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Center for Astrophysics, Space Physics and Engineering Research 2012 Baylor University RET Program
Supplemental Curriculum for the 2010 CASPER Physics Circus
Dr. Steve Rapp Linwood Holton Governor’s School
P.O. Box 1987 Abingdon, VA 24210 [email protected]
Abstract—the researcher experienced a combination of educational research and experimental research during the RET encounter. This paper will discuss the educational research segment but the experimental segment will be addressed in the final presentation. Six demonstrations presented in the 2010 CASPER Circus were used to develop a curriculum that will hopefully develop some STEM connections for both teachers and students. The six demonstrations explored the following concepts: 1. Newton’s Second Law of Motion and Kinetic Energy, 2. Conservation of Energy, Thermal Energy Transfer, Thermal Equilibrium, 3. Density, Buoyancy, Viscosity, 4. Chemical Formulas, Balancing Equations, Mass Spectrometer, 5. Electrical Circuits, Ohm’s Law, 6. Energy Transfer, British Thermal Unit. Only the first two demonstrations will be considered here. The concepts listed above were correlated with the National Science Education Standards and the Texas Standards for Science (see appendix). Demonstration 1a: Newton’s Second Law (F=ma) Students are asked to solve the mystery of who broke into the science lab. The force required to break the door was explored using the equation F=ma. However, CSI decided it would be easier to use the equation KE= ½ mv2 to determine the amount of energy the perpetrator would have to expend to break the door. The 9 minute segment of the CASPER Physics Circus video dealing with F=ma and KE=1/2mv2 may be viewed at the following link: https://www.box.com/s/69ff019fa06ba65a5f11. National Standard: CONTENT STANDARD B: As a result of their activities in grades 9‐1 2 all students should develop an understanding of: motions and forces; (demo 1b) conservation of energy and increase in disorder [1]. Texas State Standard: 112.39 (4) (D) Calculate the effect of forces on objects using Newton’s Second Law of Motion (demo 1b) (6) (B) Investigate examples of kinetic and potential energy and their transformation; (D) Demonstrate and apply the laws of conservation of energy [2]. Level 1 What is force? A force is a push or pull. Force is measured in units called newtons (N). One N = kg*m/s2.
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What is mass? The mass of an object is a fundamental property of the object; a numerical measure of its inertia; a fundamental measure of the amount of matter in the object. Units: kilogram (kg) What is acceleration? Acceleration is the rate of change of velocity. Units: m/s2. Remember: Newton’s 2nd Law declares that a net force (an unbalanced force) causes acceleration. Look at the sketch below. By knowing the mass and acceleration, one can determine the force required for Steven to move the car. What force did Steven have to apply? Answer:_________
Now look at the next sketch. Who applied the most force, Steven in the previous example, or Mary shown in the sketch? What was the force she had to apply to move the van?
Answer: ___________ Answer:___________
To learn more about force try the Lunar Lander Game by clicking on the link below: http://phet.colorado.edu/sims/lunar‐lander/lunar‐lander_en.html. Before we explore force and acceleration to a greater depth let’s look at how to graph data. On the x axis we will plot time. On the y axis we will plot speed. Here is the situation: Steven is
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trying out his new corvette and wants to see how quickly his car can get to 70 mph. He goes to the drag strip where he can test out his car in a safe manner. The following data was provided by a good friend as Steven completed his run. Plot the data in the table below to form a graph. Your graph should be very similar to the one below.
X axis (time = sec.) Y axis (speed = mph)
0 0
1 10
2 20
3 30
4 40
5 50
6 60
7 70
Notice that your data start at zero time and since the corvette is sitting still the speed is zero. So, that point on the graph is in the lower left corner at zero. This is called the origin of the graph. Our next data point is at one second on the x axis and 10 mph on the y axis. Then at 2 seconds the speed is 20 mph. The rest of the points on the graph are plotted in a similar way. Level 2 Force and acceleration are vector quantities. This means they have both magnitude (numerical value) and a direction. Acceleration is equal to change in velocity divided by change in time. By constructing a velocity‐time graph and finding the slope of the line one can determine the acceleration. So, how do you find the slope? The slope is found by using the ratio of rise over run. That means the slope is equal to change in
y divided by change in x or . Let’s consider a car moving at a constant velocity as shown below [3]:
Speed Verse Time
0
10
20
30
40
50
60
70
80
0 2 4 6 8
Time (sec)
Sp
eed
(m
ph
)
Series1
4
Here we see the data placed in a table: A velocity time graph of data would look like the one below:
Time (s)
Velocity (m/s)
0 10 1 10 2 10 3 10 4 10 5 10
Note that the line on the graph is horizontal and that the slope of the line is therefore 0 m/s/s. There is no acceleration because there is no change in velocity. So, one can say that the slope of the line is the same as the acceleration (0 m/s/s) of the car.
Now consider a car moving with a changing velocity. A car with a changing velocity will have acceleration.
Here we see the data placed in a table: Note that the line on the graph is diagonal ‐ that is, it has a slope.
Time (s)
Velocity (m/s)
0 0
1 10
2 20
3 30 4 40
5 50
The slope of the line can be calculated as 10 m/s/s. So, once again the slope of the line (10 m/s/s), is the same as the acceleration (10 m/s/s) of the car [3].
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Time for Some Practice with Newton’s Second Law Equation The Fnet = m • a equation is often used in algebraic problem solving. The table below can be filled by substituting into the equation and solving for the unknown quantity. Try it yourself and click on the question mark (?) to view the answers.
Problem # Net Force (N) Mass (kg) Acceleration (m/s/s)
1.‐ 10 2 ?
2. 20 2 ?
3. 20 4 ?
4. ? 2 5
5. 10 ? 10
Dig a little deeper: check out these links. PBS Circus Physics concerning Newton’s Laws and other topics: http://www.pbs.org/opb/circus/classroom/circus‐physics/newtons‐laws/. Newton’s Second Law and NFL Football: To better understand mass and acceleration check out the video at this link: http://www.nbclearn.com/portal/site/learn/nfl/cuecard/50974/. An excellent video on the relationship between force, mass, and acceleration can be found at the Khan Academy web site, http://www.khanacademy.org/science/physics/v/newton‐s‐second‐law‐of‐motion. Level 3
Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables ‐ the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.
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Vector quantity or scalar quantity Remember that a vector quantity is described by both a magnitude and direction (examples: force, displacement, velocity, acceleration) but a scalar quantity is described by a number only (examples: mass, speed, temperature, volume. Vectors Recall in the equation F=ma that force and acceleration are both vector quantities. Let’s focus on the following: Vector Representation – straight line with head and tail The direction is indicated by the arrow tip and the magnitude is indicated by the length of the vector.
A scale is clearly listed.
A vector arrow (with arrowhead) is drawn in a specified direction. The vector arrow has a head and a tail.
The magnitude and direction of the vector is clearly labeled. In this case, the diagram shows the magnitude is 20 m and the direction is (30 degrees
West of North).
Vector Notation
When handwritten, use an arrow: When printed, will be in bold print: A When dealing with just the magnitude of a vector in print, an italic letter will be used: A
Vector Direction – Coordinate System – Vector Diagrams Some vectors may point due East, West, North, or South but there is a convention for identifying the direction of a vector that is not due East, due West, due South, or due North. There are two ways to accomplish this:
1. The direction of a vector may be expressed as an angle of rotation of the vector about its "tail" from north, south, east, or west. For example, a vector can be said to have a
Adapted from the Physics Classroom [3]
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direction of 40 degrees West of North (meaning a vector pointing north has been rotated 40 degrees towards the westerly direction).
2. The direction of a vector may be expressed as a counterclockwise angle of rotation of the vector about its "tail" from due east. Using this convention, a vector with a direction of 30 degrees is a vector that has been rotated 30 degrees in a counterclockwise direction relative to due east. A vector with a direction of 260 degrees is a vector that has been rotated 260 degrees in a counterclockwise direction relative to due east. This is probably the most often used convention.
40o West of North 260o counterclockwise rotation from East
Vector Magnitude
The magnitude of a vector in a scaled vector diagram is indicated by the length of the arrow. The arrow is drawn an exact length in accordance with a particular scale. For example, the diagram below shows a vector with a magnitude of 20 miles. Since the scale used for constructing the diagram is 1 cm = 5 miles, the vector arrow is drawn with a length of 4 cm. That is, 4 cm x (5 miles/1 cm) = 20 miles.
Free‐Body Diagrams‐ A special kind of vector diagram Free‐body diagrams are used to show the relative magnitude and direction of all forces acting upon an object in a given situation. A free‐body diagram is a special type of vector diagram. The size of the arrow in a free‐body diagram indicates the magnitude of the force. The direction of the arrow illustrates the direction that the force is acting. Each force arrow in the diagram is labeled to show the exact type of force. It is customary in a free‐body diagram to represent the
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object by a box and to draw the force arrow from the center of the box outward in the direction that the force is acting as shown in the diagram below:
In the free‐body diagram above four forces are shown acting upon the object. However, objects may not always have four forces acting upon them. There may be situations where one, two, or three forces are represented in the diagram. So, to construct an accurate free‐body diagram it is important to know what forces are at work and in which direction they are acting.
Adapted from the Physics Classroom [3]
Vector Game: Boat to the Island: http://illuminations.nctm.org/ActivityDetail.aspx?ID=42
Demonstration 1b: Kinetic Energy (KE = ½ mv2) Level 1 Any object that is moving has kinetic energy. In the above equation, m is mass and v is velocity. The truck shown below has mass and velocity. If one knows the KE and the velocity then the mass of the truck can be determined. Likewise if one knows the KE and mass then the velocity can be calculated by using this equation: m = 2KE/v2. Also, v can be calculated by this equation: v = √2KE/m. Velocity
The equation KE= ½ mv2 shows that the kinetic energy of an object is directly proportional to the square of its velocity. Therefore, for a twofold increase in speed, the kinetic energy will increase by a factor of four. For a threefold increase in velocity, the kinetic energy will increase by a factor of nine. The kinetic energy is dependent upon the square of the velocity. One could say that the equation is a guide to thinking about the relationship between the variables. Remember that KE is a scalar quantity since it is described by a numerical value and does not involve a direction.
Adapted from the Physics Classroom [3]
Mass
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Like work and potential energy, the standard metric unit of measurement for kinetic energy is the Joule. The units involved in a Joule are shown below:
Practice Problems Let’s try a few practice problems to make sure you can calculate kinetic energy. Once you have the problem solved click on the link below the problem to check your work.
1. Determine the kinetic energy of a 600‐kg roller coaster car that is moving with a speed of 20 m/s. Check your answer here: https://www.box.com/s/82dd5ce6de42e5369586.
2. If the roller coaster car in the above problem were moving with three times the speed, then what would be its new kinetic energy? Check your answer at the site below: https://www.box.com/s/b69ae65c50dd66b478f0.
3. If a truck has a speed of 10 km/h and a kinetic energy of 20,000 J what is its mass? Check your answer here: https://www.box.com/s/41c02203e9edc3334e4d.
Build your own roller coaster at this site, http://www.jason.org/digital_library/4851.aspx to see the relationship between kinetic energy and velocity. Give it a try, it’s really fun! Potential Energy One can hardly mention Kinetic Energy without thinking about Potential Energy (PE). So, how does one calculate PE and what is it anyway? Potential energy is the stored energy of position that all objects have when they are at rest. PE has the capability to transform into KE. To determine the potential energy of an object, one can use this equation: PE = mgh where PE is potential energy, m is the mass, g is the acceleration due to gravity (generally 9.8 m/2) and h is the height of the object. Therefore, Mass: Acceleration of Gravity: Height:
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Check out the animation showing conversion of potential energy to kinetic energy by depressing the control key on your computer and then clicking on the image [3]:
Look at the great video on Law of Conservation Energy (Potential Energy Kinetic Energy) at this link: http://www.youtube.com/watch?feature=endscreen&NR=1&v=BSWl_Zj‐CZs.
Potential Energy Example Problems: 1. A cat has climbed to the top of a tree. The tree is 20 meters high and the cat has a mass of 6 kg. How much potential energy does the cat have? Solution: given m = 6 kg, h = 20 m, g = 9.8 m/s2 (gravitational acceleration of the earth) Substitute the values in the potential energy formula below: PE = m g h PE = 6 kg x 9.8 m/s2 x 20 m PE = 1176 Joules 2. On a 3m ledge, a rock sitting there has the potential energy of 120 J. What is the mass of the rock? Solution: PE = 120 J, h = 3m, g = 9.8 m/s2 (gravitational acceleration of the earth) Substitute the values in the formula below:
m= 120 J/9.8 m/s2 x 3 m m = 120 J/29.4 m/s2 *s m = 4.08 kg [4]
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Practice Problems on Potential Energy Let’s try a few practice problems to make sure you can calculate potential energy and have a good understanding of the concept. Go to this link to complete the practice problems: http://www.cstephenmurray.com/onlinequizes/physics/workandenergy/potentialenergyproblems.htm. You will be guided through the problems. Once you click on an answer you will know if you get the problem correct or not. Level 2 Work and Kinetic Energy An object’s kinetic energy can also be thought of as the amount of work the moving object could do in coming to rest. The moving hammer has kinetic energy and can do work on the nail. Since KE =1/2 mv2 one can say that: Wnet = ½ mv2f ‐ ½ mv2i = ∆ KE. Remember that work can also be calculated when a force is applied to an object and the object then moves in the same direction as the applied force. The equation here would be Work = Force x (change in displacement) or W = F ∆x. Things become more complicated when the force on the object is applied at an angle. Then the amount of work done is related to the angle at which the force is applied as in this equation: W= (F cos Ѳ) ∆x. Remember the unit for work is the joule (J); the unit for force is the newton (N) and the unit for displacement is meter (m). Work: Sample Problems
1. A bulldozer pushes a 5000 N concrete block 100 m. How much work does the bulldozer do? Start with the equation W = F ∆x. then substitute in the magnitude for the force and change in displacement: W = F ∆x = 5000 N (100 m) = 500,000
or 5.0 x 105 J
Work‐Kinetic Energy Theorem When work is done by a net force on an object and the only change in the object is its speed, the work done is equal to the change in the object’s kinetic energy.
KEKEKEW ifnet
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2. Steven likes to go fishing and he is really good at it. He is so confident that he takes a wagon to gather up all of the fish he is going to catch. Sure enough, he catches 25.0 kg of fish. The mass of the fish plus the wagon equals 100 kg. He brings his horse along to pull the wagon. The horse exerts a force on the wagon of 1.20 x 102 N with a rope angled at 30.0o. How much work does the horse do if he pulls the wagon 50.0 m?
Since an angle is involved one needs to use the work equation, W= (F cos Ѳ) ∆x. So we have W = (1.20 x 102 N x cos30o) (50.0 m) = (1.20 x 102 N x .866) (50.0 m) = 5,196 J
Go to the link listed below and scroll down to the section entitled Sample Work Problem #1.
http://www2.franciscan.edu/academic/mathsci/mathscienceintegation/MathScienceIntegatio‐11011.htm
Complete all 4 problems and check your answers. Ok it’s time to look at a sample problem that is a little more involved. Sample Problem
1. The driver of a 1.00 x 103 kg car traveling on the interstate at 35.0 m/s (about 80 mph) slams on his brakes to avoid hitting a second vehicle in front of him, which had come to a stop because of congestion ahead as shown below: vi
After the brakes are applied, a constant force of 8.00 x 103 N acts on the car (blue). At what minimum distance would the brakes have to be applied to avoid a collision with the other vehicle (yellow)? Ignore air resistance.
Strategy First, compute the net work which only involves kinetic friction (resistive force as something moves). Then set the net work equal to the change in kinetic energy. To determine the minimum distance take the final speed vf to be zero just as the braking vehicle (blue car) reaches the rear of the vehicle at rest (yellow car). Solve for the unknown, ∆x. Solution Find the minimum stopping distance.
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Apply the work‐kinetic energy theorem to the car: Wnet = ½ mv2f ‐ ½ mv2i Substitute an expression for Wnet, set vf = 0 ‐fk∆x = 0 ‐ ½ mv2i Since the net work = resistive friction force net work can be shown as ‐fk∆x: Next, substitute vi= 35.0 m/s, fk = 8.00 x 10
3 N ‐(8.00 x 103 N) ∆x = ‐1/2(1.00 x 103 kg)(35.0 m/s)2 M = 1.00 x 103 kg. Solve for ∆x: ∆x = 76.6 m Check out the diagram below showing that net work (applied force) = kinetic friction x displacement.
Let’s get Serious, Time for you to solve some work‐kinetic energy theorem problems. Problems: Work out the problems and then click on the link to check your answer.
1. What is the kinetic energy of a 2 kg ball that travels a distance of 50 meters in 5 seconds?
https://www.box.com/s/6f48404697c98435c6d1
2. In a sense we all have kinetic energy, even when we are standing still. The earth, with a radius of 6.37×106 meters, rotates about its axis once a day. Ignoring the earth's rotation about the sun, what is the kinetic energy of a 50 kg man standing on the surface of the earth?
https://www.box.com/s/298ccac7b37fb1531b59
3. A ball is dropped from a height of 10 m. What is its velocity when it hits the ground?
https://www.box.com/s/768ed571dafb8604b48b
4. A ball is thrown vertically with a velocity of 25 m/s. How high does it go? What is its velocity when it reaches a height of 25 m?
Friction is a resistive force that opposes motion. Notice that the
normal force and the gravitation force are balanced forces that are
perpendicular to the horizontal displacement and therefore are
variables that aren’t needed to solve the problem.
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https://www.box.com/s/807281afcf5037365abe
5. A ball with enough speed can complete a vertical loop. With what speed must the ball enter the loop to complete a 2 m loop? (Keep in mind that the velocity of the ball is not constant throughout the loop).
https://www.box.com/s/4d97595252523a12450a
Explore this video on work and kinetic energy: http://www.youtube.com/watch?v=‐0aizPk3a8Y&feature=relmfu. Level 3 Development of KE = ½ mv2
Work and Power
Have you thought about how the equation for kinetic energy was developed? As you can see below Newton’s Second Law of Motion has a place in the development of the equation. Also notice that the equation for computing work is also part of the derivation.
[4]
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Practice Problems: Work this out and then check your answers at the links below. Algebra Based
1. An elevator must lift 1000 kg a distance of 100 m at a velocity of 4 m/s. What is the average power the elevator exerts during this trip? https://www.box.com/s/65f57351fa80c1a36010
2. An object in free fall is said to have reached terminal velocity if the air resistance becomes strong enough to counteract all gravitational acceleration, causing the object to fall at a constant speed. The exact value of the terminal velocity varies according to the shape of the object, but can be estimated for many objects at 100 m/s. When a 10 kg object has reached terminal velocity, how much power does the air resistance exert on the object? https://www.box.com/s/89ad7c0e0a4175b6d1b5
Calculus Based
3. Derive, using the equation P = ,, an expression for the power exerted by gravity on an object that is in free fall.
https://www.box.com/s/9dd04b4377582c071167 Demonstration 2 Conservation of Energy, Thermal Energy Transfer, Thermal Equilibrium CASPER science investigators (CSI) determine the approximate time of the break in by comparing the temperature of coffee in a thermos and the temperature of coffee in a mug. A video clip of this portion of the Physics Circus can be seen at this link: https://www.box.com/s/2578554ae0e542ffe3ea. National Science Education Standard: CONTENT STANDARD B: As a result of their activities in grades 9‐1 2 all students should develop an understanding of: motions and forces; conservation of energy and increase in disorder. Texas Science Standard: 112.39 (6) (D) Demonstrate and apply the laws of conservation of energy; (F) Contrast and give examples of different processes of thermal energy transfer, including conduction, convection, and radiation. Level 1 What is the law of conservation of energy? It is a law of physics that says the total mechanical energy is the sum of the kinetic and potential energies in the system. In other words, to say a physical quantity is conserved is to say
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that the numerical value of the quantity remains constant. Another way to say that is shown in the equation below:
So, what is this equation telling us? Well, initial energy Ei = final energy Ef. This means the initial kinetic energy plus the initial potential energy = the final kinetic energy plus the final potential energy. So, energy cannot be created or destroyed. Energy is conserved, thus the total energy of the universe is constant. Remember how in the previous section we explored kinetic energy (KE= ½ mv2) and potential energy (PE = mph)? That means that we can substitute these into the equation above, so that becomes [5]:
(½ mv2)i + (mgh)I = (½ mv2)f + (mgh)f Check out this video about the Law of Conservation of Energy: http://www.youtube.com/watch?v=L24vOAXV0Cc Law of Conservation of Energy Sample Problems
1. An F/A‐18 Hornet jet fighter has a mass of 20,000 kg is flying at an altitude of 10,000 meters above the surface of the earth with a velocity of 250 m/s. In this scenario, we can calculate the total mechanical energy of the jet fighter as follows:
2. Use Conservation of energy to calculate the speed of The diver as he plunges downward due to gravity. A diver of mass m jumps off a platform that is 10.0 m above the surface of the water as shown to the right. Neglect air resistance and find the speed of the diver when he is 5.0 m above the surface.
(½ mv2)i + (mgh)I = (½ mv2)f + (mgh)f
ffii
fi
PEKEPEKE
EE
KEi = 0
PE h
KEi = 0 PEi =mghi
KEf = 1/2mv2
PEf = 0
10.0 m
5.0 m
0 m
ffii
fi
PEKEPEKE
EE
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0 + ghi = ½ vf2 + ghf
_____________ _____________________________________ Vf = √2g (hi – hf) = √2 (9.80 m/s2)(10.0 m ‐ 5.00 m) = 9.90 m/s Now, it’s time for you to try a few problems.
1. How much kinetic energy is produced when the brakes are used to bring a 1000‐kg car to rest
from a speed of 100 km/h? Click on the link below to check your answer: https://www.box.com/s/2d4c88782f8bd85ef39e
2. A 20 g pebble is put in a sling shot with a spring constant of 100 N/m and it is stretched back
to 0.7 m. Determine the maximum velocity that the stone will acquire and the speed of stone when it is shot straight up? Click on the link below to check your answer: https://www.box.com/s/109cd9515b2a4a285d82
Click on the following link to see practical applications of energy conversions: http://www.youtube.com/watch?v=VlY2Q9Q4a_g&feature=relmfu. Level 2 Some interesting problems Go to this link: http://www.wjmouat.com/Teachers/BHutchinson/Physics/Phys11/Unit%204/PS43.htm and solve the first five problems. Level 3 Practical Application of Law of Conservation of Energy Check out the problems and applications at the link listed below: http://www.physicsclassroom.com/class/energy/u5l2bc.cfm Energy Transformation But energy may be transformed into other kinds of energy, for example, electrical energy of a motor can be transformed into mechanical energy that results in turning a pulley or a gear. The chemical energy of a battery can be changed into electrical energy that charges your cell phone. Associated with energy transformation is heat transfer; more about that just a little later on. Note some of the energy conversions shown below.
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Some Examples of Energy Transformations
A television changes electrical energy into sound and light energy.
A toaster changes electrical energy into thermal energy and light.
A car changes chemical energy from fuel into thermal energy and mechanical energy.
A flashlight changes chemical energy from batteries into light energy.
When you speak into your telephone, sound energy from your voice is changed into electrical energy. The electrical energy is then converted back into sound energy on another phone, allowing someone to hear you. Light energy is converted into electrical energy using solar panels.
Campfires convert chemical energy stored in wood into thermal energy, which is useful for cooking food and staying warm.
Check out the extra cool Bill Nye the Science Guy explanation of energy conversion at this link: http://www.youtube.com/watch?v=7gX0mgzpLR8&feature=related.
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Thermal Energy Transfer (Conduction, Convection, and Radiation) Level 1 The image below illustrates the three ways that thermal energy may be transferred.
Conduction: is the transfer of heat between two objects or substances that are in direct contact with each other.
Convection: is the movement of molecules in a liquid or gas caused by heat transfer.
In the diagram at the left conduction is occurring as the heat from the pan transfers by direct contact with the handle. When the water molecules start moving around because of the heat, convection happens. By holding the pot over the stove eye, it is being heated by radiation (there is no direct contact between the stove eye and the pot).
If you hold the end of a metal rod in a flame, the end that you are holding quickly becomes hot. Energy is transferred through the rod. No part of the rod moves; energy is transferred from one atom to another within the rod. This process is known as conduction.
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The figure above shows a calculation for thermal convection in the Earth's mantle. Colors closer to red are hot areas and colors closer to blue are cold areas. A hot, less‐dense lower boundary layer sends plumes of hot material upwards, and likewise, cold material from the top moves downwards. So, convection results in the bulk movement of materials. We could say that convection is the transfer of energy between an object and its environment, due to fluid motion. Energy changes make things happen! Check out the link below: http://www.ftexploring.com/energy/energy‐1.htm Radiation: is the transfer of energy through space by means of electromagnetic waves in much the same way as electromagnetic light waves transfer light. The same laws that govern the transfer of light govern the radiant transfer of heat
Check your understanding Go to this link and take a quiz about energy and energy transformations: http://www.proprofs.com/quiz‐school/story.php?title=Energy‐Practice‐Test Level 2 Phase Changes Phase changes occur because there is a change in thermal energy—energy is either absorbed or given off. So what exactly is a phase change? The Free Dictionary (online) defines phase change as a change from one state (solid or liquid or gas) to another without a change in chemical composition. The phase changes (red arrows) shown below require energy. The reverse process (blue arrows) release energy. Melting Solid Liquid Freezing Sublimation Vaporization Deposition Condensation Gas
Unlike conductive and convective forms of heat transfer, thermal radiation can be concentrated in a small spot by using reflecting mirrors, which is exploited in concentrating solar power generation. For example, the sunlight reflected from mirrors heats the PS10 solar power tower and during the day it can heat water to 285 °C (545 °F).
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Changes between solid, liquid, and gaseous phases generally involve large amounts of energy compared to the specific heat. If heat were added at a constant rate to a mass of ice to take it through its phase changes to liquid water and then to steam, the energies required to produce the phase changes (called the latent heat of fusion and latent heat of vaporization) would lead to plateaus in the temperature verses time graph as shown below[4]:
In other words, during a phase change the temperature remains constant if the changes occur at 1 atmosphere of pressure (standard pressure). Notice the lines on the graph that are parallel to the x axis. For a really awesome video about phase changes in water see “The Water Phase Change Song” at this link: http://www.youtube.com/watch?v=8eALQ2HP45I. Level 3 Specific Heat The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius. The relationship between temperature change and heat is shown in the equation below [4]:
The equation, however, will not hold true if a phase change is encountered since the temperature remains constant during the phase change. The specific heat of water is 1 calorie/gram °C = 4.186
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joule/gram °C which is higher than any other common substance. This is significant because water plays a role in heat regulation. Listed below are some specific heats of different substances.
If you look at the table carefully, you will notice that water has a higher specific heat than any other liquid or solid in the table. We could say that different shows us that different materials store different amounts of energy as shown in the illustration below.
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Calculation of Specific Heat
1. It takes 487.5 J to heat 25 grams of copper from 25 °C to 75 °C. What is the specific heat in Joules/g∙°C? Solution: Q = cm∆T
487.5 J = (25 g)c(75 °C - 25 °C) 487.5 J = (25 g)c(50 °C) Solve for c: c = 487.5 J/(25g)(50 °C) c = 0.39 J/g∙°C Answer: The specific heat of copper is 0.39 J/g∙°C.
2. A steel strut near a ship’s furnace is 2.00 m long, with a mass of 1.57 kg and cross‐sectional area of 1.00 x 104 m2. During operation of the furnace, the strut absorbs a net thermal energy of 2.50 x 10 5 J. Find the change in temperature of the strut. Solution: Q = cm∆T
Solve for ∆T: ∆T= Q/cm ∆T= 2.50 x 10 5 J (1.57 kg)(448 J/kgoC) = 355°C Answer: The change in temperature is 355oC For a review on specific heat click on this link: http://www.brightstorm.com/science/chemistry/thermochemistry/specific‐heat/. OK, it’s time for an online specific heat simulation and quiz. Click on the link below, http://www.sciencegeek.net/Shockwave/SpecificHeat.htm, run the simulation for all the substances listed and answer the questions.
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Thermal Equilibrium Level 1
The graph below was shown in the Demonstration 2 video clip from the Physics Circus at the beginning of this exploration. One of the CASPER Science Investigators (CSI) mentioned the coffee in the cup has reached thermal equilibrium. So, what does that mean exactly?
Well, it is observed that a higher temperature object which is in contact with a lower temperature object will transfer heat to the lower temperature object. The objects will become the same temperature, and in the absence of loss to other objects, they will then maintain a constant temperature. They are then said to be in thermal equilibrium. So, in this case the cup of coffee was the same temperature as the science lab environment.
Level 2
Thermal equilibrium is the subject of the Zeroth Law of Thermodynamics. The "zeroth law" states that if two systems are simultaneously in thermal equilibrium with a third system, they are in thermal equilibrium with each other. Two objects with different temperatures can exchange energy if they are in thermal contact. When the two objects reach thermal equilibrium they are at the same temperature, thus no more energy is exchanged. Note the diagram on the next page. If object 1 and object 3 are in thermal equilibrium with object 2 then object 1 is in thermal equilibrium with object 2 [4].
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Objects in Thermal Equilibrium
Sample Thermal Equilibrium Problems
1. A 12 gram piece of aluminum (cp = .215 cal/goC) is at 70oC. It is placed in a beaker that contains 35grams of 15oC water. At what temperature will they come to thermal equilibrium?
Solution: Qobject1 = ‐ Qobject2 Qaluminum = ‐ Qwater mA x cpA x ∆TA = ‐mwcpw∆Tw mAcpA(Tf ‐ TiA) = ‐mwcpw(Tf ‐Tiw) (12 g)(.215)( Tf ‐ 70°C) = ‐ 35(1)( Tf ‐ 15°C) (putting in cpwateris 1 cal/g°C) (2.58)( Tf ‐ 70) = ‐ 35( Tf ‐ 15) (dropped units to make the algebra easier) 2.58 Tf ‐ 180.6 = ‐ 35Tf +525 37.58Tf = 705.6 Tf = 18.8°C
2. A ball of aluminum (cp = 0.897 J/g°C) has a mass of 100 grams and is initially at a temperature of 150°C. This ball is quickly inserted into an insulated cup containing 100 ml of water at a temperature of 15.0°C. What will be the final, equilibrium temperature of the ball and the water? Heat lost by Aluminum = Heat gained by water. 'T' = final temp.
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Al: 100 g x 0.897J/g/°C x (150 ‐ T) ΔT. = 13,455 ‐ 89.7T H2O: 100 g x 4.18J/g/°C x (T ‐ 15) ΔT. = 418T ‐ 6,270. 13,455 ‐ 89.7T = 418T ‐ 6,270. 13,455 + 6,270 = 418T + 89.7T. 19,725 = 507.7T T = 19,725 / 507.7 = 38.85°C = 39°C
[6]
A Challenging problem for you! What will be the final temperature of a system consisting of a 150‐g aluminum cup containing 820 g of water at 12.0o C and a 270‐g block of copper at 300ºC? Work the problem and then click on the link below to see the solution: https://www.box.com/s/5536046eb8f821772bfd.
Level 3 Stefan’s Law First we need to examine some terminology; check out the list below: Black body: A hypothetic body that completely absorbs all wavelengths of thermal radiation incident on it. Such bodies do not reflect light, and therefore appear black if their temperatures are low enough so as not to be self‐luminous [7].
Emissivity: a measure of the ability of a surface to radiate energy; the ratio of the radiant flux
emitted per unit area to that emitted by a black body at the same temperature, Symbol ε [8]. Ideal absorber: is an object that absorbs all light radiation incident on it, including infrared and
ultraviolet radiation (ε = 1) [5].
Ideal reflector: is an object that reflects all light radiation incident on it, ε = 0 [5].
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So, what in the world is Stefan’s Law all about? It is about the transfer of energy by radiation! The rate at which an object radiates energy is proportional to the fourth power of its absolute temperature. This is known as Stefan’s law, expressed in equation form as
P = ε σ A (T4 – To
4)
where P is the power in watts (or joules per second) radiated by the object, σ is the Stefan–Boltzmann constant, equal to 5.669 6 x 108 W/m2 K4, A is the surface area of the object in square meters, ε is a constant called the emissivity of the object, and T is the object’s Kelvin temperature. The value of ε can vary between zero and one, depending on the properties of the object’s surface [5]. Sample Problem A member of the Polar Bear Club, dressed only in bathing trunks of negligible size, prepares to plunge into the Baltic Sea from the beach in St. Petersburg, Russia. The air is calm, with a temperature of 5°C. If the swimmer’s surface body temperature is 25°C, compute the net rate of energy loss from his skin due to radiation. How much energy is lost in 10.0 min? Assume his emissivity is 0.900 and his surface area is 1.50 m2. Solution Remember to convert temperatures to Kelvin by adding 273 to each value in degrees Celsius! P = ε σ A (T4 – To
4) P = (5.669 6 x 108 W/m2 K4)(1.50 m2) x (0.90)[(298 K)4 – (278 K)4] P = 146 W Multiply the preceding result by the time, 10 minutes, to get the energy lost in that time due to radiation: Q = P ∆t = (146 J/s)(6.00 x 102 s) = 8.76 x 104 J
Now you are on your own: Work out the following problems and then check your answers.
1. A wall made of wood 4.00 cm thick has area of 48.0 m2. If the temperature inside is 25°C and the temperature outside is 14°C, at what rate is thermal energy transported through the wall by conduction? (a) 82 W (b) 210 W (c) 690 W (d) 1.3 x 103 W (e) 2.1 x 103 W Answer
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2. An amount of energy is added to ice, raising its temperature from ‐10°C to ‐5°C. A larger amount of energy is added to the same mass of water, raising its temperature from 15°C to 20°C. From these results, what can we conclude? (a) Overcoming the latent heat of fusion of ice requires an input of energy. (b) The latent heat of fusion of ice delivers some energy to the system. (c) The specific heat of ice is less than that of water. (d) The specific heat of ice is greater than that of water. (e) More information is needed to draw any conclusion. Answer
3. A person shakes a sealed, insulated bottle containing coffee for a few minutes. What is the change in the temperature of the coffee? (a) a large decrease (b) a slight decrease (c) no change (d) a slight increase (e) a large increase
[5]
Answer
Check out this really awesome video dealing with Blackbody Radiation: http://www.youtube.com/watch?v=KxOwqzaL7OQ&feature=related. Here is another interesting video about emissivity; Take at look at it at this link: http://www.youtube.com/watch?v=QHszoA5Cy1I&feature=related.
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References
[1] The National Academies Press. (1996). National Science Education Standards, Retrieved from http://www.nap.edu/openbook.php?record_id=4962. [2] Texas Education Agency (2012). Texas Essential Knowledge and Skills, Retrieved from http://www.tea.state.tx.us/index2.aspx?id=6148. [3] The Physics Classroom (n.d.), Retrieved from http://www.physicsclassroom.com/Class/. [4] Nave, C. R. (2012). Hyperphsics, Retried from http://hyperphysics.phy‐astr.gsu.edu/hbase/hph.html. [5] Serway, R.A., Vuille, C., and Faughn, J. S. (2009). Energy in thermal processes, College Physics (8th edition), (pp. 352‐384). Belmont, CA: Brooks/Cole. [6] Thermal Equilibrium Problems: Retrieved from: http://www.buzzle.com/articles/thermal‐equilibrium.html [7] Weisstein, E. W. Eric. (2012, July 22). Weisstein’s World of Physics, Retrieved from http://scienceworld.wolfram.com/physics/Blackbody.html. [8] Merriam‐Webster Inc. (2012), Retrieved from http://www.merriam‐webster.com/.
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Appendix: Correlation of 2010 CASPER Physics Circus Demonstrations with the National Science Education Standards and the Texas Essential Knowledge and Skills (TEKS) Standards
Demonstration National Science Education
Standards Texas Essential Knowledge &
Skills 1. Students are asked to solve a mystery of who broke into the science lab. The equation KE= ½ mv2 is used to determine the mass of the perpetrator. The force required to break the door is found with the equation F=ma.
CONTENT STANDARD B: As a result of their activities in grades 9‐1 2 all students should develop an understanding of: motions and forces; conservation of energy and increase in disorder.
112.39 (4) (D) Calculate the effect of forces on objects using Newton’s Second Law of Motion; (6) (B) Investigate examples of kinetic and potential energy and their transformation; (D) Demonstrate and apply the laws of conservation of energy
2. CASPER science investigators (CSI) determine the approximate time of the break in by comparing the temperature of coffee in a thermos and the temperature of coffee in a mug.
CONTENT STANDARD B: As a result of their activities in grades 9‐1 2 all students should develop an understanding of: motions and forces; conservation of energy and increase in disorder.
112.39 (6) (D) Demonstrate and apply the laws of conservation of energy; (F) Contrast and give examples of different processes of thermal energy transfer, including conduction, convection, and radiation.
3. CSI ask students to help them figure out the density (D=m/v) of a non‐Newtonian fluid (slime) that was found in the science lab where the principal met his demise. Volume found via Archimedes Principle. Properties of buoyancy and viscosity also explored.
CONTENT STANDARD A: As a result of activities in grades 9 ‐ 1 2 , all students should develop abilities necessary to do scientific inquiry and understandings about scientific inquiry
112.38 (6) (C) Analyze physical and chemical properties of elements and compounds such as color, density, viscosity, buoyancy, boiling point, freezing point, conductivity, and reactivity.
4. An unknown compound is found in the science lab. CSI are asked to figure out what the compound is (function of mass spectrometer explained). The chemical formula of the compound is found on a piece of paper. This leads to discussion of balancing equations.
CONTENT STANDARD B: As a result of their activities in grades 9‐1 2 , all students should develop an understanding of structure and properties of matter, chemical reactions, and interactions of energy and matter.
112.35 (8) (D) Use the law of conservation of mass to write and balance chemical equations; 112.39 (6) (B) Investigate examples of kinetic and potential energy and their transformations;
112.38 (5) Recognize multiple forms of
energy and knows the impact of energy transfer and energy conservation in everyday life.
5. It appears the power went out in the lab since the lights and the alarm system was off; this causes the CSI to wonder if any other circuits were affected. This leads to a discussion of series and parallel circuits as well as Ohm’s Law where the Resistance = Voltage/Current.
CONTENT STANDARD B: As a result of their activities in grades 9‐1 2 , all students should develop an understanding of electricity and magnetism and realize that they are two aspects of a single electromagnetic force. Moving electric charges produce magnetic forces, and moving magnets produce electric forces.
112.38 (5) (F) Evaluate the transfer of electrical energy in series and parallel circuits and conductive materials. 112..39 (5) (F) Design, construct, and calculate in terms of current through, potential difference across, resistance of, and power used by electric circuit elements connected in both series and parallel combinations
6. Since the previous demonstration involved series and parallel circuits, one of the CSI wondered if heat would be an additional form of energy (besides light) that was produced from energy transfer. The British Thermal Unit (BTU) is discussed.
CONTENT STANDARD B: As a result of their activities in grades 9‐1 2 , all students should develop an understanding that all energy can be considered to be either kinetic energy, which is the energy of motion or potential energy, which depends on relative position; or energy contained by a field, such as electromagnetic waves.
112.39 (6) (F) Contrast and give examples of different processes of thermal energy transfer, including conduction, convection, and radiation. 112.38 (6) (B) investigate examples of kinetic and potential energy and their transformations.
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