Seat:
PHYS 1500 (Fall 2012) Exam #3, V1
5 pts
1. From book Prob 11.24 You own a heat engine that exhausts 5 J of energy into the
the cold reservoir for every 2.5 J of work you get out. What is the efficiency of this
engine?
(a) 0.25
5 pts
Name:
(b) 0.33
(c) 0.50
(d) 0.75
2. From book Concep Ques 10.12 A 0.5 kg mass on a 2-m-long string swings in a
circle on a horizontal, frictionless table at a steady speed of 3 m/s. How much work
does the tension in the string do on the mass during one revolution? Explain.
The tension in the string does no work on the mass. The reason is that
the tension acts perpendicular to the direction of motion. A force that acts
perpendicular to the motion doesn’t do any work: W = F d cos(90◦ ) = 0.
5 pts
3. From book HWK Prob 12.30 The aliens who live on the planet Breeble III use
aluminum for their walkways. The walkway is constructed from segments as shown.
The length of each segment is 2.90 m, the width is 1.10 m, and the thickness is 2.0 cm.
If the walkway is laid when the temperature is 10.0 ◦ C, how large do the gaps between
the segments need to be if the Breeblans don’t want the walkway to buckle if the
temperature gets as high as 45.0 ◦ C?
This is a problem about the expansion of materials when the temperature
changes. The governing relationship is ∆L = αLi ∆T where α is the linear
expansion coefficient, Li is the starting length, and ∆T is the change in
temperature. In this problem, you’ve been given these three quantities and
you’re being asked for ∆L
∆L = 2.3 × 10−5 ◦ C−1 2.9 m 35 ◦ C = 2.33 mm
(V 1)
∆L = 1.7 × 10−5 ◦ C−1 2.5 m 30 ◦ C = 1.28 mm
(V 2)
∆L = 1.2 × 10−5 ◦ C−1 2.3 m 40 ◦ C = 1.10 mm
(V 3)
10 pts
4. Based on book HWK Prob 13.27 Styrofoam has a density of 300 kg/m3 and is in
the shape of a sphere with a diameter of 50.0 cm. You tie a hunk of aluminum to the
styrofoam. What is the maximum mass of aluminum that can be tied to the styrofoam
without sinking it? Do not assume the volume of the aluminum is negligible compared
to that of the sphere.
Set up the force on the styrofoam sphere (tension down from string, buoyant
force up on styrofoam, and weight of styrofoam) and the force on the metal
(tension up from string, buoyant force up on aluminum, and weight of
aluminum). Solving for T shows the total weight of the styrofoam plus
metal needs to be balanced by the total buoyant force: FB = ρf Vf g where ρf
is the density of the fluid (1000 kg/m3 for water), and Vf is the volume of
fluid displaced. The volume of fluid displaced is the volume of the styrofoam
plus the volume of metal. This gives
Ms g + Mm g = (ρs Vs + ρm Vm )g = ρf (Vs + Vm )g
where Ms = ρs Vs is the mass, density, volume of styrofoam and Mm = ρm Vm
is the mass, density, volume of metal. The volume of the styrofoam is the
volume of a sphere Vs = 4πrs3 /3. The strategy is to use this relationship to
find the volume of metal Vm and use that to find the mass of the metal.
ρs Vs + ρm Vm = ρf (Vs + Vm ) = ρf VS + ρf Vm
→
ρm Vm − ρf Vm = ρf Vs − ρs Vs
4π(0.25 m)3
= 0.0654 m3
3
(
)
)
(
kg
kg
kg
kg
2700 3 − 1000 3 Vm = 1000 3 − 300 3 0.0654 m3 = 45.8 kg
m
m
m
m
45.8 kg
kg
Vm =
= 0.0269 m3
Mm = Vm ρm = 0.0269 m3 2700 3 = 72.8 kg
3
1700 kg/m
m
(V 1)
Vs =
4π(0.30 m)3
= 0.113 m3
3
(
)
(
)
kg
kg
kg
kg
2700 3 − 1000 3 Vm = 1000 3 − 400 3 0.113 m3 = 67.9 kg
m
m
m
m
67.9 kg
kg
Vm =
= 0.0399 m3
Mm = Vm ρm = 0.0399 m3 2700 3 = 108 kg
3
1700 kg/m
m
(V 2)
Vs =
4π(0.20 m)3
= 0.0335 m3
3
(
)
(
)
kg
kg
kg
kg
2700 3 − 1000 3 Vm = 1000 3 − 200 3 0.0335 m3 = 26.8 kg
m
m
m
m
26.8 kg
kg
3
3
Vm =
=
0.0158
m
M
=
V
ρ
=
0.0158
m
2700
= 42.6 kg
m
m
m
1700 kg/m3
m3
(V 3)
Vs =
5 pts
5. From book Mult Choice 12.36 Steam at 100 ◦ C causes worse burns than liquid
water at 100 ◦ C. This is because
(a)
(b)
(c)
(d)
5 pts
the steam is hotter than the water.
heat is transferred to the skin as steam condenses.
steam has a higher specific heat than water.
evaporation of liquid water on the skin causes cooling.
6. From book Concep Ques 13.3 You are given an irregularly shaped chunk of material
and asked to find its density. List the specific steps that you would follow to do so.
To find the density, you need to figure out its mass and its volume. To find
the mass, you can use a scale to weigh it and then divide by g to get the
mass. To find the volume, you can put it into a cylinder that contains a
fluid (like water) so that it is completely submerged. You then measure the
change in the volume of the water. This is the volume of the object. There
are many other possible ways. For example, measure the tension in a string
that holds only the object and then measure the tension when the object
is submerged. The ratio of the tensions is the density of the object divided
by the difference in the density of the fluid. From this you can solve for the
density of the object...
5 pts
7. Based on book HWK Prob 12.53 You have a copper pot that has radius of 6.00 cm
and a thickness of 8.00 mm. Inside the pot is water with a temperature of 45.0 ◦ C.
The pot is sitting on a heater that is at a temperature of 55.0 ◦ C. How much heat flows
through the bottom of the pot in one minute?
This problem is about heat flow due to conduction. The relevant relationship is Q/∆t = kA∆T /L where k is the thermal conductivity, A is the area
through which the heat is flowing, ∆T is the difference in temperature from
the cold to the hot end, and L is the distance through which the heat is
flowing. In this problem, you’ve been given all of these and ∆t and asked
for Q.
Q = kA∆T∆t/L = 400
W
π(0.06 m)2 10 ◦ C 60 s/(0.008 m) = 3.39 × 105 J (V 1)
m·K
Q = kA∆T∆t/L = 72
W
π(0.07 m)2 15 ◦ C 60 s/(0.009 m) = 1.11 × 105 J
m·K
(V 2)
Q = kA∆T∆t/L = 14
W
π(0.05 m)2 20 ◦ C 60 s/(0.007 m) = 1.88 × 104 J
m·K
(V 3)
10 pts
8. From book HWK Prob 10.57 A 75.0 g aluminum cube can slide without friction
up and down a 30.0◦ slope. The cube is pressed against a spring at the bottom of the
slope, compressing the spring 8.00 cm. The spring constant is 35.0 N/m. When the
cube is released, what distance will it travel up the slope before reversing direction?
This is a conservation of energy problem since there is no friction. The idea
is to find the initial energy and set it equal to the final energy. The initial
energy is only from the compressed spring: Us = (1/2)kx2 . The final energy
is only from gravity due to the change in height of the cube: Ug = mgy.
Once you find the change in height you can compute how far up the incline
using geometry: sin(θ) = y/distance = height/distance.
1 N
Ei = 35
(0.08 m)2 = 0.112 J
2 m
0.112 J
Ei
=
= 0.152 m
Ef = mgy = Ei
y=
mg
0.075 kg 9.8 m/s2
dist =
0.152 m
= 0.305 m
sin(30◦ )
(V 1)
1 N
Ei = 30
(0.07 m)2 = 0.0735 J
2 m
Ei
0.0735 J
Ef = mgy = Ei
y=
=
= 0.115 m
mg
0.065 kg 9.8 m/s2
dist =
0.115 m
= 0.273 m
sin(25◦ )
(V 2)
1 N
Ei = 40
(0.09 m)2 = 0.162 J
2 m
Ei
0.162 J
Ef = mgy = Ei
y=
=
= 0.301 m
mg
0.055 kg 9.8 m/s2
dist =
0.301 m
= 0.524 m
sin(35◦ )
(V 1)
5 pts
9. From book Mult Choice 10.23 You and a friend each carry a 15 kg suitcase up
two flights of stairs, walking at a constant speed. Take each suitcase to be the system.
Suppose you carry your suitcase up the stairs in 30 s while your friend takes 60 s.
Which of the following is true?
(a)
(b)
(c)
(d)
5 pts
You did more work, but both of you expended the same power.
You did more work and expended more power.
Both of you did equal work, but you expended more power.
Both of you did equal work, but you expended less power.
10. From book Prob 12.36 (similar HWK Prob 12.35) Which, if any, of the refrigerators in the figure below violate (a) the first law of thermodynamics or (b) the second
law of thermodynamics? Explain.
The first law of thermodynamics is that the energy is conserved. The only
case where this is not true is (c). The second law of thermodynamics
gives the COPmax = Tc /(TH − TC ) where COP = Qc /Win . For (a) COP = 2
and COPmax = 2; for (b) COP = 4 and COPmax = 3; for (c) COP = 1.5 and
COPmax = 3. Therefore, only (b) violates the second law of thermodynamics.
5 pts
11. From book Prob 13.11 You have a swimming pool that 4.00 m wide, 5.00 m long,
and 2.00 m deep. What is the downward force of the water on the bottom of the pool?
You need to find the pressure at the bottom of the pool. You then use
F = p A to find the force. The pressure at the bottom is given by p = p0 +ρgd
where p0 = 1.01×105 Pa is atmospheric pressure, ρ = 1000 kg/m3 is the density
of water and g = 9.8 m/s2 .
kg
m
9.8 2 2 m = 1.21 × 105 Pa
3
m
s
5
2
F = 1.21 × 10 Pa 20 m = 2.41 × 106 N
(V 1)
kg
m
9.8
3 m = 1.30 × 105 Pa
3
2
m
s
5
2
F = 1.30 × 10 Pa 30 m = 3.91 × 106 N
(V 2)
kg
m
9.8 2 2.5 m = 1.26 × 105 Pa
3
m
s
5
2
F = 1.26 × 10 Pa 42 m = 5.27 × 106 N
(V 3)
p = 1.01 × 105 Pa + 1000
p = 1.01 × 105 Pa + 1000
p = 1.01 × 105 Pa + 1000
10 pts 12. From book HWK Prob 12.45 Weirdly, you’ve been given 300 g of hot tea at a
temperature of 75.0 ◦ C, but you want cold tea. The cup is well insulated so hardly any
heat goes in or out. You take ice out of your glass of ice water and add it to your tea.
What mass of ice should you add to obtain a final temperature of 15.0 ◦ C?
This is a conservation of energy problem using heat. The heat added to the
ice must equal the heat removed from the tea. This leads to the relationship:
Mice Lf,ice + Mice cice ∆Tice + Mtea ctea ∆Ttea = 0
where Lf,ice = 3.33 × 105 J/kg, cice = 4190 J/(kg ◦ C (this is the specific heat of
water because once the ice melts it is water), and ctea = 4190 J/(kg ◦ C (this
is the specific heat of water because tea is almost completely water). Since
you took the ice cube from your glass of ice water, the initial temperature
of the ice is 0 ◦ C like you had in lab.
(
Mice
)
J
J
J
◦
3.33 × 10
+ 4190
15
C
+
0.3
kg
4190
(−60◦ C)
◦
◦
kg
kg C
kg C
5
= Mice 3.96 × 105
Mice =
(
75420 J
= 0.191 kg
3.96 × 105 J/kg
(V 1)
)
J
J
J
+ 4190
20 ◦ C + 0.25 kg 4190
(−50◦ C)
3.33 × 10
◦
kg
kg C
kg ◦ C
5
Mice
= Mice 4.17 × 105
Mice =
(
Mice
J
− 75420 J = 0
kg
J
− 52375 J = 0
kg
52375 J
= 0.126 kg
4.17 × 105 J/kg
(V 2)
)
J
J
J
3.33 × 105
+ 4190
18 ◦ C + 0.35 kg 4190
(−70◦ C)
◦
kg
kg C
kg ◦ C
= Mice 4.08 × 105
Mice =
J
− 1.03 × 105 J = 0
kg
1.03 × 105 J
= 0.251 kg
4.08 × 105 J/kg
(V 3)
Equations
Basic Mathematic Formulas
cos θ = A/H tan θ = O/A H 2 = O2 + A2
sin θ = O/H
Ccirc = 2πr
Vsph =
4π 3
r
3
Acirc = πr2
√
−b ± b2 − 4ac
x=
2a
Asur of sph = 4πr2
Chapter 2
∆x = xf − xi
(vx )f = (vx )i +ax ∆t
āx = [(vx )f − (vx )i ]/(tf − ti ) = ∆vx /∆t
1
1
∆x = [(vx )i +(vx )f ]∆t
xf = xi +(vx )i ∆t+ ax ∆t2
(vx )2f = (vx )2i +2ax ∆x
2
2
v̄x = ∆x/∆t
Chapter 3
Ax = A cos θ
(vy )f = (vy )i +ay ∆t
Ay = A sin θ
A=
√
A2x + A2y
tan θ = Ay /Ax
⃗vav = ∆⃗d/∆t
⃗aav = (∆⃗v)/∆t
∆⃗d = ⃗df − ⃗di
1
1
∆y = [(vy )i +(vy )f ]∆t
yf = yi +(vy )i ∆t+ ay ∆t2
2
2
2
f = 1/T
v = 2πR/T
a = v /R
⃗ net
F
Chapter 4
⃗1 +F
⃗2 +F
⃗ 3 + ... = m⃗a
=F
(vy )2f = (vy )2i +2ay ∆y
⃗ 12 = −F
⃗ 21
F
Chapter 5
w = mg
fs max = µs n
fk = µk n
fr = µ r n
Chapter 6
s
θ=
r
ωav
θf − θi
∆θ
=
=
tf − ti
∆t
Fg = G
m1 m2
r2
ω = (2π rad)f
G = 6.67259 × 10−11
N · m2
kg2
v = ωr
T2 =
v2
a=
= ω2r
r
4π 2 3
r
GM
Chapter 7
ωf − ωi
∆ω
x1 m1 + x2 m2 + x3 m3 + ...
αav =
=
at = αr
xcg =
tf − ti
∆t
m1 + m2 + m3 + ...
1
1
∆θ = ωi ∆t + α∆t2
ω 2 = ωi2 + 2α∆θ
ω = ωi + α∆t
∆θ = (ωi + ω)∆t
2
2
∑
∑
τ = rF⊥ = r⊥ F = rF sin ϕ
I≡
mr2
τ = Iα
vobj = ωR
aobj = αR
∑
Fx = 0
∑
Chapter 8
Fy = 0
∑
τ =0
Fsp = −k∆x
Chapter 9
⃗J = F
⃗ avg ∆t
⃗p = m⃗v
⃗ avg ∆t = ∆⃗p = m⃗vf −m⃗vi
F
∑
L ≡ Iω
τ=
m1⃗v1i +m2⃗v2i = m1⃗v1f +m2⃗v2f
∆L
∆t
Chapter 10
W = ∆E = ∆K + ∆Ug + ∆Us + ∆Eth + ∆Echem + ...
1
Ktrans = mv 2
2
1
Krot = Iω 2
2
W = F∥ d = (F cos θ)d
1
Us = kx2
2
Ug = mgy
P =
∆E
∆t
P = Fv
Chapter 11
what you get
QH − QC
TC
emax = 1 −
∆Eth = W + Q
e=
what you pay
QH
TH
Qc
TC
QH
TH
COP =
COPmax =
COP =
COPmax =
Win
TH − TC
Win
TH − TC
e=
Chapter 12
n = N/NA = M (in grams)/Mmol
9
TC = T − 273
TF = TC + 32
5
Q = M c∆T
∑
Qk = 0
Q = ±M L
p = F/A
pV = N kB T = nRT
∆L = αLi ∆T
∆V = βVi ∆T
Q
Th − Tc
= kA
∆t
L
Q
= σAe(T 4 − T04 )
∆t
Chapter 13
ρ=
m
V
p=
F
A
p = p0 +ρgd
FB = ρf Vf g
A1 v1 = A2 v2
1
1
p1 + ρv12 +ρgy1 = p2 + ρv22 +ρgy2
2
2
Chapter 14
Fs = −kx
1
Us = kx2
2
√
T = 2π
v = −Aω sin(ωt)
m
k
f=
1
T
a = −Aω cos(ωt)
2
ω = 2πf
x = A cos(ωt)
√
T = 2π
L
g
Some Properties of Materials
Material
Lead
Iron/Steel
Copper
Aluminum
Water
Ice
density
11000 kg/m3
7900 kg/m3
8900 kg/m3
2700 kg/m3
1000 kg/m3
917 kg/m3
Linear Expansion Coef
2.9 × 10−5 ◦ C−1
1.2 × 10−5 ◦ C−1
1.7 × 10−5 ◦ C−1
2.3 × 10−5 ◦ C−1
N.A.
2.8 × 10−5 ◦ C−1
Specific Heat
128 J/(kg ◦ C)
448 J/(kg ◦ C)
387 J/(kg ◦ C)
900 J/(kg ◦ C)
4190 J/(kg ◦ C)
2090 J/(kg ◦ C)