5. Determine and graph an 18% increase in Velocity for each vehicle - Show how
the Times would be affected by the increase in speed.
*Include pictures and brief description of each mode of transportation
a. Fastest runner - Usain Bolt, an Olympic runner from Jamaica
i. with 18% increase= 33 mph: 50 hrs -
1. T=D/V
2. T=1651/33
3. T=50
b. Model T Ford - a Ford model created in 1908 and remained in production until
1927
i. with 18% increase= 53 mph: 31 hrs -
1. T=D/V
2. T=1651/53
3. T=31
c. Hindenburg - a German blimp that caught fire and was destroyed in 1937
i. with 18% increase= 99 mph: 17 hrs -
1. T=D/V
2. T=1651/99
3. T=17
d. Tesla Top Speed - an energy conserving vehicle created in 2003 in California
i. with 18% increase= 183 mph: 9 hrs -
1. T=D/V
2. T=1651/183
3. T=9
e. Fastest Train - is nicknamed the bullet train and is located in Japan
i. with 18% increse= 441 mph: 3 1/2 hrs -
1. T=D/V
2. T=1651/441
3. T=3 1/2
f. F35 Fighter Jet - a war aircraft with one engine, is designed to withstand air and
ground attacks
i. with 18% increase= 1415 mph: 1 hr -
1. T=D/V
2. T=1651/1415
3. T=1
g. Electric Scooter - a scooter powered by a small motor
i. with 18% increase= 47 mph: 35 hrs -
1. T=D/V
2. T=1651/47
3. T=35
If the speed was increased, the time in which I would be travelling would decrease.
18% Increase
Transport Time (hrs)
Fastest Runner 50
31
Model T Ford 17
Hindenberg 9
Tesla (Top
Speed)
Fastest Train 3 1/2 1
F35 Fighter Jet 35
Electric Scooter
6. Use a math calculation to show how long it would take the F35 Fighter Jet to
get to
A. Sun - 92.96 mil mi
B. Saturn - 74.56 bil mi
C. Neptune - 2.7 bil
F35 - 1199 mph
A. T=D/V
T=9 2,960,000/1199
T=77531 = 7.75*104
B. T=D/V
T=74,560,000,000/1199
T=62185 = 6.2*104
C. T=D/V
T=2,700,000,000/1199
T=2251 = 2.25*104
(Use scientific notation)
Velocity Practice Problems
Unit 1: Uniform Motion Name Nicole Calbo, Nicole Lockwood, Faryal Chaudhary
Worksheet 8 Date 2 /22/17 Period S 7
Speed and Velocity Problems
1. What is the average speed of a cheetah that sprints 100 m in 4 s? How about if it sprints
50 m in 2 s?
V=D/T
V=100 m/4 s, V=50 m/2 s
V=25 m/s, V= 25 m/s
2. If a car moves with an average speed of 60 km/hr for an hour, it will travel a distance of
60 km. How far will it travel if it continues this average rate for 4 hrs?
D=V • T
D=60km/hr • 4hrs
D=240 km
3. A runner makes one lap around a 200m track in a time of 25.0 s. What was the runner's
average speed? Answer: 8.0 m/s
V=D/T
V=200m/25s
V=8 m/s
4. Light and radio waves travel through a vacuum in a straight line at a speed of very nearly
3.00 × 108 m/s. How far is light year (the distance light travels in a year)? Answer: 9.50
× 1015 m.
D=V • T
D=3 • 108 m/s
D= 9.50 • 101 5 m.
5. A motorist travels 406 km during a 7.0 hr period. What was the average speed in km/hr
and m/s? Answers: 58 km/hr, 16 m/s.
V=D/T
V=406km/7hr
V=58 km/h, 16 m/s
6. A bullet is shot from a rifle with a speed of 720 m/s. What time is required for the bullet
to strike a target 3240 m away? Answer: 4.5 s.
T=D/V
T=3240m/720m/s
T= 4.5 s
7. Light from the sun reaches the earth in 8.3 minutes. The speed of light is 3.0 × 108 m/s.
In kilometers, how far is the earth from the sun? Answer: 1.5 × 108 km.
D=V • T
D=1.5*108 km
D=1.5 x 108 km
8. *An automobile travels at a rate of 25 km/hr for 4 minutes, then at 50 km/hr for 8
minutes, and finally at 20 km/hr for 2 minutes. Find the total distance covered in km and
the average speed for the complete trip in m/s. Answers: 9 km, 10.7 m/s.
V=D/T
V=25km/r/4m, 50km/hr/8m, 20km/hr/2m
V= 9km, 10.7 m/s
9. *If you traveled one mile at a speed of 100 miles per hour and another mile at a speed of
1 mile per hour, your average speed would not be (100 mph + 1 mph)/2 or 50.5 mph.
What would be your average speed? (Hint: What is the total distance and total time?)
Answer: 1.98 mph.
T=D/V
T=100m/1 hr
T=1.6mph
T=D/V
T=1m/ 1 hr
T=1mph
1.3 mph
10. *What is your average speed in each of these cases?
a. You run 100 m at a speed of 5.0 m/s and then you walk 100 m at a speed of 1.0
m/s.
b. You run for 100 s at a speed of 5.0 m/s and then you walk for 100 s at a speed of
1.0 m/s. Answers: 1.7 m/s, 3.0 m/s.
V=D/T
V= 100m/ 5.0m/s/1.0m/s
V=1.7 m/s, 3 m/s
11. *A race car driver must average 200 km/hr for four laps to qualify for a race. Because of
engine trouble, the car averages only 170 km/hr over the first two laps. What average
speed must be maintained for the last two laps?
170=85% of 200
+15% increase to 200
15% of 200=30
230 km/hr
Gravitational Potential Energy Project
Define and make note cards for the following words:
Energy - the ability to Joules - a universal Chemical Potential Law of Conservation
do work, can have unit to measure Energy - a form of of Energy - the total
energy of their motion energy chemical energy that energy of an isolated
or position can be absorbed or system that remains
released in a constant
chemical reaction or
phase transition
Kinetic Energy - Kilojoules - equal to Elastic Potential Gravity - the force
energy that 1000 joules Energy - potential that attracts things
somethign posseses energy of an elastic towards the center of
by being in motion object, caused by its the Earth
deformation
Potential Energy - Gravitational Mechanical Energy -
energy produced by Potential Energy - the energy due to
something based amount of energy an something’s motion
upon its position, object posseses or position
relative to others. because of its
position in a
gravitiational field
Resource: http://www.physicsclassroom.com/class/energy/Lesson-1/Potential-Energy
Gravitational Potential Energy
Determine the Gravitational Potential Energy (GPE) of 3 different masses (g) at 3 different
heights.
3 objects: You, gallon of milk, television (research the masses)
* 2.2 lbs = 1 kg
Object 1 - average person
- 120 lbs
- 264 kg
Object 2 - gallon of milk
- 8.6 lbs
- 18.92 kg
Object 3 - television
- 30 lbs
- 66 kg
Data Table:
Object Mass (kg) Gravity (m/s2) Height (m) GPE (joules)
Average 264 9.8 1 2587.2
Person
264 9.8 5 12936
Average
Person 264 9.8 10 25872
Average
Person
Object Mass (kg) Gravity (m/s2) Height (m) GPE (joules)
Gallon of Milk 18.92 9.8 1 185.42
Gallon of Milk 18.92 9.8 5 927.08
Gallon of Milk 18.92 9.8 10 1854.16
Object Mass (kg) Gravity (m/s2) Height (m) GPE (joules)
TV
TV 66 9.8 1 646.8
TV
66 9.8 5 3234
66 9.8 10 6468
Your data table will need: Object, mass, gravity, height, GPE
Videos: http://www.youtube.com/watch?v=x5JeLiSBqQY
*Video shows you how to use the GPE equation.
Determine the GPE of one of the masses on the following planets:
“Kotulskian” - 17% greater than Earth’s Gravity
9.8m/s2* 17%/100
166.6m/s2/100
1.67m/s2+9.8m/s2= 11.47m/s2
“Danuzzitopia” - 39% less than Earth’s Gravity
9.8m/s2 * 39%/100
382.2m/s2 / 100
3.82m/s2 , 9.8m/s2=3.82m/s2
“Carlucciville” - 82% greater than Earth’s Gravity
9.8m/s2 *82%/100
803.6m/s2 / 100
8.04m/s2+9.8m/s2 =1 7.84m/s2
“Cheshire” - 63% less than Earth’s Gravity
9.8m/s2 *63%/100
617.4m/s2/100
6.17m/s2, 9.8m/s2-6.17m/s2= 3.63m/s2
Calculations: Object - gallon of milk
Change the names below
A. Planet Kotulskian:
GPE=mgh
GPE=18.92kg*11.47m/s2 * 1m
GPE=217.01 joules (1m)
=1085.07 (5m)
=2170.12 (10m)
B. Planet Danuzzitopia:
GPE=mgh
GPE=18.92kg*5.98m/s2 *1m
GPE=113.14 joules (1m)
=565.7 joules (5m)
=1131.4 joules (10m)
C. Planet Carlucciville:
GPE=mgh
GPE=18.92kg*17.84m/s2*1m
GPE=337.53 joules (1m)
=1687.66 joules (5m)
=3375.33 joules (10m)
D. Planet Cheshire:
GPE=mgh
GPE=18.92kg*3.63m/s2 *1m
GPE=68.68 joules (1m)
=343.4 joules (5m)
=686.8 joules (10m)
Data Table:
Planet GPE (joules)
Kotluskian 217.01
Danuzzitopia 113.14
Carlucciville 337.53
Cheshire 68.68
Graph:
Critical Thinking Questions:
1. What factors affect Gravitational Potential Energy?
Gravity, mass, height
2. Why did the GPE change on the other planets?
Because the other planets had different amounts of gravity
3. Which planet would you be able to hit a golf ball further? Explain using data.
Cheshire because there is the smallest amount of gravity and the smallest amount of
resistence, therefore the golf ball will go the furthest.
4. How does GPE relate to Chemical Potential Energy?
Because in GPE, its the amount of energy potentially there in a planets gravity. With CPE, it is
the same, however, only with a possible amount of energy in a chemical experiment, rather than
an entire planet.
5. How do Energy companies use GPE to generate Electrical Energy? Give an example
They use it by, for example, in a pumped storage hydroelectric power station, storing the water
underground. Eventually this transfers into potential energy.
6. What happens to the GPE when the object falls to the ground? Describe the Energy
transformations along the way. Use a diagram.
When an object hits the ground, its GPE increases. This is because each time it hits the ground,
it is getting closer to the center of the earth, therefore, there is more of a gravitational pull. Along
the way, the potential energy transfers into kinetic energy. At the top, all of the energy is
potential, in the middle of the fall, the energy is evenly divded between kinetic and potential, and
once the object hits the ground, all of the energy is kinetic.
Kinetic Energy Project
Problem Statement:
1. Design an experiment to test how changing the angle of a ramp affects Kinetic
Energy.
2. What is the velocity of a roller coaster at the bottom of the hill?
Hypothesis:
If you increase the angle of the ramp, the ramp with the highest angle will have the highest
final velocity.
Independent Variable:
Angle
Dependent Variable:
Speed/velocity of the car
*Use the following angles: 8, 18, 28
Example: Suppose you want to make a ramp with a 23 degree angle. Look up 23
degrees on the Angle Chart and find the sine (decimal).
Sin 23 = 0.39
Substitute Sin 23 with 0.39
0.39 = opposite
hypotenuse
0.39 = opposite (height of ramp)
180 cm
solve for opposite (height of ramp)
*A ngle Chart
*Include diagrams of the 3 Angles
*Include math examples for determining the angles and KE.
- SinB=opp/hyp
8°=opp/1.21m
0.14=opp/1.21m
.169m=opp
GPE=KE
mgh=0.5mv2
33.3g*9.8m/s2* .169m=0.5*33.3g*v2
55.15g/s2= 16.65g*v2
3.31s2 =v2
1.82=v
- SinB=opp/hyp
18°=opp/1.21m
0.3=opp/1.21m
.363m=opp
GPE=KE
mgh=0.5mv2
33.3g*9.8m/s2*0.363m=0.5*33.3g*v2
118.46g/s2= 16.65g*v2
7.11s2= v2
2.67=v
- SinB=opp/hyp
28°=opp/1.21m
0.47=opp/1.21m
.569m=opp
GPE=KE
mgh=0.5mv2
33.3g*9.8m/s2* 0.569m=0.5*33.3g*v2
185.69g/s2=16.65g*v2
11.15s2 =v2
3.34=v
Angle (degree) Velocity (m/s2)
Angle 8 1.82
Angle 18 2.67
Angle 28 3.34
Video Resources:
1. How to solve KE problems: https://www.youtube.com/watch?v=tryiwu4RhSM
2. PE/KE: h ttps://www.youtube.com/watch?v=Je8nT93dxGg
3. PE/KE: h ttps://www.youtube.com/watch?v=BSWl_Zj-CZs
4. PE/KE: h ttps://www.youtube.com/watch?v=7K4V0NvUxRg
Data Analysis - W rite your data analysis paragraph here
It has been found in the experiment that as the angle of the ramp increases, as does the
height of the ramp. For example, the ramp with the 8° angle had the smallest opposite. This
was 0.169m, and because of this, it had to lowest amount of velocity- 1.82m/s2. Also, the ramp
with the highest angle of 28°, had the greatest velocity, of 3.34 m/s2. Therefore, because of
this, the objects with the greatest amounts of velocities will, in turn, become the objects with
the highest amounts of Kinetic Energy at the bottom of the ramp. Therefore, the data has
proven that the higher the angle of the ramp, the larger the velocity will be, and because of
this, have the largest amount of kinetic energy on the bottom of the ramp.
Inclined Plane Project
1. Define the following vocabulary: Use pgs. 124 - 153
Simple Machine- A Mechanical Work- Transfer of Input Force- The
machine that does Advantage- The ratio energy that occurs force that is applied
work with only one of the output force to when a force makes to the machine. Fi n
movement of the the input force. an object move.
machine.
Compound Machine- Ideal Mechanical Power- The amount Output Distance-
Two or more simple Advantage- Can be of work done in one Distance output force
machines that calculated for any second. It is a rate - is exerted through.
operate together. machine by dividing the rate and which
the input distance by work is done.
the output distance.
(IMA)
Efficiency- Measure Actual Mechanical Input Distance- Output Force- The
of how much of the Advantage- The force Distance the force force applied by the
work put into the that a machine can acts through. machine. Fo ut
machine is changed multiply while
into useful output subtracting losses
work by the machine. from the machine
having to overcome
friction. Machine
multiplies applied
force (AMA).
First Class Lever- Second Class Lever- Third Class Lever- Energy- When you do
The fulcrum is The output force is The input force is work on the machine,
located between the located between the applied between the you transfer energy
input and output input force and the output force and the to the machine.
forces. fulcrum. fulcrum.
Block and Tackle Fixed Pulley- the Movable Pulley- a
Pulley- a system of wheel and axel stay pulley that moves an
two or more pulleys in the same place object
that work together
2. Experiment: How does the angle of an inclined plane affect:
A. Ideal Mechanical Advantage
B. Actual Mechanical Advantage
C. Efficiency
*Think about the scientific Method
DATA TABLE
***Why is the Actual Mechanical Advantage always less than the Ideal Mechanical
Conclusion:
*Write your OWN CONCLUSION HERE!
When comparing the data, one can see that the lower the angle of the ramp, the
higher the efficiency; because in the data you can see that the angle of 30 degrees
had a lower efficiency, and the anlge of 25 degrees had the highest efficiency.
Wherefore, our hypothesis was not supported.
3. Critical Thinking (Include in Presentation):
A. How much WORK would be done to lift a 350 kg Piano to the top of the Empire State
Building using a ramp with an angle of 35 degrees?
1. What is the length of the ramp to the top of the Empire State Building?
>A=90 degrees
>B=55 degrees
>C=35 degrees
SinC=opp/hyp
Sin35=opp/hyp
Sin35=443m/hyp
Sin35*hyp=443m
Sin0.573*hyp=443m
hyp=773.12m
The length of the ramp is 773.12m.
2. Suppose the Ideal Mechanical Advantage (IMA) is 3.2
IMA=effort/height
3.2=effort/443m
1417.6=effort
3. The Actual Mechanical Advantage (AMA) is 2.5.
AMA=output force/effort
2.5=output force/1417.6
3544=output force
4. What is the Efficiency of this Machine?
input work=input force*input distance
input work=1417.6*3.2
input work=4536.32
output work=output force*output distance
output work=3544*2.5
output work=8860
efficiency=(work output/work input)*100
efficiency=(8860/4536.32)*100
efficiency=1.95*100
efficiency=195
5. Provide a diagram of this example.
4. Explain how the Ideal Mechanical Advantage and Actual Mechanical Advantage is
determined for the following simple machines:
A. Inclined Plane
- Force can be exerted over a longer distance that requires a smaller input and output
force
IMA- use formula IMA=distance in/distance out
AMA- use formula AMA=force out/force in
B. Lever
- Can pull multiple forces at a time
IMA- use formula IMA=effort distance fulcrum/resistance distance fulcrum
AMA- use formula AMA=resistance force/effort force
C. Pulleys
- Able to move weight equal to the amount of force being used
IMA- mainly calculated by counting the number of strings
AMA- use formula AMA= force resistance/force effort
D. Wheel and Axle
- Can lift heavy weights with just a string/rope and wheel
IMA- use formula IMA=radius of wheel/radius of axle
AMA- use formula AMA=diameter of wheel/diameter of axle
QUIZ: Inclined Plane
Directions: A nalyze the Inclined Plane Data Table that is shared on
Classroom and determine which machine has the greatest Actual
Mechanical Advantage (AMA).
Problem Statement:
How does the angle of an inclined plane affect the Mechanical
Advantage?
Hypothesis: (Use proper form!)
If the angle of the inclined plane increases, its mechanical advantage will increase as
well.
Diagrams of Inclined Planes: (Label Diagrams)
Angle Chart: https://drive.google.com/open?id=0B4RmhXJlHvo1YXZhcDNMSDNSMXc
Calculations ( Examples):
IMA=Di n/Do ut AMA=Fo ut/ Fi n Efficiency=Wo ut/ Win*100
=150m/30m =7/2 =210/300*100
=5 =3.5 =0.7*100
=70%
IMA=Di n/ Dout AMA=Fout/ Fin
=90m/30m =7/3 Efficiency=Wout/Win*100
=3 =2.33 =210/270*100
=0.78
IMA=Di n/ Dout AMA=Fo ut/Fin =78%
Efficiency=Wout/ Win* 100
=50m/30m =7/4 =210/200*100
=1.67 =1.75 =1.05*100
=105%
Graph: ( Angle and Mechanical Advantage)
Conclusion:
Option #1 Write a Conclusion.
***Your conclusion must also address which machine would be impossible
and why?
In conclusion, when comparing efficiencies between different inclined
planes, it will be most efficient taking an object up the ramp of 19
degrees. This is because although the angle of 37 degrees has a
seemingly higher efficiency than the other planes, it is impossib. This is
because its efficiency is over 100%; 105%. Therefore, the most efficient
inclined plane is that of 19 degrees.
Option #2 Building
Compare 2 famous buildings based on the following Inclined Plane Data.
A builder wanted to get a 60 kg bag of concrete to the top of the 2
buildings.
Space Needle- 184m to tip
Eiffel Tower- 300m to tip
Fo ut=588kg (mass*gravity)
1. What would the length of the ramp be if the machine has:
IMA - 5.9
AMA - 3.7
IMA=Din/Do ut IMA=Di n/Dout
5.9=Di n/ 184m 5.9=Di n/300m
1085.6m=Din 1770m=Di n
The length of the ramp to the top of the Space Needle is 1085.6m.
The length of the ramp to the top of the Eiffel Tower is 1770m.
2. What are the angles of the ramps?
>A=90 degrees >A=90 degrees
>B=opp/hyp >B=opp/hyp
=184m/1085.6m =300m/1770m
=0.17 =0.17
=10 degrees =10 degrees
>C=80 degrees >C=80 degrees
3. Using Work Input and Output, what is the efficiency of the machines?
AMA=Fout/ Fin AMA=Fout/ Fin
3.7=588N/Fin 3.7=588N/Fin
3.7Fin= 588N 3.7Fi n=588N
Fin=158.92N Fi n=158.92N
Win=Fi n*Din Win=Fi n*Di n
=158.92N*1085.6m =158.92N*1770m
=172523.55J =281288.4J
Wout= Fout*Do ut Wout=Fo ut*Dout
=588N*184m =588N*300m
=108192J =176400J
Efficiency=Wout/Wi n* 100 Efficiency=Wout/ Wi n*100
=108192J/172523.55J*100 =176400J/281288.4J*100
=0.63 =0.63
=63% =63%
In comparing the efficiency of taking a 60kg bag of cement up an
inclined plane either leading to the top of the Space Needle or Eiffel
Tower, both of their efficiencies would be the same.
Heat Project - Specifc Heat Data Table
Obje Mass Mass Δ Temp H20 Δ Temp Metal Heat Gain Heat Lost SH
ct Metal Water H20 Metal Metal
22 C - 21 C = 1 49.5 C - 22 C = 100 cal 100 cal
H 28.1 g 100 g C 27.5 C 150 cal 150 cal 0.13cal/g
200 cal 200 cal C
I 28.95 g 100 g 21.5 C - 20 C = 56 C - 21.5 C = 300 cal 300 cal
1.5 C 34.5 C 50 cal 50 cal 0.15cal/g
J 29.2 g 100 g 250 cal 250 cal C
27 C - 25 C = 2 61 C - 27 C = 34 100 cal 100 cal
A 72.1 g 100 g C C 100 cal 100 cal 0.2cal/g
C
B 28.8 g 100 g 26 C - 23 C = 3 66.5 C - 26 C =
C 40.5 C 0.1
C 72.65 g 100 g cal/gC
24 C - 23.5 C = 63 C - 24 C = 39
D 29.5 g 100 g 0.5 C C 0.04
cal/gC
F 28.8 g 100 g 25.5 C - 23 C = 69 C - 25.5 C =
2.5 C 43.5 C 0.08cal/g
C
24 C - 23 C = 1 72 C - 24 C = 48
C C 0.02
cal/gC
24 C - 23 C = 1 65.5 C - 24 C =
C 41.5 C 0.08cal/g
C
Specific Heat Practice - 5/15/17
SH (cal/gC) Temperature 2 Temperature 1
(C) (C) Change (C)
0.35 51.57 23 28.57
0.76 36.16 23 13.16
0.09 134.11 23 111.11
Heat Project
1. Vocabulary - Define and make note cards
Conduction- t he Heat- the quality of Insulator- a Calorie- t he energy
process by which being hot; high substance that needed to raise the
heat or electricity is does not readily temperature of 1
directly transmitted temperature. allow the passage gram of water
through a of heat or sound. through 1 °C (now
substance when usually defined as
there is a difference 4.1868 joules).
of temperature or of
electrical potential
between adjoining
regions, without
movement of the
material.
Convection- t he Temperature- t he Second Law of Turbine- a
movement caused degree or intensity Thermodynamics- turbomachine with
within a fluid by the of heat present in a the branch of at least one moving
tendency of hotter substance or physical science part called a rotor
and therefore less object, especially that deals with the assembly, which is
dense material to as expressed relations between a shaft or drum with
rise, and colder, according to a heat and other blades attached.
denser material to comparative scale forms of energy Moving fluid acts on
sink under the and shown by a (such as the blades so that
influence of gravity, thermometer or mechanical, they move and
which consequently perceived by touch. electrical, or impart rotational
results in transfer of chemical energy), energy to the rotor.
heat. and, by extension,
of the relationships
between all forms
of energy.
Radiation- t he Heat Engine- a Specific Heat- the Generator- a thing
emission of energy device for heat required to that generates
as electromagnetic producing motive raise the
waves or as moving power from heat, temperature of the something, in
subatomic particles, such as a gasoline unit mass of a
especially engine or steam given substance by particular.
high-energy engine. a given amount
particles that cause (usually one
ionization. degree).
First Law of Conductor- Kinetic Energy-
Thermodynamics- materials that energy that a body
the branch of permit electrons to possesses by virtue
physical science flow freely from of being in motion
that deals with the particle to particle
relations between
heat and other
forms of energy
(such as
mechanical,
electrical, or
chemical energy),
and, by extension,
of the relationships
between all forms
of energy.
2. Provide a diagram showing molecular motion in Solids, Liquids, and gases.
*How are they different?
Liquid Solid Gas
Gas’s motion is more sparatic and has more movement, liquid is slightly closer together
and has less motion, and solid barely has any motion and its particles are very close
together.
3. Discuss the energy needed to change a 15 gram ice cube into steam. Use a
graph and one calculation from our unit on Phase Changes.
H=m*CiT*SH
H=15g*100C*1cal/gC
H=1500cal
H=m*HoV
H=15g*2257C
H=33855gC
H=m*HoF
H=15g*333C
H=4995gC
H=345350cal/gC
4. What is the difference between Heat and Temperature? Provide a definition,
picture and video link to help you review.
Heat- the quality of being hot, or holding a high temperature
Temperature- the degree or intensity of heat present in a substance
The difference between heat and temperature is that heat holds a certain temperature,
whereas tempterature is the amount of heat a substance holds.
https://www.youtube.com/watch?v=wTi3Hn09OBs
5. Construct a graph showing the average monthly temperatures in Hartford, CT.,
a city on the equator and a city in the Southern Hemisphere.
Hartford:
Month Temperature
January (F)
February
March 40
April 44
May 42
June 63
July 62
August 82
September 89
October 89
November 79
December 65
54
41
Quito, Ecuador:
Month Temperature
January (F)
February
March 70
April 70
May 69
June 72
July 69
August 72
September 71
October 76
November 76
December 73
75
73
Sao Paulo, Brazil:
Month Temperature
January (F)
February
March 83
April 83
May 82
June 79
July 73
August 72
September 72
October 75
November 75
December 78
79
81
Questions:
1. What do you notice about the temperatures?
The average temperatures get higher as the months continue in Hartford (Northern
Hemisphere). The average temperatures in cities on the equator (Quito, Ecuador)
usually stay around the same temperature, although there are some flunctuating
months. Lastly, cities located in the Southern Hemisphere (Sao Paulo, Brazil) have a
drop in average temperature as the months continue.
2. How is heat transferred throughout the Earth?
Convection of liquids, conduction in solids, and radiation through anything that allows
radiation to pass through.
4. How is Steam used to create electricity in Power Plants?
A. Coal Plant
The steam produced flows into a turbine which spins a generator, therefore creating
electricity.
B. Natural Gas Plant
The steam travels through a turbine and runs the generator.
C. Nuclear Plant
The steam goes through a process called fission, and produces electricty.
D. Where did Fossil Fuels originate?
The three main fossil fuels are coal, oil, and natural gas. They come from things formed
hundreds of years ago in the time of the dinosaurs, thus the name f ossil feuls.
E. What is the difference between Renewable and NonRenewable forms of energy?
Renewable forms of energy are able to be renewed and reused. NonRenewbale forms
of energy are energy.
Part II - Water, Orange Juice and Vegetable Oil
1. Conduct an experiment to determine the Heat Gained by 20 g of each substance
2. You must measure the mass of Orange Juice and Vegetable Oil.
3. Research the Specific Heats of Orange Juice and Vegetable Oil in Calories/g C not in
Joules.
4. Make a data table
5. Construct a 3 Line graph for 2 minutes of data collection - 1 pt every 10 seconds
6. Write a conclusion about your results.
Critical Thinking Questions
1. What happens to the molecules in each of the beakers as heat is added?
2. Which substance showed the greatest temperature change? Least? Use data
3. Which substance does research say should show the greatest temperature increase?
Least? Why? How does this relate to Specific Heat?
4. How does Average Kinetic Energy relate to this experiment?
5. Why is water a great substance to put into a car engine radiator?
Practice Calculation
1. How much heat was gained by a 50 g sample of Orange Juice that increased its
temperature from 35 C to 75 C?
2. How much heat was gained by a 350 g sample of Vegetable oil that increased its
temperature from 24 C to 95 C?
Lopez Lab
Water (32 - 23) Oil (39-23)
http://www.kentchemistry.com/links/Energy/SpecificHeat.htm
Use this to help solve problems
6. Lab Experiment:
*Conduct an experiment that tests 3 different cups for their ability to insulate.
A. Conduct experiment
B. Create Data Table - Include Specific Heat
C. Write short conclusion paragraph that relates your data to research about the
effectiveness of the 3 materials to provide insulation.
Critical Thinking - Choose 2 out of 3 to research
Provide pictures
1. How did NASA protect the astronauts in their space vehicles from the harmful
radiation from space?
2. How is your home insulated? Research the “R” value system for insulation.
3. How does the atmosphere act as an insulator?
7. Lab Experiment: April 28-30
Metals change temperature very easily because they have very low specifc heat.
*Conduct an experiment to determine the Specific Heat of 3 different metals.
A. LAB TEMPLATE S pecific Heat of water=1 cal/gC, 100mL of water=100g
Specific Heat of different metals=unknown, need to know the
mass of the metals - Calories lost (in metal)=Calories gained (in water)
Example:
Heat=m*CiT*SH
1. Heat Gain(H2O)= m*CiT*SH
=100g*(T2- T1)*SH
=100g*(23C-21C)*1cal/gC
=100g*2C*1cal/gC
=200cal
2. Heat Lost(Metal)=m*CiT*SH
200=65g*(67C-23C)*SH
200=65g*44C*SH
200=2860cal/gC*SH
0.07cal/gC=SH
Mass Mass Δ Temp H20 Δ Temp Metal Heat Gain Heat Lost SH Metal
Object Metal Water H20 Metal 0.13
H 28.1 g 100 g cal/gC
I 28.95 g 100 g 22 C - 21 C = 1 49.5 C - 22 C = 100 cal 100 cal 0.15
J 29.2 g 100 g C 27.5 C cal/gC
A 72.1 g 100 g
B 28.8 g 100 g 21.5 C - 20 C = 56 C - 21.5 C = 0.2 cal/gC
C 72.65 g 100 g
D 29.5 g 100 g 1.5 C 34.5 C 150 cal 150 cal 0.1 cal/gC
F 28.8 g 100 g 0.04
27 C - 25 C = 2 61 C - 27 C = 34 cal/gC
0.08
CC 200 cal 200 cal cal/gC
0.02
26 C - 23 C = 3 66.5 C - 26 C = 300 cal 300 cal cal/gC
C 40.5 C 0.08
cal/gC
24 C - 23.5 C = 63 C - 24 C = 39
0.5 C C 50 cal 50 cal
25.5 C - 23 C = 69 C - 25.5 C =
2.5 C 43.5 C 250 cal 250 cal
24 C - 23 C = 1 72 C - 24 C = 48
CC 100 cal 100 cal
24 C - 23 C = 1 65.5 C - 24 C = 100 cal 100 cal
C 41.5 C
In this experiment, the specific heats of different metals were tested. Furthermore, we
tried to determine which metals were which based upon their specific heats. The data
shows that by finding the temperature change of a metal and using its mass, one is able
to calculate both the metals’ specific heats and heat energy inside.
B. LAB RUBRIC - Focus on DATA ANALYSIS SECTION
8. SPECIFIC HEAT WORKSHEET
WORKSHEET LINK - Use this worksheet and show your work
Specific Heat
Due: Wednesday May 13
Video Lessons:
1. https://www.youtube.com/watch?v=4RkDJDDnIss
2. https://www.youtube.com/watch?v=2uHQLZ3gJAc
3. https://www.youtube.com/watch?v=tU-7gQ1vtWo
DIRECTIONS: Heat = mass * change in temperature * Specific Heat
1. A 15.75-g piece of iron absorbs 1086.75 joules of heat energy, and its temperature changes from
25°C to 175°C. Calculate the specific heat capacity of iron.
Heat = mass * change in temperature * Specific Heat
1086.75J=15.75g*150°C*SH
1086.75J=2362.5g°C*SH
0.46J/g°C=SH
The specific heat capacity of iron is 0.46J/g°C.
2. How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to
55°C, if the specific heat of aluminum is 0.90 J/g°C?
Heat = mass * change in temperature * Specific Heat
Heat=10g*33°C*0.9J/g°C
Heat=297J
Aluminum would need 297J of heat to undertake a 33°C change.
3. To what temperature will a 50.0 g piece of glass raise if it absorbs 5275 joules of heat and its
specific heat capacity is 0.50 J/g°C? The initial temperature of the glass is 20.0°C.
Heat = mass * change in temperature * Specific Heat
5275J=50g*(T2- 20°C)*0.5J/g°C
105.5J/g=(T2-20°C)*0.5J/g°C
211°C=T2- 20°C
231°C=T2
The piece of glass will raise to 231°C.
4. Calculate the heat capacity of a piece of wood if 1500.0 g of the wood absorbs 6.75×104 joules of
heat, and its temperature changes from 32°C to 57°C.
Heat = mass * change in temperature * Specific Heat
6.75x104J =1500g*25°C*SH
6.75x104J =37500g°C*SH
1.8J/g°C=SH
The heat capacity of the piece of wood is 1.8J/g°C.
5. 100.0 mL of 4.0°C water is heated until its temperature is 37°C. If the specific heat of water is
4.18 J/g°C, calculate the amount of heat energy needed to cause this rise in temperature.
Heat = mass * change in temperature * Specific Heat
Heat=100g*33°C*4.18J/g°C
Heat=13794J
The water needs 13794 joules of heat energy.
6. 25.0 g of mercury is heated from 25°C to 155°C, and absorbs 455 joules of heat in the process.
Calculate the specific heat capacity of mercury.
Heat = mass * change in temperature * Specific Heat
455J=25g*130°C*SH
455J=3250g°C*SH
0.14J/g°C=SH
The specific heat of the mercury is 0.14J/g°C.
7. What is the specific heat capacity of silver metal if 55.00 g of the metal absorbs 47.3 calories of
heat and the temperature rises 15.0°C?
Heat = mass * change in temperature * Specific Heat
47.3cal=55g*15°C*SH
47.3cal=825g°C*SH
0.06cal/g°C=SH
The specific heat of the silver metal is 0.06cal/g°C.
8. If a sample of chloroform is initially at 25°C, what is its final temperature if 150.0 g of
chloroform absorbs 1000 joules of heat, and the specific heat of chloroform is 0.96 J/g°C?
Heat = mass * change in temperature * Specific Heat
1000J=150g*(T2- 25° C)*0.96J/g°C
6.67J/g=(T2-25°C)*0.96J/g°C
6.94°C=T2 -25°C
31.94°C=T2
The chloroform will reach a final temperature of 31.94°C.
1. How much energy must be absorbed by 20.0 g of water to increase its temperature from 283.0 °C
to 303.0 °C? (Cp of H2O = 4.184 J/g °C)
Heat = mass * change in temperature * Specific Heat
Heat=20g*20°C*4.184J/g°C
Heat=1673J
The water needs to absorb 1673 joules of heat energy.
10. When 15.0 g of steam drops in temperature from 275.0 °C to 250.0 °C, how much heat energy is
released?
(Cp of H2O = 4.184 J/g °C)
Heat = mass * change in temperature * Specific Heat
Heat=15g*25°C*4.184J/g°C
Heat=1569J
The steam will release 1569 joules of heat energy.
11. How much energy is required to heat 120.0 g of water from 2.0 °C to 24.0 °C? (Cp of H2 O = 4.184
J/g °C)
Heat = mass * change in temperature * Specific Heat
Heat=120g*22° C*4.184J/g°C
Heat=11045.76J
The water needs 11045.76 joules of heat energy.
12. How much heat (in J) is given out when 85.0 g of lead cools from 200.0 °C to 10.0 °C? (Cp of Pb
= 0.129 J/g °C)
Heat = mass * change in temperature * Specific Heat
Heat=85g*90°C*0.129J/g°C
Heat=986.85J
There will be 986.85 joules of heat given out.
13. If it takes 41.72 joules to heat a piece of gold weighing 18.69 g from 10.0 °C to 27.0 °C, what is
the specific heat
of the gold?
Heat = mass * change in temperature * Specific Heat
41.72J=18.69g*17°C*SH
41.72J=317.73g°C*SH
0.13J/g°C=SH
The specific heat is 0.13J/g°C.
14. A certain mass of water was heated with 41,840 Joules, raising its temperature from 22.0 °C to
28.5 °C. Find the mass of the water, in grams. (Cp of H2 O = 4.184 J/g °C)
Heat = mass * change in temperature * Specific Heat
41840J=m*6.5°C*4.184J/g°C
41840J=m*27.2J/g
1538.24g=m
The mass is 1538.24 grams.
15. How many joules of heat are needed to change 50.0 grams of ice at -15.0 °C to steam at 120.0°C?
(Cp of H2 O = 4.184 J/g °C)
Heat = mass * change in temperature * Specific Heat
Heat=50g*105° C*4.184J/g°C
Heat=21.97J
The steam needs 21.97J of heat.
16. Calculate the number of joules given off when 32.0 grams of steam cools from 110.0 °C to ice at
-40.0 °C.
(Cp of H2O = 4.184 J/g °C)
Heat = mass * change in temperature * Specific Heat
Heat=32g*78°C*4.184J/g°C
Heat=10443.26J
The ice will give off 10443.26 joules of heat.
17. The specific heat of ethanol is 2.46 J/g o C. Find the heat required to raise the temperature of 193g
of ethanol from 19o C to 35oC .
Heat = mass * change in temperature * Specific Heat
Heat=193g*16oC*2.46J/go C
Heat=7596.48J
The amount of heat required is 7596.48 joules.
18. When a 120 g sample of aluminum (Al) absorbs 9612 J of energy, its temperature increases from
25o C to 115o C. Find the specific heat of aluminum.
Heat = mass * change in temperature * Specific Heat
9612J=120g*90oC *SH
9612J=10800goC*SH
0.89J/goC=SH
The specific heat is 0.89J/goC .
NUT LAB
Nut Mass (g) Mass (g) Temp. Change Heat (cal)
Almond- 0.7 (C) SH (cal/gC) 1 100
Cashew- 0.9 100 24 - 23 = 1 1 100
Cashew- 1.1 100 23 - 22 = 1 1 200
Almond- 0.8 100 24 - 22 = 2 1 200
Walnut - 0.3 100 25 - 23 = 2 1 300
100 25 - 22 = 3
The purpose of this experiment was to determine the specific heats of different nuts,
based upon their masses and temperature changes.
Use this website for examples
http://www.kentchemistry.com/links/Energy/SpecificHeat.htm
9. TEST REVIEW
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