2

F3

F4

O

A plum

ber pushes straight dow

n on the end of a long w

rench as show

n. What is the

magnitude of the

torque he applies about the pipe at low

er right?

A. (0.80 m

)(900 N)sin 19°

B. (0.80 m

)(900 N)cos 19°

C. (0.80 m

)(900 N)tan 19°

D. none of the above

Q10.3

A. (0.80 m

)(900 N)sin 19°

B. (0.80 m

)(900 N)cos 19°

C. (0.80 m

)(900 N)tan 19°

D. none of the above

A10.3

A plum

ber pushes straight dow

n on the end of a long w

rench as show

n. What is the

magnitude of the

torque he applies about the pipe at low

er right?

A. m

2 g = T2 = T

1

B. m

2 g > T2 = T

1

C. m

2 g > T2 > T

1

D. m

2 g = T2 > T

1

E. none of the above

Q10.5 A

glider of mass m

1 on a frictionless horizontal track is connected to an object of m

ass m2 by a m

assless string. The glider accelerates to the right, the object accelerates dow

nward, and the string rotates

the pulley. What is the relationship am

ong T1 (the tension in the

horizontal part of the string), T2 (the tension in the vertical part of

the string), and the weight m

2 g of the object?

A. m

2 g = T2 = T

1

B. m

2 g > T2 = T

1

C. m

2 g > T2 > T

1

D. m

2 g = T2 > T

1

E. none of the above

A10.5 A

glider of mass m

1 on a frictionless horizontal track is connected to an object of m

ass m2 by a m

assless string. The glider accelerates to the right, the object accelerates dow

nward, and the string rotates

the pulley. What is the relationship am

ong T1 (the tension in the

horizontal part of the string), T2 (the tension in the vertical part of

the string), and the weight m

2 g of the object?

A lightw

eight string is wrapped

several times around the rim

of a sm

all hoop. If the free end of the string is held in place and the hoop is released from

rest, the string unw

inds and the hoop descends. H

ow does the tension in the string

(T) compare to the w

eight of the hoop (w

)?

A. T = w

B. T > w

C. T < w

D. not enough inform

ation given to decide

Q10.6

A lightw

eight string is wrapped

several times around the rim

of a sm

all hoop. If the free end of the string is held in place and the hoop is released from

rest, the string unw

inds and the hoop descends. H

ow does the tension in the string

(T) compare to the w

eight of the hoop (w

)?

A. T = w

B. T > w

C. T < w

D. not enough inform

ation given to decide

A10.6

A solid bow

ling ball rolls dow

n a ramp.

Which of the follow

ing forces exerts a torque on the bow

ling ball about its center?

A. the w

eight of the ball

B. the norm

al force exerted by the ramp

C. the friction force exerted by the ram

p

D. m

ore than one of the above

E. T

he answer depends on w

hether the ball rolls without

slipping.

Q10.7

A solid bow

ling ball rolls dow

n a ramp.

Which of the follow

ing forces exerts a torque on the bow

ling ball about its center?

A10.7 A

. the weight of the ball

B. the norm

al force exerted by the ramp

C. the friction force exerted by the ram

p

D. m

ore than one of the above

E. T

he answer depends on w

hether the ball rolls without

slipping.

76

5

77

A 2.00-kg frictionless block is attached to an ideal spring w

ith force constant k = 315 N/m

. initially the spring is neither stretched nor com

pressed, but the block is moving in the negative direction at 12.0

m/s, and undergoes a sim

ple harmonic m

otion (SHM

). Let’s characterize the SHM

of the block. Find: (a)(5 pts) the period (T) in seconds (b)(5 pts) the m

aximum

speed (vm

ax ) in m/s

(c)(5 pts) the amplitude (A) of the m

otion in meters

(d)(5 pts) the maxim

um m

agnitude of the force (in N) on the

block exerted by the spring during the m

otion.

Visualizing SHM

[Bonus (10 pts)] If you have tim

e, sketch x-t and v-t graphs ofthis m

otion.

xm k

ax

�

�x

kF

x

��v

max , a

x =0

Mechanical Energy Conservation

SSH

M to

Wave M

otio

n

78

[Q] H

ow can you describe the shape of the rope? [A

]

x = R0 cos ���

where �� = � t

x(t) = R0 cos (� t)

Kin. Equation of SHM

Simple H

armonic O

scillator (SHO

)

T = 2� / �

A, ��

Figure 15.4

and TIME dependence…

79

vwave = ��/ T�

vwave = ��(� /2�)�

vwave = ��f�

SSta

ndin

g Wave

Wave M

otion

80

(a)F

T if � = 40.0 g/m and f1 = 20.0 H

z? (b)

f2 and wavelength of second

harmonic?

(c)f2 and w

avelength of second overtone (or 3

rd harmonic)?

f2 = 2 f1 f3 = 3 f1

Wave M

otion

Exam

ple 2 A

transverse traveling wave on a cord is represented by

y(x, t) = 0.48 sin(0.56x + 84t)

where y and x are in m

eters and t in seconds. For this wave,

determine:

(a)the am

plitude, (b)

wavelength, frequency, velocity (m

agnitude and direction), (c)

maxim

um and m

inimum

speeds of particles of the cord, and (d)

maxim

um acceleration (m

agnitude) of the particles. [A

] …

81

The wave function for a sinusoidal w

ave m

oving in the +x-direction is

y(x, t) = A sin(� t – k x), w

here k = 2π/� , � = 2�f, ��= v T …

Oscillations

Graphs

882

Which of the follow

ing wave functions describe a w

ave that m

oves in the –x-direction?

A. y(x,t) = A sin (–kx – �t)

B. y(x,t) = A sin (kx + �t)

C. y(x,t) = A cos (kx + �t)

D. both B

. and C.

E. all of A., B

., and C.

Q15.2

83

Which of the follow

ing wave functions describe a w

ave that m

oves in the –x-direction?

A. y(x,t) = A sin (–kx – �t) = A sin ( – (kx + �t) )

B. y(x,t) = A sin (kx + �t)

C. y(x,t) = A cos (kx + �t)

D. both B

. and C.

E. all of A., B

., and C.

A15.2

84

Wave M

otion

A transverse w

ave pulse travels to the right along a string with

speed v = 2.0 m/s. A

t t = 0, the shape of the pulse is given by the function y = 0.45 cos(3.0x) w

here y and x are in meters and t in

seconds. For this wave, determ

ine: (a)

the wavelength, frequency, and am

plitude, (b)

maxim

um and m

inimum

speeds of particles of the string, and (c)

maxim

um and m

inimum

accelerations (magnitudes) of the

particles. [A

] …

Wave function is:

y(x, t) = 0.45 cos(3.0x �� 6.0t)

Example 3

85

The wave function for a sinusoidal w

ave m

oving in the +x-direction is

y(x, t) = A sin(� t – k x), w

here k = 2π/� , � = 2�f, ��= v T …

UUn

dersta

ndin

g Pro

blem

86

2dB)

10(

02

1dB)

10(

01

Pat

10)

(

Pat

10)

(2 1

/ /

Ir

I

Ir

I� �

��

��

Exam

ple 12.9

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blem

87

12.42: "By w

hat factor must the sound intensity be increased

to increase the sound intensity level by 12.5 dB?”

�

I2 /I1 = ? = 10^{�2 /10 – �

1 /10 } = 10^{ (�2 – �

1 ) / 10} �

12.5 dB = �

2 – �1

Exam

ple 12.9

2dB)

10(

02

1dB)

10(

01

Pat

10)

(

Pat

10)

(2 1

/ /

Ir

I

Ir

I� �

��

��

�Shifts in observed frequency can be caused by m

otion of the source, the listener, or both. Exam

ples 12.10-12.13.

SL

fv

vf

m/s)

30(

)0(

��

��

L

SL

fv

vf

m/s)

30(

)0(

��

��

L

fS = 300 Hz

v = 340 m/s

observed frequency can be caused by mmotion o

motion

Th

e Doppler E

ffect

SL

vv

��

)0( m/s)

30

(� ��

�S

Lv

v�

�)0

( m/s)

30(� �

��

88

TTh

e Doppler E

ffect

L S

A

SB

�Shifts in observed frequency can be caused by m

otion of the source, the listener, or both. Exam

ples 12.10-12.13 and P.12-53,54,60

89

9

1

6

HHow

to S

tudy C

hap. 1

4

1)Heat transfer,

equilibrium, tem

perature: P.14-5, 24, 27, 53, 56, 64, 74, 82

2)Therm

al expansion: P.14-15, 16, 73

3)Phase change, calorim

etry: P.14-32, 44, 49

92

LT

TkA

dtdQ

H

Tm

cQ

TL

L

LowH

igh ��

� � ���

�

0

1 1 2 3

HHea

t Capacity /

Calo

rimetry

�Substances have an ability to “hold heat” that goes to the atom

ic level.

Q

= m c ��T

[J] = [kg] [?] [K]

93

�c = specific heat capacity [J / (kg * K)]

�c

water = 4.19 x 103 J/(kg*K) vs. c

copper = 0.39 x 103 J/(kg*K)

�W

hat we see in life? �

One of the best reasons to spray

water on a fire is that it suffocates com

bustion. But, another reason is that water has a huge heat capacity. Stated differently, it has im

mense therm

al inertia. In plain term

s, it’s good at cooling things off because it’s good at holding heat.

�Taking a copper frying pan off the stove with your bare hands is an awful idea because m

etals have small heat capacity. In plain

terms, m

etals give heat away as fast as they can. �

Examples 14.6 and 14.7 ; Exam

ples 14.8 and 14.9

Ph

ase C

han

ges �

The steam contains the energy (heat of vaporization) that it took to

become a gas. This is 2.3 M

ILLION

joules per kg of water.

Q/m

= Lv = 2.26 x 10

6 J/kg Q

/m= L

f = 3.34 x 105 J/kg

94

�The ice needs to absorb the latent heat of fusion to becom

e a liquid.

TTh

ermal E

xpan

sion

� : The expansion is proportional to the original length and the tem

perature change (for reasonable �T). (Table 14.1)

95

��: Volum

e expansion (Table 14.2; Example 14.4)

SStress o

n a

Spacer

�Consider a alum

imun spacer

(L0 =10 cm

) at 17,2 oC. �

Thermal Expansion

�

Stress

�Therm

al Stress

�

Example: Road Expansion and

Contraction

96

)(

0 0

YA

/F

L/L

A/F

L/L

Y

T T

��

��

�� �

TL/

LT

LL

� �

�

��

��

0 0

YT

A/F

YA

/F

TT

T

��

��

��

��

� )

(

Example 14.5 Pa

10

700

(alminum

)

K 10

42

(aluminum

)11

15

��

��

��

.Y

.

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blem

97

TL

L�

�

0�

Td

dd

d�

�

00�

��

At �78 oC

A

t ��� oC

UUn

dersta

ndin

g Pro

blem

98

LT

TkA

dtdQ

dtdQ

Lm

dtdQ

Lm

Q

LowH

ighiron

water

fice

waterf

icewater

��

�

��

�sec

600

UUn

dersta

ndin

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blem

99

Q/m

= Lv = 2.26 x 10

6 J/kg

Tc

mQ

Qiron

ironstep

step�

��

21

Phase Change

CCh

ap. 1

5 : p

V =

nR

T

10

0

Key N

umbers and Equations

10

1

Details of K

inetic Property (II)

10

2

V, N, T, p

oN

= Num

ber of particles in volum

e V

oN

umber of particles per unit

volume is N

/V

opV = n R T �

pV = N k T

�p = pf,i – p

i,x = (040 kg)(+30 m

/s) – (0.40 kg)(-30 m/s)

=2 (Mass) x |v

x |

vi,x = 30 m

/s

vi,x = -30 m

/s

Chap. 8

Container

Kav =

(3/2) k T

(3/2) nR T = N

Kav =

Ktrans

Kav =

(1/2) m v

2

10

3

� In sim

ple terms, “the

energy added to a system

will be distributed between heat and work”.

�“W

ork” is defined differently than we did in earlier chapters, here it refers to a p�v (a pressure increasing a volum

e).

The first law of therm

odynamics

Q = n C

V ��T + p (V2 – V

1 ) �U

�U

Q = �U

+ W

Q

Ktrans = N

Kav = (3/2) n R T

Thermodynam

ic processes

�A process can be adiabatic and have no heat transfer

in or out of the system

�A process can be isochoric and have no volum

e change. �

A process can be isobaric and have no volum

e change. �

A process can be isotherm

al and have no temperature

change.

10

4

You heat a sample of air to twice its original

temperature in a constant-volum

e container. The average translational kinetic energy of the m

olecules is

A. half the original value.

B. unchanged.

C. tw

ice the original value. D

. four times the original value.

10

5

Q = n C

V ��T + p (V2 – V

1 )

Q = �U

+ W

Ktrans = N

Kav = (3/2) n R T

You heat a sample of air to twice its original

temperature in a constant-volum

e container. The average translational kinetic energy of the m

olecules is

C. tw

ice the original value.

10

6

You compress a sam

ple of air slowly to half its original volum

e, keeping its temperature constant. The internal

energy of the gas

A. decreases to half its original value.

B. rem

ains unchanged. C

. increases to twice its original value.

10

7

Q = n C

V ��T + p (V2 – V

1 )

Q = �U

+ W

Ktrans = N

Kav = (3/2) n R T

You compress a sam

ple of air slowly to half its original volum

e, keeping its temperature constant. The internal

energy of the gas

B. rem

ains unchanged.

10

8

10

9