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
UUn
dersta
ndin
g Pro
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
.
UUn
dersta
ndin
g Pro
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
g Pro
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