Student’s Name: Vickie Gray Date of Experiment: 21 Oct 2013
Date Report Submitted: 27 Oct 2013
Title: EXPERIMENT 11: Pendulum and the Calculation of g
Purpose:
The objective of this experiment is to let the students make an experimental calculation of acceleration to gravity (g) by observing and getting some parameters of pendulum.
Procedure:
A weight bag of mass 25.1 grams was connected to a fixed support via a 101 cm string attached to the bag that acted as a bob for the pendulum set-up. Protractor was placed behind the string for amplitude measurement. The meter was placed horizontally with the 50 cm mark behind the bob at rest. The bob was pulled out at different angles (5, 10,15,20,25 and 30) and the time for 5 cycles was measured for each of those angles. These trials were done 3 times for each angle. The average time and period were measured. On the next set-up, the weight bag was replaced with another with mass two times its weight (52.1 kg). Three trials were made at 10 degrees and 101 cm and the average time and period were computed. On the last set-up, the original 25.1 grams weight bob was used as the final bob. By pulling it out at 5 degrees, three trials were made for varying length at 0.25, 0.5, 0.75 and 1 meters and the average time and period were computed. The experimental values of g were obtained with g = 4L / T2and the result was evaluated with the theoretical value of g equal to 9.81 m/s2.
Data Tables:
|
DATA TABLE 1 |
Length of string: 101 cm = 1.01 m |
Mass of bob = 25.1 g = 0.0251 kg |
|
|
|
|
|
Amplitude |
Amp. |
Trial 1 -seconds |
Trial 2 -seconds |
Trial 3-seconds |
Avg. Time |
Period |
|
Degrees |
cm |
5 cycles |
5 cycles |
5 cycles |
5 cycles |
1 cycles |
|
5 |
100.6 |
10.1 |
10.3 |
10.4 |
10.267 |
2.054 |
|
10 |
99.5 |
10.5 |
10.4 |
10.4 |
10.433 |
2.087 |
|
15 |
97.5 |
10.9 |
11 |
11.1 |
10.967 |
2.193 |
|
20 |
94.9 |
11.1 |
11.3 |
11.5 |
11.3 |
2.26 |
|
25 |
91.5 |
11.5 |
11.7 |
11.9 |
11.7 |
2.34 |
|
30 |
87.5 |
12 |
12.2 |
12.3 |
12.167 |
2.43 |
|
DATA TABLE 2 |
Length of string: 101 cm = 1.01 m |
Amplitude: 10 deg |
|
|
|
|
Bob weight |
Trial 1 |
Trial 2 |
Trial 3 |
Avg. Time |
Period |
|
Grams |
|
|
|
|
|
|
50.2 |
10.5 |
10.6 |
10.6 |
10.567 |
2.113 |
|
DATA TABLE 3 |
Mass of bob: 25.1 g |
Amplitude: 5 deg |
|
|
|
|
Length(m) |
Trial 1-sec |
Trial 2 -sec |
Trial 3 -sec |
Avg. Time |
Period |
|
0.25 |
5.1 |
5.2 |
5.1 |
5.133 |
1.027 |
|
0.5 |
7.2 |
7.3 |
7.4 |
7.3 |
1.46 |
|
0.75 |
8.6 |
8.5 |
8.7 |
8.6 |
1.72 |
|
1 |
10.1 |
10.3 |
10.4 |
10.267 |
2.053 |
Observations:
All measurements are recorded above onthe Data Tables section. On Data Table 1, it can be observed that as the amplitude is increased, the period also increased by a negligible amount. On Data Table 2, it can be observed that doubling the mass at 50.2 grams has negligible effect on the period. On Data Table 3, increasing the length increases the period significantly.
With all of these having said, it is wise enough to say that angles at small values and mass did not affect the period. It was the length that affected the period.
The student would like to add the following computation for the experimental value of g at varying length.
For Data Table 3 (varying length)
T = 2√(L/g)
g = 4L / T2
at L = 0.250 m T = 1.027
g = 4L / T2
g = 4 (0.250) / (1.027)2
g = 9.357 m/s2
at L = 0.500 m T = 1.46
g = 4L / T2
g = 4 (0.500) / (1.46)2
g = 9.260 m/s2
at L = 0.750 m T = 1.720
g = 4L / T2
g = 4(0.750) / (1.720)2
g = 10 m/s2
at L = 1 m T = 2.053
g = 4L / T2
g = 4 (1) / (2.053)2
g = 9.367 m/s2
We will just use the average of the g for 4 trials as a basis for our comparison with the theoretical value of g which is 9.810.
Average = (9.357 + 9.260 + 10.000 + 9.367) / 4
Average = 9.496 m/s2
% Error = | Experimental – Actual| /Actual
% Error = | 9.496 – 9.810 | / 9.810
% Error = 3.2 %
Questions:
A. How did the change in the weight of the bob affect the resulting period and frequency?
Adding the weight increases the period by a very small amount in the experiment. But this difference is negligible (0.0264 seconds) as theoretically mass should not affect the period in accordance with T = 2pi√ (L/g). Thus it is enough to say that the mass has no bearing with period.
B. How did the change in amplitude affect the resulting period and frequency?
It did not affect the resulting period and frequency that much. As the amplitude increased in degrees, the period also increased a little bit. Theoretically speaking, as angle increases the period obtained by using the equation L = 2pi√(L/g) is less reliable.
C. How did the change in length of the pendulum affect the period and frequency?
D. What would happen if you used very large amplitudes? Check your hypothesis by
trial. What amplitude did you use? What is the result?
Hypothesis: The period increases since the angle is increases.
At 80 degrees, L =101 cm, and m=0.251 kg, the period is 2.9 seconds which is way higher than the theoretical value of T = 2*pi√(1.01/9.8) = 2.02 seconds
E. Hypothesize about how a magnet placed directly under the center point would affect
an iron bob? Try it and find out. Did your trial verify your hypothesis?
Hypothesis: A magnet would make the travelling of the bob faster as it would have the same effect as increasing the gravitational force.
The result of the trial conformed to my hypothesis. The period decreased by a small amount which indicate faster travel time.
F. How close was your calculation of the value of g at your location? What might be afew sources for error in your experimental data and calculations?
The experimental and theoretical values of g were close enough. The error was 3.2 %. There are actually several reasons but probably the main contributor was air resistance. Equipment limitation and inaccuracy of measurements are amongst other reasons.
G. What would you expect of a pendulum at a high altitude, for example on a high mountain top? What would your pendulum do under weightless conditions?
At high altitude the acceleration due to gravity is decreased. In this case the gravitational force also is decreased. Having said these, we know that when these parameters are decreased, the period will increase as they are inversely proportional.
At weightless condition, the period becomes infinite as the force of gravity approaches zero. In this case the bob will remain as where it is raised.
Conclusions:
The acceleration due to gravity is 9.81 m/s2. This value can be approximated by observing and measuring the value of length of the string and the period of the pendulum according to the equationg = 4L / T2. Rearranging this equation, period is T = 2√ (L/g). Thus, period of the pendulum is not dependent on its mass and depends primarily on the length. However the approximation given by T = 2pi√ (L/g) is accurate only at small amplitude. Increasing the amplitude increases the inaccuracy of the calculated period. This independence of period with mass and small angles and the dependence to length were all verified in the experiment.
Student’s Name: Vickie Gray Date of Experiment: 21 Oct 2013
Date Report Submitted: 27 Oct 2013
Title: EXPERIMENT 11: Pendulum and the Calculation of g
Purpose:
The objective of this experiment is to let the students make an experimental calculation of acceleration to gravity (g) by observing and getting some parameters of pendulum.
Procedure:
A weight bag of mass 25.1 grams was connected to a fixed support via a 101 cm string attached to the bag that acted as a bob for the pendulum set-up. Protractor was placed behind the string for amplitude measurement. The meter was placed horizontally with the 50 cm mark behind the bob at rest. The bob was pulled out at different angles (5, 10,15,20,25 and 30) and the time for 5 cycles was measured for each of those angles. These trials were done 3 times for each angle. The average time and period were measured. On the next set-up, the weight bag was replaced with another with mass two times its weight (52.1 kg). Three trials were made at 10 degrees and 101 cm and the average time and period were computed. On the last set-up, the original 25.1 grams weight bob was used as the final bob. By pulling it out at 5 degrees, three trials were made for varying length at 0.25, 0.5, 0.75 and 1 meters and the average time and period were computed. The experimental values of g were obtained with g = 4L / T2and the result was evaluated with the theoretical value of g equal to 9.81 m/s2.
Data Tables:
|
DATA TABLE 1 |
Length of string: 101 cm = 1.01 m |
Mass of bob = 25.1 g = 0.0251 kg |
|
|
|
|
|
Amplitude |
Amp. |
Trial 1 -seconds |
Trial 2 -seconds |
Trial 3-seconds |
Avg. Time |
Period |
|
Degrees |
cm |
5 cycles |
5 cycles |
5 cycles |
5 cycles |
1 cycles |
|
5 |
100.6 |
10.1 |
10.3 |
10.4 |
10.267 |
2.054 |
|
10 |
99.5 |
10.5 |
10.4 |
10.4 |
10.433 |
2.087 |
|
15 |
97.5 |
10.9 |
11 |
11.1 |
10.967 |
2.193 |
|
20 |
94.9 |
11.1 |
11.3 |
11.5 |
11.3 |
2.26 |
|
25 |
91.5 |
11.5 |
11.7 |
11.9 |
11.7 |
2.34 |
|
30 |
87.5 |
12 |
12.2 |
12.3 |
12.167 |
2.43 |
|
DATA TABLE 2 |
Length of string: 101 cm = 1.01 m |
Amplitude: 10 deg |
|
|
|
|
Bob weight |
Trial 1 |
Trial 2 |
Trial 3 |
Avg. Time |
Period |
|
Grams |
|
|
|
|
|
|
50.2 |
10.5 |
10.6 |
10.6 |
10.567 |
2.113 |
|
DATA TABLE 3 |
Mass of bob: 25.1 g |
Amplitude: 5 deg |
|
|
|
|
Length(m) |
Trial 1-sec |
Trial 2 -sec |
Trial 3 -sec |
Avg. Time |
Period |
|
0.25 |
5.1 |
5.2 |
5.1 |
5.133 |
1.027 |
|
0.5 |
7.2 |
7.3 |
7.4 |
7.3 |
1.46 |
|
0.75 |
8.6 |
8.5 |
8.7 |
8.6 |
1.72 |
|
1 |
10.1 |
10.3 |
10.4 |
10.267 |
2.053 |
Observations:
All measurements are recorded above onthe Data Tables section. On Data Table 1, it can be observed that as the amplitude is increased, the period also increased by a negligible amount. On Data Table 2, it can be observed that doubling the mass at 50.2 grams has negligible effect on the period. On Data Table 3, increasing the length increases the period significantly.
With all of these having said, it is wise enough to say that angles at small values and mass did not affect the period. It was the length that affected the period.
The student would like to add the following computation for the experimental value of g at varying length.
For Data Table 3 (varying length)
T = 2√(L/g)
g = 4L / T2
at L = 0.250 m T = 1.027
g = 4L / T2
g = 4 (0.250) / (1.027)2
g = 9.357 m/s2
at L = 0.500 m T = 1.46
g = 4L / T2
g = 4 (0.500) / (1.46)2
g = 9.260 m/s2
at L = 0.750 m T = 1.720
g = 4L / T2
g = 4(0.750) / (1.720)2
g = 10 m/s2
at L = 1 m T = 2.053
g = 4L / T2
g = 4 (1) / (2.053)2
g = 9.367 m/s2
We will just use the average of the g for 4 trials as a basis for our comparison with the theoretical value of g which is 9.810.
Average = (9.357 + 9.260 + 10.000 + 9.367) / 4
Average = 9.496 m/s2
% Error = | Experimental – Actual| /Actual
% Error = | 9.496 – 9.810 | / 9.810
% Error = 3.2 %
Questions:
A. How did the change in the weight of the bob affect the resulting period and frequency?
Adding the weight increases the period by a very small amount in the experiment. But this difference is negligible (0.0264 seconds) as theoretically mass should not affect the period in accordance with T = 2pi√ (L/g). Thus it is enough to say that the mass has no bearing with period.
B. How did the change in amplitude affect the resulting period and frequency?
It did not affect the resulting period and frequency that much. As the amplitude increased in degrees, the period also increased a little bit. Theoretically speaking, as angle increases the period obtained by using the equation L = 2pi√(L/g) is less reliable.
C. How did the change in length of the pendulum affect the period and frequency?
D. What would happen if you used very large amplitudes? Check your hypothesis by
trial. What amplitude did you use? What is the result?
Hypothesis: The period increases since the angle is increases.
At 80 degrees, L =101 cm, and m=0.251 kg, the period is 2.9 seconds which is way higher than the theoretical value of T = 2*pi√(1.01/9.8) = 2.02 seconds
E. Hypothesize about how a magnet placed directly under the center point would affect
an iron bob? Try it and find out. Did your trial verify your hypothesis?
Hypothesis: A magnet would make the travelling of the bob faster as it would have the same effect as increasing the gravitational force.
The result of the trial conformed to my hypothesis. The period decreased by a small amount which indicate faster travel time.
F. How close was your calculation of the value of g at your location? What might be afew sources for error in your experimental data and calculations?
The experimental and theoretical values of g were close enough. The error was 3.2 %. There are actually several reasons but probably the main contributor was air resistance. Equipment limitation and inaccuracy of measurements are amongst other reasons.
G. What would you expect of a pendulum at a high altitude, for example on a high mountain top? What would your pendulum do under weightless conditions?
At high altitude the acceleration due to gravity is decreased. In this case the gravitational force also is decreased. Having said these, we know that when these parameters are decreased, the period will increase as they are inversely proportional.
At weightless condition, the period becomes infinite as the force of gravity approaches zero. In this case the bob will remain as where it is raised.
Conclusions:
The acceleration due to gravity is 9.81 m/s2. This value can be approximated by observing and measuring the value of length of the string and the period of the pendulum according to the equationg = 4L / T2. Rearranging this equation, period is T = 2√ (L/g). Thus, period of the pendulum is not dependent on its mass and depends primarily on the length. However the approximation given by T = 2pi√ (L/g) is accurate only at small amplitude. Increasing the amplitude increases the inaccuracy of the calculated period. This independence of period with mass and small angles and the dependence to length were all verified in the experiment.
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EXPERIMENT 12: Crystal Growing and the Rock Cycle
Note: Part One of this lab should be performed at least 10 days before your report due date. Read the entire experiment and organize time, materials, and work space before beginning.
Remember to review the safety sections and wear goggles when appropriate. Objectives: To grow synthetic crystals from a supersaturated solution by evaporation,
To measure the interfacial angles of minerals, To make sugar “glass,” To understand the role of evaporation in mineral growth, and To determine the dissolution point of certain crystals.
Materials: Student Provides: Pan, small Spoon or blunt knife Cup saucer Stovetop burner Refrigerator 50 g sugar
From LabPaq: Tweezers Protractor Ruler
Magnifying hand lens Digital scale
100-mL Beaker 3 Petri dishes, large Thermometer Set of 18 numbered minerals Igneous rock sample #19 Sedimentary rock sample #36 Metamorphic rock sample #47
Epsom salt: Magnesium Sulfate Heptahydrate, MgSO4 · 7H2O
Alum: Aluminum Potassium Sulfate Dodecahydrate, KAI(SO4) 2 · 12 H2O
Discussion and Review: The textbook definition of a mineral is “a homogeneous, naturally occurring, solid substance with a definable chemical composition and an internal structure characterized by an orderly arrangement of atoms in a crystalline structure” (from Earth; Portrait of a Planet; Stephen Marshak (Norton, 2005). A crystal grown in a lab is not a true mineral since it did not form by geologic processes. However, crystals grown in a lab are virtually identical to true minerals in many other
Hands-On Labs SM-1 Lab Manual
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aspects: they are solid, inorganic, homogeneous, and have a definite chemical composition and an ordered structure. By growing crystals in a laboratory setting you will be able to investigate the different properties that define a mineral. In addition, growing synthetic minerals can offer insight into the factors that affect the crystal growing process in a true geologic setting. By “watching” your crystals grow, you’ll be able to better understand how crystal faces develop in rocks and what influences them, plus you won’t have to wait through geologic time to view the results! Since you will know the composition of the material that went into your crystals and guided the crystal growth from beginning to end, you’ll be able to identify your crystals easily. However, you may have picked up a “pretty rock” at one point or another in your life and wondered how to go about identifying it. Since all rocks are composed of a mixture of one or more minerals, the first step in identifying that pretty rock is identifying the minerals of which it is composed. Minerals are identified on the basis of several different physical characteristics which can be determined through tests, and you will perform some of these tests with the crystals you grow. This will help to increase your confidence about identifying minerals. Eventually you’ll be able to identify each mineral in your LabPaq by using those same tests, and this knowledge will then help you identify rocks. The Law of Interfacial Angles, first described by Nicolaus Steno in 1669, is one characteristic or test used to distinguish between minerals. This law states that the angles between the faces of a crystal are constant for that particular mineral, no matter how large or small the faces are. Thus, if you know the interfacial angles for a particular mineral, you can identify it by this rule. Crystal form is another physical characteristic used to identify minerals. Some of your larger crystals should show a distinctive geometric shape. Look at crystals displayed by several different minerals in your LabPaq and you’ll notice their different shapes. A crystal’s shape is indicative of how the individual molecules are arranged in the crystal and is an excellent means of identification in minerals that exhibit good crystal form. Unfortunately, as you will see in growing your own crystals, good crystal form is rare due to competition for space during the growth process. Your crystals’ edges and corners may be inter-grown with other crystals, and the bottoms will be flat due to the crystals growing upward from the flat surface of a petri dish. However, most of your crystals will tend to have the same general shape: perhaps a cube, an octahedron, or a rectangular form. This is the mineral’s crystal form. Crystal form can be distorted into needles, flattened, or elongated into prisms. Competition for space can cause a mineral’s crystals to not have any discernable geometric shape at all; this is shown well in igneous rocks which are composed of several different minerals. To verify how competition for space distorts their shape, examine your LabPaq’s igneous rock sample #19 with your hand lens and try to discern the geometric shapes of the crystals it contains.
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The Rate of Crystal Growth forms a basis for identifying igneous rocks since all igneous rocks are formed from the cooling of magma. Igneous rocks are said to be phaneritic if the individual minerals are large enough to see and aphanitic if the individual minerals are too small to see with the naked eye. This size of the mineral crystals directly relates to how rapidly the rock cooled. In some cases, the lava cools so rapidly that its molecules do not have sufficient time to organize into crystals and instead form into a volcanic or natural glass like obsidian. Volcanic glass is essentially a super-cooled liquid blob with no crystals. This lab simulates creating natural glass with molten sugar. Chemical Sedimentary Rocks can form from the evaporation of seawater which concentrates the chemicals in solution to the point where the chemicals will eventually precipitate out of the water and form minerals. The two most common minerals to form in this way are gypsum and halite. Gypsum and halite are names for both minerals and rocks; if the mineral deposit is large in scale, it is designated a rock unit as opposed to a smaller mineral occurrence. Mineral sample #9 and sedimentary rock sample #36 are examples of a material in both its mineral and rock forms. An example of a halite sedimentary rock unit is the salt flats around the Great Salt Lake in Utah. An example of a halite mineral occurrence would be if several halite crystals were found in the crevice of an igneous rock along the ocean shoreline. In lab we will grow Epsom salt and alum crystals by the evaporation method. The crystals you grow will be sufficiently small in scale and so would be considered mineral occurrences if they were formed by natural processes. Metamorphic Rocks are formed by heat, pressure, and/or the introduction of fluids like water and magma that alters the preexisting rock. Metamorphic rocks sometimes crystallize new minerals as part of the metamorphism process; garnet crystals are a good example of this. A metamorphic rock must stay at least partially solid at all times and never transform into a liquid during metamorphism; otherwise it becomes an igneous rock. However, chemicals in solution can be dissolved out of the host rock, transported to another place, and crystallize into new minerals during metamorphism. This lab will show how metamorphic rocks crystals can dissolve and reform naturally. In all crystal formation cases, the crucial component is a liquid solution. This lab will determine the temperatures at which our newly formed crystals will dissolve into solution. Dissolved minerals will precipitate out of a saturated solution at temperatures below their melting point. As you can see, many aspects of crystal growing can have applications to the different types of rock. Hopefully by completing this lab you will gain a better appreciation of the rock cycle. PROCEDURES: Part One: First, you will investigate the process of mineral crystallization by allowing Epsom salt and alum crystals to form by precipitation from evaporating water.
1. Create Epsom salt crystals as follows:
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A. Turn on and open the digital scale’s lid. With an empty beaker on its weighing
surface, press the scale’s tare or zero button so it reads “0” grams. Add Epson salt to the beaker until the scale reads exactly 30 grams. Pour the Epson salt into a small pan.
B. Measure 50 mL of warm tap water into a beaker and add the water to the pan of Epsom salt. Stir well. If the Epsom salt does not fully dissolve, place the pan on a stovetop burner and turn the heat on low. Constantly stir the contents as the pan slowly heats until you reach the point, but not beyond, where the Epsom salt is completely dissolved. Do NOT boil.
C. Remove the pan from the heat and pour its saturated solution into a petri dish. A
saturated solution is one that contains as much dissolved solute as it can under existing conditions and no additional quantity of that solute can be dissolved in it.
D. Wash the pan and beaker with dish soap, rinse, and dry well for their next use. E. Prepare a label by writing on the top half of a 4x4” sheet of paper: Science
Experiment – Not Dangerous – Please Do Not Disturb. Fold the paper in half so that a petri dish can sit on one half and the writing will face out and be immediately noticeable.
F. Set the petri dish, uncovered, with its label in a safe place where it will not be
disturbed, jiggled, bumped, or otherwise moved and where natural evaporation can take its course, preferably in a sunny, dry area. Within a few days you should see crystals beginning to form in the petri dish. Depending upon the humidity in your area, your dish should be fully evaporated and your crystals fully formed within 6 to 9 days. Don’t worry if for several days there is nothing in the Petri dish. Once crystallization begins, it will proceed quite rapidly.
G. Once the water has completely
evaporated, allow the crystals to dry out for 24 hours and then examine them closely with your hand lens. There may still be some residual water underneath the crystals. Record your observations and draw a sketch of the specimen’s crystal form. The photo at right is of Epsom salt crystals grown by the author. As you can see from the picture, they are not museum-worthy specimens. Due to competition for space in the Petri dish and the supersaturated conditions, the crystals are very crowded together.
H. Do NOT discard this petri dish or its contents; they are used again in this lab.
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2. Create alum crystals by following the above
instructions, A through H, except this time you will measure and dissolve only 10 grams of alum in 50 mL of water. The photo at right is of alum crystals. Again, competition for space and rapid growth in a supersaturated solution leads to crowding of crystals.
Part Two: Measure the interfacial angles of different mineral crystals. 1. Select the best four minerals from the LabPaq’s 18 numbered minerals that show
good crystal form plus one of the best Epsom salt and alum crystals grown in Part One.
2. Prepare a data table similar to Table 1 below, and record the numbers of the LabPaq minerals selected for measuring.
Data Table1:1 Interfacial angle measurements
Mineral No. Angle 1 Angle 2 Angle 3 Angle 4 Average # # # #
Epsom Salt Alum
3. Measure interfacial angles of the four selected mineral crystals. It will not be easy to
find good crystal interfacial angles as you must try to differentiate between the external form the mineral takes when it grows and the form it may show when it breaks. Your samples will probably contain many parts of broken crystals, but they should also show a few crystals that are reasonably well-formed. A further complication is that the crystals of most minerals have more than one set of interfacial angles. Try to recognize and measure the interfacial angles of similar faces for similar crystals at the same adjacent crystal faces. To do this:
A. Hold the straight edge of a ruler against the line of one
crystal face and the protractor against the adjacent crystal face as shown in the following illustration. The straight-edge ruler should intersect the protractor at an angle as shown. Tweezers and a hand lens will be helpful in this process. Record the angle in your table.
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B. Try to find at least two, but preferably four, interfacial angles in each mineral to measure. Record each measurement in your data table.
C. For your home-grown crystals, disregard the angles between the bottoms of the crystals where they adhered to the petri dish. Since your crystals were prohibited from growing into the petri dish, they will not show interfacial angles or crystal form in all directions. Measure the angles as best you can, using your hand lens and tweezers to assist you if needed. You may have trouble due to the small size of the crystals. Record any measuring problems you encounter in your lab notes.
D. Compute and record the average interfacial angles of the crystals for each mineral and your home grown crystals. You need to compute the average for the measurements taken because the method of measuring is too imprecise to discover the actual constant angles. To compute an average, add together the measured angles of each mineral and divide this total by the total number of measurements taken. For example, assume the angle measurements for mineral #8 were 102, 104, 100, 101, 103, and 102 which totals 612. Divide 612 by 6 and you arrive at an average angle of 102.
Part Three: Now you will observe how a crystal growth is affected by its cooling rate. 1. Forming crystals through rapid cooling:
A. Look closely at the sugar crystals in your packet. Use your hand lens to observe
them carefully and make a sketch of the sugar crystals’ shape.
B. Place a clean, dry petri dish on a saucer.
C. Measure 50 grams of table or raw sugar and place it in a small dry saucepan. Heat and stir over very low heat until, but not beyond, the point where all of the sugar is completely melted into a liquid solution. Do not add any water. Also, do not overheat the sugar or you’ll soon have caramel candy! Continuously stir the sugar over very low heat just until it is completely melted into a liquid. BE CAREFUL! MOLTEN SUGAR CAN CAUSE NASTY BURNS!
D. Once the sugar has completely melted, immediately pour it into a petri dish
resting on the saucer which will guard against spills and potential burns while transporting the melted sugar. Also, take care because the Petri dish may melt! Place the saucer with petri dish of sugar solution into your refrigerator and leave it undisturbed until it cools completely. This should take only a few minutes. What you are doing is trying to replicate lava solidifying while it cools in the air after being shot out of a hot volcano.
2. Once the sugar has cooled, take the petri dish out of the refrigerator and inspect it.
Hopefully you have just made sugar glass, where the cooling rate was so fast that no sugar crystals were able to form.
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3. Use your magnifying hand lens to examine the sugar glass and observe if there are
any crystals in the sugar glass. 4. Save the petri dish of sugar glass in your LabPaq to observe in a future experiment. Part Four: In this step, you will re-dissolve the alum and Epsom salt crystals you grew in Part One and find their melting temperatures which by inference should be the same as their maximum crystallization temperatures.
1. Re-dissolve the Epsom salt crystals by doing the following:
A. First, remove from the dish of Epsom salt any large crystals you might like to
save or try to grow into larger crystals (optional instructions follow). B. Place 50 to 100 mL of tap water into a pan. Measure the water’s temperature
with a thermometer and record it in your notes. [Note: The quantity of water used is not important here, but use the smaller amount if you want to have a well saturated solution for further crystal growing. See Optional Fun below.]
C. With a spoon or blunt knife, scrape all of the crystals from the petri dish into the
water. Stir and observe. Do the crystals dissolve? D. If the crystals do not dissolve in Step 1C, heat the pan’s contents slowly while
stirring. When the crystals are completely dissolved, promptly remove the pan from the heat source and measure the liquid’s temperature with a thermometer.
E. After you have finished this and any chosen optional experiments (see below),
discard your crystal solution by washing it down the sink with water. 2. Re-dissolve the alum by following Steps 1A through 1E above using alum crystals. Optional Fun Experiments with Crystals: 1. You can pour the saturated solution liquids from Part Four back into their petri
dishes or other containers and let them evaporate again! If you wish, you can crystallize and dissolve your crystals over and over again this way.
2. Sprinkle a few grains of table salt in your hand, examine them with your hand lens,
and note their perfectly cubical shape. You can grow larger salt crystals in a petri dish or other container similar to the way you grew crystals in Part One. To make a saturated solution of salt water, continue to add and stir in small amounts of salt into any quantity of warmed water until an additional amount will no longer completely dissolve. That’s when you know your solution is fully saturated.
Hands-On Labs SM-1 Lab Manual
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3. To grow a larger crystal, first allow the saturated solution to cool to room temperature and then pour it into a narrow glass or container. Tie a piece of thread around a single crystal you’ve already grown. Suspend the crystal in the middle of the solution. Tie the opposite end of the string to a spoon or knife placed across the top of the glass. This seed crystal should be suspended in the middle of the solution and not be touching the glass. Observe the crystal daily for the next week and you should see it grow much larger! You may need to periodically adjust the string’s length to ensure the crystal is always submerged in the saturated solution.
Note: Your LabPaq contains sufficient alum and Epsom salt to perform Part One’s instructions twice in case you have a problem on your first try. Thus, you should have ample materials left over to make extra solutions and grow very nice large crystals. It’s fun to see how large a crystal you can grow!
4. You can also use your new knowledge about crystal growing to make rock candy.
Prepare a saturated solution of sugar water and suspend a line of string or a thin wooden skewer into the water as you would to grow a seed crystal. Within a few days crystals will begin growing along the string or skewer and you’ll eventually produce a nice sweet snack of rock sugar crystal candy. To prepare a saturated sugar solution, continue to add and stir in small amounts of sugar into any quantity of hot water until an additional amount will no longer completely dissolve. That’s when you know your solution is fully saturated.
Results and Conclusions: Include in your formal report all tables, graphs, calculations, lists, sketches, charts, etc. that you produced in this lab. Questions: (Please give thoughtful and complete responses to the questions following this lab and all future labs as required.) A. Describe in detail the crystal form of Epsom salt, alum, and sugar.
B. Regarding your sugar glass, how can you relate what you did to the way in which
obsidian forms? Is your sugar glass phaneritic, aphanitic, porphyritic, or glassy? Explain your answer.
C. Do evaporate minerals tend to show good crystal form? Why or why not? Did the
Epsom salt or alum show better crystal form? D. Review your table of interfacial angle measurements. Does the Law of Interfacial
Angles hold true for your mineral samples and the crystals you grew? E. At what temperatures did the Epsom salt and alum dissolve? In a metamorphic
process where there was an introduction of water into a surface-temperature rock, would these two minerals probably be among the first or last to dissolve? What if the rock was buried deep in the earth and heat was also part of the metamorphic process?
- SM-1 Manual COLOR 105 08-17-07.pdf
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