Transcript Physics Semester I Final Exam Review • Write the following information in your notebook • It is to be turned in the day.

Physics
Semester I Final Exam Review
• Write the following information in your
notebook
• It is to be turned in the day of final exams as a
major grade
• We are to cover the Conservation of Energy,
Conservation of Momentum, and Thermal
Energy in this block of instruction.
• This is most of your review for the final exam
Science Physics Unit 01: Laboratory
Management
1. A student carefully opens the gas valve and attempts to
light a Bunsen burner using a striker. After several attempts,
the student is unable to light the burner. The student
believes the striker is broken. Which is the best course of
action for the student?
• A. Disconnect the Bunsen burner from
the gas valve.
• B. Open the gas valve further and continue trying to
light the burner.
• C. Attempt to light the burner using another lit
Bunsen burner.
• D. Close the gas valve and look for another striker.
2. You are using a concentrated solution of copper (II)
sulfate to generate a current for electroplating. When
finished, you need to dispose of the solution. Which of
the following is the best method of disposal?
• A. Pour the solution down the drain and
run the water for five minutes.
• B. Pour the solution into a sealed bottle
and place in the storeroom.
• C. Discard the solution in accordance
with the school's waste disposal plan.
• D. Take the solution outside and pour it
out on a grassy area.
3. In the drawing, approximately how many
volts does the voltmeter indicate?
•A. 0.07 V
•B. 0.7 V
•C. 6.7 V
•D. 67 V
4. Which of the following is the BEST sequence for the
Laboratory Report format?
• A. Title, Purpose/Problem, Introduction, Materials,
Procedures, Results/Analysis, Discussion,
Conclusion
• B. Title, Introduction, Purpose/Problem,
Procedures, Materials, Results/Analysis, Discussion,
Conclusion
• C. Title, Purpose/Problem, Introduction, Materials,
Procedures, Results/Analysis, Conclusion, Discussion
• D. Title, Purpose/Problem, Discussion, Introduction,
Materials, Procedures, Results/Analysis, Conclusion
Science Physics Unit 07: Conservation of Energy and
Energy Transformations
5. A 100 kg gymnast comes to a stop after
tumbling. Her feet do 5,000 J of work to stop
her. Which of the following was the girl's
velocity when she began to stop?
• A. 5 m/sec
• B. 10 m/sec
• C. 50 m/sec
• D. 100 m/sec
• KE = 1/2 mv2
• V = (2KE/m)1/2 = (10000/100) 1/2
6. Which of the following does not contribute to
the potential energy of an object?
• A. height of the object above the Earth's
surface
• B. gravitational constant or g of the Earth
• C. velocity of the object
• D. mass of the object
• GPE = mgh
• 7. A 1000 kg truck moving at 10 m/s runs into a
concrete wall. It takes 1.0 sec for the truck to
completely stop. What is the magnitude of
force exerted on the truck during the
collision?
Given m, vi, vf, and t
• W = F * d = KE F = W/d
F = (KE)/d
• KE = 1/2 mv2 = (.5)(1000 kg)(10 m/s)2
= 50,000 J
• d = vt = (10 m/s)(1.0 s) = 10 m
• F = (KE)/d = (50,000 J)/(10 m) = 5,000 N
8. A football player pushes a 300 N coach riding a
blocking sled up a 6.0 m long ramp using a force of
150 N. The coach was raised a vertical distance of 3.0
m. The actual work done by the worker was —
• A. 300 N
• B. 600 N
• C. 900 N
• D. 1200 N
• W = Fd
Win = Wout
9. Which of the following is the best example of
kinetic energy being transformed into
potential energy?
• A. Winding a grandfather clock
• B. Bending a paper clip
• C. Coasting down hill on a bicycle
• D. Starting an automobile engine
10. Which of the following is not associated with
mechanical energy?
• A. A ball rolling down a ramp
• B. A diver on a spring board waiting to dive
• C. A loaded spring
• D. A heat lamp
11. A student designs a machine that he claims
has at least a 95% efficiency. His machine
based on a simple lever requires 100 N of
force applied through a distance of 3 m to
raise a 50 N object 5.8 m. Can you support
his claim? Show your calculations
• eff = Wout/Win x 100 =
• [(50N)(5.8m)]/[(100N)(3m)] x 100 = 96.7 %
• YES
12. If a power lifter raises a 1000 N weight a
distance of 2.0 meters in 0.5 seconds, what is
his power output?
• P = work/time W = Fd P = Fd/t
• = (1000N)(2m)/0.5s = 4000Watts
13. An ice skater glides across a skating rink, slowly
coming to a stop. The total work done by friction to
stop this skater is known. Which of the following
information is needed to calculate the initial velocity of
the skater?
• A. The mass of the skater
• B. The acceleration of the skater
• C. The force that starts the skater sliding
• D. The amount of time it takes the skater to stop
• W = KE = ½ mv2
v = (2KE/m)1/2
14. For the lever apparatus shown, the weight raises 2 cm for each
20 cm that the end of the bar raises. If the weight is 100 N, how far
would the end move to do 40 J of work on the ball?
•A. 1 m
•B. 2 m
•C. 3 m
•D. 4 m
W = Fd
d =W/F = 40J/ 100N = .4m
Class II lever mechanical advantage = dout/din = 20cm/2cm
.4m x 20cm/2cm = 4m
dout
din
15. The speed for an object to escape the Earth's gravity is
called the escape velocity. This speed is calculated from
energy considerations. An object must have enough
kinetic energy to overcome the gravitational potential
energy. For a 20 kg satellite in a circular orbit around the
Earth at a distance of 7x105 m from the center, we find
the orbital speed to be 6.75x103 m/sec. From that
location calculate the gravitational PE, the kinetic energy
and the escape velocity. m1 = 20kg m2 = 6x1024 d = 7x105
MEarth = 6x1024 kg
G = 6.67x10-11
• GPE = -G(m1)(m2)/d =
• -[(6.67x10-11) (6x1024)(20)] /(7x105)= 1.14x1010 J
• Orbital KE = ( mv2) = (.5)(20)(6.75x103)2 =4.56x108 J
• v = [2(GPE)/m] 1/2 = [2(1.14x1010)/20] 1/2 =3.38x104 m/sec
Science Physics Unit 08: Momentum and Energy in
Collisions
Momentum is defined as p = mv where p is the symbol for
momentum, m is the symbol for mass, and v is the
symbol for velocity. Momentum is a vector quantity and
points in the direction of the direction of the velocity.
• Momentum has units of kg m/sec. A large mass moving at a
high velocity has a large momentum.
Impulse is defined as J = Fav Δt where J is the impulse, Fav is
the average force over an event (collision) over
a time period Δt. The time interval is normally short but that
is not a real restriction. Impulse uses the same
units as momentum and, in normal usage, an impulse is
given to an object producing a change in momentum.
• Thus J = Δp for an object involved in a collision type event.
Conservation of Momentum is a law regarding the total momentum of a system
when external forces can be
neglected. The most common use of this law is in describing features of
collisions. The simplest of these events is a head on collision between two
objects. In that situation the total momentum of the colliding objects before
the collision is equal to the total momentum after the collision.
• Totally Elastic Collision is a collision where kinetic energy is conserved. This is
an ideal situation because some kinetic energy is almost always lost in a
collision.
• Totally Inelastic Collision is a collision where the two (or more) colliding
objects stick together after the collision.
• Coefficient of Restitution (e) is a term sometimes used to indicate the
relative elasticity of collisions. It is defined as the ratio of the departing
relative velocity of colliding object to the approaching relative velocity. The
values range from 1 for a perfectly elastic collision to 0 for a perfectly inelastic
collision. Kinetic energy will be lost in this type collision.
16. A 2 kg cart rolls across the floor and runs into the wall
and sticks.. Before the collision the cart has a velocity
of 4 m/sec in the x direction. The collision took place in
0.1 sec. The bold quantities below are vectors and
require a direction as part of the answer.
• 1. What was the cart’s momentum before the
collision? m = 2kg, v = 4m/s, t = .1s
• P=mv=2kg x 4m/s = 8 kg m/sec in the x direction
• 2. What was the cart’s KE before the collision?
• KE= 1/2mv2 = ½ (2kg x 42) = 16J
• 3. What was the cart’s momentum after the collision?
• 0
• 4. What was the change in the cart’s momentum?
• 8 kg m/sec in the negative x direction
• 17. What was the impulse delivered to the car by
the wall? ? m = 2kg, v = 4m/s, t = .1s
• J = Ft = mat = m x v/t x t = 2kg x 4m/s/.1s x .1s =
8 kg m/sec in the negative x direction
• 18. What was the average force exerted on the car
by the wall?
• F = J/t = 8kgm/s / .1s = 80 N in negative x direction
• 19. What was the average force exerted on the wall
by the car?
• 80 N in the positive x direction
• 20. Describe the collision as partially elastic, totally
elastic, totally inelastic.
• totally inelastic
21. A 1 kg mass moving at 1 m/sec has a totally inelastic collision with a 0.7 kg
mass. What is the speed of the resulting combined mass after the collision? What
is the ratio of final kinetic energy to the initial total kinetic energy for the
collision?
m1 = 1 kg
v1i = 1 m/sec
m2 = 0.7 kg
m1v1i = (m1 + m2) vf
Vf = (m1v1i) / (m1 + m2) =
1 / 1.7 = .59m/s
Vf = m1V1i/m1 + m2 =
.59m/s
• KEi = ½ m1v1i2 = ½ (1)(1) = .5 J
• KEf = ½ (m1 + m2)Vf2 =
½ (1.7)(.59)2 = .3J
• Ratio = .3/.5
22. An excellent golfer can provide an impulse of nearly 2 kg m/sec
impulse to a 0.045 kg ball. What momentum and speed result from
such a hit?
• J = 2 kg m/sec
• m = 0.045 kg
• The impulse is equal to the
momentum thus p=2 kg m/sec
• p=mv
• v=p/m =2/0.045 =44.4m/s
• P = mv = .045kg x 44.4m/s =
2 kg m/s
23. A car with a mass of 1000 kg moves at 20 m/sec. What
braking force is needed to bring the car to a halt in 10 sec?
•
•
•
•
•
•
•
m = 1000 kg
vi = 20 m/sec
Δt = 10 sec
FΔt = Δp and F = Δp/ Δt
Δp = mv
F = mv/ Δt
F = (1000kg)(20m/s)/10s = 2000N
24. Air bags help to minimize injuries by providing a large
area for the force of impact to be exerted. In terms of
the impulse momentum theory, what is another
advantage of an air bag?
• A. Decreases the momentum of the
passenger
• B. Increases the amount of time the force
is applied
• C. Decreases the net force applied by the air bag
• D. Increases the impulsive force applied to the
passenger
25. A billiard ball moving at 0.5 m/s strikes another
identical billiard ball, which is at rest, in an elastic headon collision on a level table. Which of the following
statements best describes the billiard balls after the
collision?
• A. The moving ball rebounds directly back at 0.5 m/s
and the other ball remains at rest.
• B. The balls both move off together in the same
direction at 0.25 m/s.
• C. The moving ball continues to move at a 0.2 m/s in its
original direction and the other ball moves at 0.4 m/s in
the
same direction.
• D. The moving ball stops and the other ball moves off at
0.5 m/s in the original direction of the moving ball.
26. A 7 kg bowling ball traveling at 2.0 m/s collides with a
stationary 0.5-kg beach ball in an elastic collision. The bowling
ball leaves the collision with a velocity of 1.5 m/s traveling in
the same direction as the beach ball. Calculate the speed of
the beach ball.
Momentum before the collision equals momentum after the
collision.
• mb = momentum of bowling
• me = momentum of beach ball
• pi = pf
• mbvbi + mevei = mbvbf + mevef
• (7kg)(2 m/s) + (0.5 kg)(0 m/s) = (7kg)(1.5 m/s) + (0.5 kg)(vef)
• vef = (3.5)/(0.5) = 7 m/s
27. Which of the following has the least
momentum?
• A. .01 kg mass moving at 200 m/s
• B. 1 kg mass moving at 35 m/s
• C. 100 kg mass moving at 3 m/s
• D. 2000 kg mass moving at 0.1 m/s
• P = mv
28. A hammer strikes a nail with a 10 N force for
0.01 seconds. Calculate the impulse of the
hammer.
• J = Ft =
• (10 N)(0.01 s) = 0.1 kg m/s
29. A force of of 500 N is exerted on a baseball
by the bat for .001 s. What is the change in
momentum of the baseball?
• A. 0.5 kg m/s
• B. 5 kg m/s
• C. 50 kg m/s
• D. 400 kg m/s
• J = Ft
30. Two students on roller skates stand face to face and push each
other away. One student(A) has a mass of 90 kg and the other (B)
has a mass of 60 kg.What must the resulting speed of student (A)
be relative to student (B) in order for momentum to be conserved?
(Assume that the momentum before the push is zero.) In
otherwords, looking for a ratio of student B to student A
•vA = velocity of student one
vB = velocity of student two
mA = mass of student one = 90kg
mB = mass of student two = 60kg
•mAvA = mBvB or vA/vB = mB/mA = 60/90
= .667
31. A student using an air track with two carts records the
result of an inelastic collision. One cart has a mass of 0.2 kg
and the other a mass of 0.4 kg. They moved off together at
0.2 m/s after the collision. The 0.4 kg cart was initially at
rest. m1 = .2kg, m2 = .4kg, Vf = .2m/s
• -What was the final momentum of the carts?
•
The final momentum Pf was (m1 + m2)(Vf) = (0.2 + 0.4 )kg
(0.2 m/s) = 0.12 kg m/s.
• -What was the initial velocity of the 0.2 kg cart?
• m1v1 + m2v2 = (m1 + m2)(Vf) so (.2kg) v1 = (0.12kg m/s)
v1 = (0.12kg m/s) / .2kg = 0.6 m/s.
• -If the collision lasted 0.02 s, what force did
each cart exert on the other?
• The force between the carts is obtained by dividing the
change in momentum by the time.(0.12 kg m/s)/ .02 s = 6 N.
32. A baseball player hits a 2.5-kg ball with a force of
20 N. The duration of the force was .05 s.
m = 2.5kg, F = 20N, t = .05s
• -Calculate the impulse delivered to the ball.
J =Ft = (20N)(.05s) = 1 kg m/s
• -What was the change in momentum of the ball?
1 kg m/s
• -What was the velocity of the ball as it left the
bat? P =J = mv, v = J/m = (1 kg m/s)/ (2.5 kg) = .4 m/s
In an auditorium with a 8.5-m ceiling, a large pendulum was constructed using
a 7.0-kg bowling ball suspended on an 8.0-m steel wire. The bowling ball
was pulled to the side of its lowest hanging point. It was raised a distance
2.0 m vertically above its lowest point and released. m = 7kg, h = 2m
33. -What was its initial potential energy?
The initial potential energy is mgh = (7.0 kg)(9.8 m/s2)(2.0 m) = 137 J.
34. After dropping 1.0 m vertically, what is the bowling ball's velocity?
The ball's velocity after falling 1.0 m is calculated from 1/2 mv2 = 137 J.
This gives 4.4 m/s.
35. What is the bowling ball's maximum possible velocity?
KE = PE = 1/2 mv2 , v = (2PE/m)1/2 = (2(137)/7) 1/2 = 6.3 m/s.
36. At what height above its lowest point is the bowling ball's velocity
one-half of the maximum velocity?
The velocity would be 3.15 m/s at 1.5 m above the lowest point.
Substitute into mgh = 1/2 mv2 h = 1/2 mv2 / mg = ½ v2/g
h = ½(3.15m/s)2/9.8m/s/s = .50m
2.0m - .5m = 1.5m above its lowest point
A 5,000-kg railroad car moving at 2 m/s collides and
connects to another identical car initially at rest.
37. What is the momentum of the moving train
car before the collision?
p = mv = (5000 kg)(2 m/s) = 10,000 J
38. What is the final velocity of the two connected
train cars?
• P = mv
mf = m1 + m2
vf = p/(m1 + m2) = 10,000 J / 10,000 kg = 1 m/s
At scout camp, a 1.00-kg arrow is shot at 50m/s. It hits
and is embedded into a 2.00-kg block of wood that
slides to a stop.
39. Calculate the kinetic energy of the arrow.
KE= 1/2 mv2
= .5(1.00 kg)(50 m/s)2 =1250 J
40. Calculate the momentum of the arrow before it
hits the block of wood.
p= mv = (1.00 kg)(50 m/s) = 50 J
41. Calculate the initial velocity of the arrow and wood
after the inelastic collision.
v = p/(ma + mw) = 50/3kg = 16.7 m/s
42. Calculate the energy of the block and arrow after
the collision.
• KE= 1/2 mv2 = .5(3.00 )(16.7 m/s)2 = 418.3 J
Science Physics Unit 09: Thermodynamics
Equations and Constants
1. KEave = (½ m v2)ave = 3/2 kT: k = 1.38 x 10-23J/K
2. 0C = ( 0F -32) (5/9): 0F = (0C)9/5 + 32 : 0C = K - 273
3. Q = m c ΔT
4. Work = PΔV
5. 1 cal = 4.186 J = 3.97 x 10-3 BTU
6. Cwater = 1 cal/gram oC
Latent Heat Fusion = 80 cal/g0C
Latent Heat Vaporization = 540 cal/g0C
1. Temperature: how hot or cold
Temperature is related to average KE of molecules:
KEave = (1/2 mv2)average = 3/2 kT :
k = 1.38 x 10-23 J/K:
T is Kelvin temperature.
• Concept –higher temperature (macroscopic) is
faster molecules (microscopic)
• Three popular temperature scales:
Fahrenheit scale
Celsius scale –0 freezing ice, 100 boiling water, value
is Kelvin + 2730
Kelvin scale –0 K is absolute coldest.
• Absolute temperature scale
2. Thermal Energy: energy of temperature
U –Internal energy of a system (like an engine) is the total
of KE and PE of molecules –formally more than the
thermal energy but often the same. Official term in
engines and gasses.
Q –Heat –transfer of energy due to a temperature
difference – officially only called heat in transit
Heat "flows" from high temperature to low temperature
until thermal equilibrium (same temperature) is reached.
Conduction –requires matter but mass does not move,
similar to electricity, electrons carry energy
Convection –requires matter- mass moves carrying energy
Radiation –does not require matter –electromagnetic
waves carry energy –black absorbs and emits better.
Mirror –white reflect. Hot emits more energy
Heat units are calorie, Calorie, Joule, BTU (British
Thermal Unit).
1 calorie = heat to raise temperature of 1 gram of
water by 1o Celsius
1 Calorie = 1000 cal (diet calorie)
1 cal = 4.186 J = 3.97 x 10-3 BTU
Heat as a form of energy was established by James
Joule in an experiment where paddles
converted mechanical motion into heating water.
3. Specific Heat Capacity: how efficient at
absorbing, storing, or giving up internal energy
C –defined by the equation Q = m c ΔT: cal/gram
oC is unit –or other appropriate units
m = mass, c = specific heat, T = temperature
Materials with high heat capacity require lots of
energy to raise the temperature much
Water has a relatively high heat capacity of 1
cal/gramoC
Is one of the primary variables in heat transfer
and calorimetry problems
4. Expansion: most substances expand as temperature rises:
solids least, gasses most
Used in thermometers, thermostats
ΔL = αL0 ΔT –change in length proportional to length L0 and
ΔT: α is constant depending upon type of material. Bimetallic strip had metals of different α : linear coefficient of
expansion
ΔV = 3αV0 ΔT –change in volume is proportional to original
volume V0 and ΔT: 3α is volume expansion coefficient – for
gasses and liquids β substitutes for 3α in the equation
Water's behavior is unusual since it has maximum density
(minimum volume) at 40C. Thus as it gets near freezing
temperature it floats and freezes –from the top down, most
things freeze from the bottom up. Water also expands upon
freezing –forming hollow ringed structures
5. Phase Change: Energy is required to change a solid to a
liquid and from a liquid to a vapor
Phases are solid, liquid, and vapor: for water it takes 80
calories per gram to melt or freeze water and 540 calories to
vaporize water. Since this energy is hidden (doesn’t change
the temperature) it is called latent heat of –fusion,
vaporization
The heat of fusion and vaporization are basically the energy
necessary to break the solid structure bonds for fusion and
the bonds between molecules for vaporization. This is
analogous to breaking through the surface tension energy
for evaporation. In a related thought, boiling involves
forming bubbles in the depth of the liquid. This temperature
increases with increased air pressure. Water boils at lower
temperature in the mountains
43. Kinetic energy is a microscopic concept. The
equivalent macroscopic concept is —
• A. mass
• B. entropy
• C. temperature
• D. heat
44. The Second Law of Thermodynamics has a number of
equivalent definitions, all of which give an indication of
which way nature flows (in a time fashion). Which of these
statements is not generally derivable from the concept
that a particular engine is the most efficient engine
possible and that it is reversible?
• A. Heat will not flow spontaneously from a cold reservoir
to hot reservoir.
• B. An engine or combination of engines alone cannot
make heat flow from a cold reservoir to a hot reservoir.
• C. Heat will flow spontaneously from a cold reservoir to
hot reservoir.
• D. It is not possible to convert 100% of heat into work.
Practice: Thermochemistry
45. Convert 500C to Kelvin temperature.
add 273 and get 323 Thus 500C = 323 Kelvin
46. Helium gas atoms have a mass of 6.8 x 10-27 kg. What
is the speed of a He atom in a gas at 00C?
(1/2 mv2)average = 3/2 kT : .5 (6.8 x 10-27)(v2) = 1.5(1.38 x
10-23)(273)
vave = 1.3 km/sec: a heavier atom would be slower, and at
a higher temperature faster.
47. Sand has a specific heat of 0.5cal/gram0C. How many
calories does it take to raise the temperature of 100 g of
sand from 600C to 700 C?
• Q = m c ΔT Q = 100(.5)(10) = 500 calories.
48. How much would this amount of heat energy
(example 3) raise the temperature of 100
grams of water?
Q = m c ΔT : ΔT = Q / (m c) = 500/100 = 5 0C
49. How much energy does is required to take 10 grams
of ice at 00C and produce 10 grams of steam at 1000C.
For 1 gram it takes 80 calories to melt, 100 calories to
heat the water to 1000C and 540 calories to change it
into steam. Thus, for 1 gram it takes 720 calories. For
10 grams it would take 7,200 calories or 7.2 Calories
(with a capital C).
50. Heat flows from a hotter source to a cooler
region by which methods? Mark all that apply.
• a. convection
• b. radiation
• c. conduction
• d. expiration
• e. reduction
51. Water has a higher specific heat capacity than
sand. Which will cool down faster in the evening
and warm up faster in the morning Sun?
Sand heats and cools faster.
How does this influence the direction of the
winds on the beach in the day and night?
Hot air rises, so in the morning as the sand
temperature increases the beach air rises
pulling in air from the ocean. So the wind blows in
from the ocean in the day and out at night
52. Water boils at a lower temperature in the
mountains than at sea level. The primary
reason is that
• a. The temperature in the mountains is cooler
• b. The air pressure is less in the mountains
• c. The air is cleaner with less pollution
• d. The water is purer in the mountains than
from the faucet
53. Compare the temperature of a glass of ice
water (stirred completely) nearly full of ice
compared to a glass of ice water (stirred
completely) half full of ice.
• a. The nearly full of ice glass is colder
• b. The half full of ice glass is colder
• c. They are at the same temperature
• Explain:
• The ice water mixture will be at the meltingfreezing temperature of water.
54. The temperature of helium gas is directly
related to the kinetic energy of the gas. If a
quantity of helium in a closed container at 100 0C
has the kinetic energy of the helium atoms
doubled, what is the resulting temperature of the
gas?
• a. 200 0C
• b. 746 Kelvin = 2(273 + 100)
• c. 400 0C
• d. 540 Kelvin
55. Calculate the energy to convert 50g of ice at
00C to water at 500C.
• M = 50 grams ΔT = 500C
• Q = (Latent Heat of fusion )( mass) + m c ΔT =
80(50) + 50 (1) (50) = 6500 calories = 6.5 kcal
56. A bag of Peanut M&M’s has 220 Calories. If
this was used to heat the body of a 50 kg girl
(assume all water), how many degrees Celsius
would it raise her body temperature? (These
are kilocalories.)
• Q = 220 Calories (big calories) Q = m c ΔT
• thus ΔT = Q/mc = 220,000cal /(50,000 g)(1)
• ΔT = 4.4 0C
57. What is the final temperature when 100g of water
at 250C is mixed with 75g of water at 400C ?
• Since no latent heats, we can consider this a
straight conservation of energy. The number of
calories add to a total. Q1 + Q2 = Qtotal
• Q = m c T and c is 1 for water
• m1 = 100 g T1 = 25 0C
• m2 = 75 g T2 = 40 0C
m1T1 + m2T2 = (m1 + m2 ) Tfinal
• (100) (20) + (75)( 40) = 175 (Tfinal)
• Tfinal = 31.4 0C
58. Steel expands 1 part in 100,000 for each 10C increase
in temperature (α = 1 x 10-5 /0C). If the 1.5-km main span
of a steel suspension bridge has no expansion joints, how
much longer will it be for a temperature increase of 200C?
α = 1 x 10-5 /0C
ΔL = αL0 ΔT
ΔT = 200C
ΔL = αL0 ΔT = (1 x 10-5) (1500m) 20 = 0.3 m
59. Calculate the energy to convert
50g of ice at 00C to water at 500C
m = 50 g
heat = melt + raise temp
Q = [80 (m) + mc ΔT] = m(80 + 50)(1)
50[ 80 + 50] = 6500 calories
c =1
60. In one cycle, an engine burning a mixture of
air and alcohol absorbs 525 J and expels 415 J.
What is the engine’s efficiency?
• Qhot = 525 J
• Qcold = 415 J
• Although not asked, we need the Work = 525415 = 110 J
• efficiency = Work/Qhot = 110/525 = .21