Directions: Analyze 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? Is there a machine that is impossible? Explain using
data.
Hypothesis: (Use proper form!)
If the angle of the inclined plane increases, then the mechanical advantage will
decrease.
Diagrams of Inclined Planes: (Use DRAWING - Label Diagrams)
Angle Chart: https://drive.google.com/open?id=0B4RmhXJlHvo1YXZhcDNMSDNSMXc
Calculations (Examples):
IMA AMA Efficiency
IMA = id / od AMA = of / if E = wo / wi X 100%
IMA = 300 / 70 AMA = 12 / 4 E = 840 / 1200 X 100%
IMA = 4.29 AMA = 3 E = 0.7 X 100%
E = 70%
IMA AMA Efficiency
IMA = id / od AMA = of / if E = wo / wi X 100%
IMA = 200 / 70 AMA = 12 / 6 E = 840 / 1200 X 100%
IMA = 2.86 AMA = 2 E = 0.7 X 100%
E = 70%
IMA AMA Efficiency
IMA = id / od AMA = of / if E = wo / wi X 100%
IMA = 100 / 70 AMA = 12 / 8 E = 840 / 800 X 100%
IMA = 1.43 AMA = 1.5 E = 1.05 X 100%
E = 105%
Data Table: ( Located on Google Classroom)
Trial Output Force (N) Output Dist. (m) Output Work (J) Input Force Input Dist. Input Work IMA AMA Efficiency
Angle = 13 12 70 840 4 300 1200 4.29 3 70%
Angle = 21 12 70 840 6 200 1200 2.86 2 70%
Angle = 45 12 70 840 8 100 800 1.43 1.5 105%
Graph: ( Angle and Mechanical Advantage)
Research:
1. Is there a machine that is impossible? Explain using research on
the Law of Conservation of Energy.
The Law of Conservation of Energy states that “in a closed system,
i.e., a system that isolated from its surroundings, the total energy of the
system is conserved”1. Perpetual motion machines are machines that
are impossible. They never have worked and they most likely never will.
These machines are machines that work on their own and they work
forever. According to the Law of Conservation of Energy, energy cannot
be created or destroyed. To make a perpetual motion machine work,
you would have to produce work without any energy therefore this
violates the Law of Conservation of Energy because a machine needs to
have input energy.2 To relate to the experiment, the inclined plane with
an angle of 45° is not possible because it breaks the Law of
Conservation of Energy and has an efficiency over 100%.
Option #1 Write a Conclusion.
***Your conclusion must also address which machine would be
impossible and why?
1 "Law of conservation of energy - NYU."
http://www.nyu.edu/classes/tuckerman/adv.chem/lectures/lecture_2/node4.html. Accessed 11 Apr. 2018.
2 "Perpetual Motion Machines: Working Against Physical Laws." 30 Aug. 2016,
https://www.livescience.com/55944-perpetual-motion-machines.html. Accessed 12 Apr. 2018.
2. Discuss purpose
3. Restate hypothesis - angle and mechanical advantage
4. Data to support hypothesis
5. Is there a machine that is impossible? Explain using research on
the Law of Conservation of Energy (Support with research - Use
Explore Tool research - INLINE CITATIONS )3
6. Use this source to explain the relationship of this machine to
Newton’s First Law of Motion.
TEXTBOOK REVIEW pg. 152-153 (1-28) Study these
Rubric
Lab Rubric - Data Analysis Sections
1 2 3 4
Data/ ____Data is poorly ____Data is ____Data is ___Data is clearly
Observatio organized or missing represented in a
ns altogether. table or graph, represented in and accurately
No mention of but it is
observations incomplete or the table or graph represented in a
there are major
errors. Some with minor errors. table or graph.
discussion of
observations More complete Observations
discussion of include
observations. discussion of
both qualitative
and quantitative
observations.
____No conclusion is ____Somewhat ____Adequately ____Clearly
Conclusion written in this report or explains whether explains whether explains whether
/ it is very brief. No data or not the or not the or not the
Analysis is cited. hypothesis was hypothesis was hypothesis was
____No analysis is supported. supported. Data supported. Data
included or it is ____ Possible is cited to support was cited to
extremely brief no sources of error hypothesis. support the
sources of error are are somewhat ____Possible hypothesis.
explained. explained. sources of error ____Possible
____No discussion of ____ No are adequately sources of error
patterns or trends in discussion of explained. are clearly
the data patterns or trends ____Some explained.
discussion of
patterns or trend ____Trends and
in the data Patterns in the
data are clearly
3 "Introduction to mechanical advantage (video) | Khan Academy."
https://www.khanacademy.org/science/physics/work-and-energy/mechanical-advantage/v/introduction-to-
mechanical-advantage. Accessed 9 Apr. 2018.
discussed.
Thermal (Heat) Energy Project
Chapter 6 (pg. 156-180)
DUE: Friday May 16th
1. Vocabulary - Define and make note cards or quizlet
Conduction Heat Insulator Calorie
The transfer of Thermal energy A material in which a measure of
thermal energy by that flows from heat flows slowly energy in food,
collisions between something at a specifically the
particles in matter higher measure of heat
temperature to needed to raise a
something at a kilogram or a gram
lower temperature of water by one
degree Celsius
Convection Temperature Second Law of Turbine
The transfer of A measure of the Thermodynamics a machine f4or
It is impossible for producing
thermal energy in average kinetic heat to flow from a continuous power in
a fluid by the energy of the which a wheel or
rotor, typically fitted
movement of particles in the cool object to a with vanes, is made
warmer object to revolve by a
warmer and cooler object unless work is fast-moving flow of
done
fluid from place to
place
water, steam, gas,
air, or other fluid
Radiation Heat Engine Specific Heat Generator
The transfer of A device that
energy by converts heat into The amount of A device that uses
electromagnetic work
waves heat that is electromagnetic
needed to raise induction to
the temperature of convert
1 kg of some mechanical energy
material by 1℃ to electrical
energy
First Law of Conductor Kinetic Energy
Energy a moving
Thermodynamics object has
The increase in A material through because of its
thermal energy of
a system equals which electric motion on the
the work done on
the system plus current can pass mass and speed of
the heat
transferred to the the object
system
2. Provide a diagram showing molecular motion in Solids, Liquids, and gases.
*How are they different?
3. Discuss the energy needed to change a 15 gram ice cube into steam. Use one
calculation from our unit on Phase Changes.
HEAT= m*heat of fusion
HEAT=15g*80cal/g
HEAT= 1,200 calories
HEAT= m*change in temperature*SH
HEAT= 15g * 100 C * 1 cal/g C
HEAT= 1,500 calories
HEAT= m*heat of vaporization
HEAT= 15g * 540 cal/g
HEAT = 8,100 calories
Total: 10,800 calories
Scientific Notation: 1.08 x 104
4. What is the difference between Heat and Temperature? Provide a definition,
picture and video link to help you review.
The hotter an object is, the faster the motion of the molecules inside it. Therefore,
the heat of an object is the total energy of all the molecular motion inside that
object. Temperature, on the other hand, is a measure of the average heat or thermal
energy of the molecules in a substance.
VIDEO LINK: h ttps://youtu.be/uk76zwQHdtU
5. Construct a graph showing the average monthly temperatures in Hartford, CT., a
city on the equator and a city in the Southern Hemisphere.
Questions:
1. What do you notice about the temperatures?
a. I noticed that Fortaleza, Brazil, the city near the equator, stays hot all
year long because places closest to the equator are the hottest and
stay hot throughout the year.
b. I also noticed that in Sydney, Australia has opposite temperatures than
Hartford, Connecticut because it is on the opposite side of the earth.
2. How is heat transferred throughout the Earth?
Heat is transferred to the surface of the Earth from the hot Earth's core by
conduction and from radiation from the Sun.
4. How is Steam used to create electricity in Power Plants?
A. Coal Plant
a. When water is turned into steam, and drives turbine generators to produce
electricity.
b.
B. Natural Gas Plant
c. Natural gas is used in steam turbines and gas turbines to generate electricity.
d.
C. Nuclear Plant
e. Nuclear energy originates from the splitting of uranium atoms – a process called
fission. This generates heat to produce steam, which is used by a turbine
generator to generate electricity.
f.
D. Where did Fossil Fuels originate?
a. Fossil fuel is a general term for buried combustible geologic deposits of organic
materials, formed from decayed plants and animals that have been converted
to crude oil, coal, natural gas, or heavy oils by exposure to heat and pressure in
the earth's crust over hundreds of millions of years.
E. What is the difference between Renewable and NonRenewable forms of energy?
Part II - Water, Vinegar and Salt Water
1. Conduct an experiment to determine the Heat Gained by 20 g of each substance
2. You must measure the mass of saltwater, vinegar, and Vegetable Oil.
3. Research the Specific Heats of saltwater, vinegar, and Vegetable Oil in Calories/g C
not in Joules.
Liquid Specific Heat in Calories/g°c
Water 1.0cal/g°c
Salt water 0.932/g°c
Vinegar 490/g°c
4. Make a data table Vinegar Oil 25
25 25 26
25 26 28
Time (seconds) Saltwater 26 28 30
0 28 30 33
10 29 31 36
31 32 39
20 32 34 41
30 33 35
40
50
60
70
80 34 37 43
90 36 38 47
100 37 40 49
110 39 41 51
120 40 42 53
5. Construct a 3 Line graph for 2 minutes of data collection - 1 pt every 10 seconds
Critical Thinking Questions
1. What happens to the molecules in each of the beakers as heat is added?
● As heat is added the molecules and atoms vibrate faster
2. Which substance showed the greatest temperature change? Least? Use data.
● Oil showed the greatest temperature change because on the graph all of the
liquids started in the same place at 25 degrees but oil had the greatest
temperature change and ended at 53 degrees which is a 28 degrees
temperature change. Saltwater showed the least temperature change because
it started at 25 degrees and ended at 40 degrees which is only a 15 degrees
temperature change.
3. Which substance does research say should show the greatest temperature
increase? Least? Why? How does this relate to Specific Heat?
● The substance that research says should show the greatest temperature
increase is salt water because is has the lowest specific heat. The substance
that research says should show the lowest temperature increase is vinegar
because is has the lowest specific heat.
● This relates to specific heat because
4. How does Average Kinetic Energy relate to this experiment?
● The molecules in a substance have a range of kinetic energies because they
don't all move at the same speed. As a substance absorbs heat the particles
move faster so the average kinetic energy and therefore the temperature
increases.
5. Why is water a great substance to put into a car engine radiator?
● Water is a great substance to put in a car engine radiator because water-cooled
engines have over air-cooled. This not only reduces the potential for
catastrophic damage but also greatly simplifies the overall design of the engine.
Critical Thinking - Choose 2 out of 3 to research
Provide pictures
2. How is your home insulated? Research the “R” value system for insulation.
Insulation reduces the heat loss by conduction. The material also prevents air
circulating inside the cavity, therefore reducing heat loss by convection.
3. How does the atmosphere act as an insulator?
It absorbs the heat from the sun and keeps the heat from the Earth’s core to keep the
Earth’s surface warm.
7. Lab Experiment: April 28-30
*Conduct an experiment to determine the Specific Heat of 3 different metals.
Name: A very DePodesta
Class: S 4
Teacher: Mr. Lopez
Date: May 7, 2018
Investigation Title: Specific Heat of Metals
I. Investigation Design
A. Problem Statement:
Identify the specific heats of metals
B. Hypothesis: (Hint: Something about comparing metals to water - use increase or
decrease)
If the unknown metals specific heat is found, then the metal can be determined.
C. Independent Variable: x
Levels of IV
*What metals did you use?
Copper Iron Aluminum
D. Dependent Variable:y
Specific Heat
Specific Heat
E. Constants:
Amount of Water Hot Plate Temperature
F. Control:
*What substance makes good control in many labs?
Water
G. Materials: (List with numbers)
1. 1 triple beam balance
2. One beaker
3. 1 graduated cylinder
4. two thermometers
5. one pair of tongs
6. 1 hot plate
7. 1 foam coffee cup
8. 1 copper
9. 1 iron
10. 1 aluminum
H. Procedures: (List with numbers and details)
1. Gather materials
2. Measure mass of metal on triple beam balance to nearest tenth of gram and record.
3. Fill Calorimeter Cup (Foam coffee cup) with exactly 100 grams of water.
4. Record temperature of water in calorimeter cup to nearest tenth of degree Celsius
5. Fill glass beaker halfway with hot water and submerge metal in beaker.
6. Leave metal in hot water until the temperature stops rising.
7. Record the hot water temperature after temperature stops rising. - M etal Initial
Temp.
8. Use tool to remove metal from hot water and carefully place into calorimeter cup
and close lid with thermometer placed in spout.
9. Record Final Temperature for Metal and Water after the water temperature stops
rising.
10. Perform the calculations using the examples discussed class - Record Specific Heat
for the metal.
A. Heat Gained Water = mass of water * Change in temp of water * Specific Heat of
Water
Copper:
100g * 2 * 1
200
Iron:
100g * 2 * 1
200
Aluminum:
100g * 2 * 1
200
B. Heat Lost Metal = Mass of metal * Change in Temp of Metal * Specific Heat of Metal
Copper:
200 = 68.5g * 52 * X
200 = 3 562 * X
3562 3562
0.056 = X
ACTUAL = 0.092
Iron:
200 = 50.3g * 54 * X
200 = 2716.2 * X
2716.2 2716.2
0.074 = X
ACTUAL = 0.11
Aluminum:
200 = 20.2g * 48 * X
200 = 969.6 * X
969.6 969.6
0.2063 = X
ACTUAL = 0.022
II. Data Collection
A. Qualitative Observations: (Describe the metals using characteristics)
- Copper
- reddish-orange color
- Soft
- Malleable
- ductile metal
- very high thermal and electrical conductivity
- Iron
- by mass the most common element on Earth
- forming much of Earth's outer and inner core
- Aluminum
- Silvery-white
- Soft
- Nonmagnetic
- Ductile metal
B. Quantitative Observations: (Key data)
1. Data Table
Mass Mass Δ Temp Δ Temp Heat Gain Heat Lost SH
Metal Water H20 Metal
Object Metal H20 Metal
Example 65 100 27-21 = 6 75-27 = 48 600 Use
600 notes
Copper 68.5 100 25-23=2 77-25=52 200 200 0.056
Iron 50.3 100 23-21=2 77-23=54 200 200 0.074
Aluminum 20.2 100 27-25=2 75-27=48 200 200 0.022
Metal Specific Heat Actual Specific Heat
Copper 0.056 0.092
Iron 0.074 0.11
Aluminum 0.2603 0.21
2. Graph - Metal and Specific Heat
3. Calculations - Show examples of how you solved for specific heat (2 or 3 examples)
A. Heat Gained Water = mass of water * Change in temp of water * Specific Heat of Water
Copper:
100g * 2 * 1
200
Iron:
100g * 2 * 1
200
Aluminum:
100g * 2 * 1
200
B. Heat Lost Metal = Mass of metal * Change in Temp of Metal * Specific Heat of Metal
Copper:
200 = 68.5g * 52 * X
200 = 3562 * X
3562 3562
0.056 = X
ACTUAL = 0.092
Iron:
200 = 50.3g * 54 * X
200 = 2 716.2 * X
2716.2 2716.2
0.074 = X
ACTUAL = 0.11
Aluminum:
200 = 20.2g * 48 * X
200 = 969.6 * X
969.6 969.6
0.2063 = X
ACTUAL = 0.022
IV. Research
1. How does Specific Heat relate to a real life application?
(Land/Sea Breezes, Cooking, Mercury in Thermometers?, Water in engines, think of
others…)
2. Include 2 sources for evidence
Land has lower heat capacity than sea water. Therefore, in day time, the temperature
of the land increases faster than the sea. Hot air (lower density) above the land rises.
Cooler air from the sea flows towards land and hence produces sea breeze.
https://www.brisbanehotairballooning.com.au/sea-land-breezes/
http://spmphysics.onlinetuition.com.my/2013/07/phenomena-related-to-specific-hea
t.html
8. SPECIFIC HEAT WORKSHEET
WORKSHEET LINK - Use this worksheet and show your work
DIRECTIONS: Heat = mass * change in temperature * Specific Heat
1. A 15.75-g p iece 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
1086.75 = 15.75 g * 150C * X
1086.75 J = 2362.5 X
0.46 = Specific Heat of Iron
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 = 10 g * 33 C * 0.9 J/gC
Heat = 297 Joules
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
5275 = 50g * X * 0.5 J/g°C
5275 = 25 * X
211 = X
ΔT = Tf - Ti
211 = Tf - 20.0°C
Tf = 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
67500 = 1500g * 25°C * C
67500 = 37500 * C
1.8J/g°C = 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
E = 100.0mL * 4.18J/g°C * (37°C - 4.0°C)
E = 13,794 J
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 = 25.0g * C * (155 - 25)
455J = 3250 * C
0.14J/g°C = 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
47.3 = 55.0g * C * 15.0°C
47.3 = 825 * C
0.057cal/g°C = 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
1.0 KJ = 1000J
1000 J = 150g * 0.96J /g°C * ΔT
1000J = 144 * ΔT
6.94 = ΔT
ΔT = Tf - Ti
6.94 = Tf - 25
Tf = 31.94°C
Use this website for examples
http://www.kentchemistry.com/links/Energy/SpecificHeat.htm
9. TEST REVIEW