Physics

We have worked with hot plates and thermometers in other labs before, so we know how to be safe. I wonder what a graph of the changing temperatures over time between the phases would look like. Hypothesis Based on prior learning, I predict that the graph of the water phase changes will not be one straight slope, but instead will plateau, or be flat, and then slope between the change of ice to water, plateau and slope once more between the change of water to steam, and then plateau once more.

If we record the temperature of the H2O at regular Intervals, then we will be able to make the predicted graph. Procedure List of Materials: 1 thermometer (Tot 100 OIC 1 beaker (250 or 400 ml), 1 hot plate, CE (enough to fill the beaker), tap water (about 25 ml 1 graduated cylinder (optional), safety goggles, lab apron, at least 1 rag (to clean up spills), pencil/pen, notebook/paper, graph paper, and access to an electrical outlet. Steps: 1. Prepare a data table to collect measurements from the lab.

Make sure to put the dependent and independent variables in the correct place. 2. Fill the beaker with ice and add a small amount of water (about 25 ml) 3. Place the beaker on the hot plate with the hot plate TURNED OFF. 4. Take two temperature measurements of the Ice/ water slurry 30 seconds apart. 5. Do not let go of the thermometer. 6. Do not let the thermometer touch the beaker’s sides or bottom. 7. Turn the hot plate on to about 6 (about halfway). 8. Collect temperature measurements every 30 seconds.

The plateaus were not completely flat and fluctuated between 2 degrees of temperature. We didn’t get to record measurements after reaching 100 co because we ran out of time and our outlet wasn’t working. Analysis 2. Phase changes are happening to the water during the “flat” areas of the graph. 3. The water is heating up during the slanted areas of the graph. 4. More than one phase of water is present in the “flat” areas of the graph. 5. The heat from the hot late is converted to energy when the water is changing from ice to liquid. 6.

The heat from the hot plate is changing the temperature of the water when it is only liquid. 7. The graph would have a slower change in temperature over time if we had twice the amount of water, so the graph would be longer horizontally. My hypothesis was confirmed because our data that showed minuscule fluctuating temperatures for the plateaus and the rise/Jump in temperature for the slopes. However, I had not initially accounted for the slight drop of temperature in the beginning due to the ice lolling the water because the water was room temperature.

I learned that plateau points on the graph do not stay only one temperature but instead fluctuate. We did have a possible margin of error due to possible inaccurate temperature readings, not exact temperature readings, inconsistent time, and not exact time intervals. If we do this experiment again, we can try changing the temperature of the initial water added or the amount of water added to see if we produce any noticeable differences. Watching H2O change phases took longer than I though!

Let There Be Light Lamp Shade Company serves an upscale local market and is currently placing a bid for several public buildings in Asia. A total of 5,400 identical lights will be installed and delivered to the foreign port where the buyer would take possession. Let There Be Light Lamp Shade Company has three styles of lampshades. In order to determine the best possible bid, the company will determine how many lampshades can fit in the intermodal container and the total cost of delivering for style A, B, and Clamp shades.

The interior dimensions of the intermodal container are 8 feet wide by 8. 5 feet high by 40 feet long and can hold up to 44,000 pounds per loaded container. The intermodal container could hold 2,560 styles. Lampshades because it holds 2,720 cubic feet, however, the top six inches cannot be used. The style B shades can be stacked two packages high with the square foot on the bottom. Each column could hold 12 shades and there can be a total of 320 (8 x 40) columns of 12. In terms of style B, the intermodal container could hold 3,840 lamp shades without exceeding the weight. Style C shades can be stacked the same ways as style B and a container could hold 320 columns of 20, totaling 6,400 lamp shades. However, this would exceed the 44,000-pound weight limit. To stay under the weight limit, the number of lampshades the container could hold is 4,356 lampshades (10. 1 the weight of one lampshade divided by 44,000). The total cost of delivering each style of lampshades can be calculated by adding the cost of the lampshade being manufactured, packaged, shipped, insurance, and ocean freight rates. For style A lampshade, the cost of 5,400 lamp shades to be manufactured is $21,600 ($4 x 5,400).

Packaging style A lampshade is $0. 60 per lamp shade for a total of $3,240. The lampshades will need to be shipped to the Port of Oakland, which will cost $3,000 ($1,000 per load). The cost of insurance for shipping style A lampshade is $556. 80 because the total cost of the company at this point is $27,840 times 2% of the value of the shipment. The cost of ocean freight rates is $2,970. Adding the figures together brings the total costs of delivering for style A shades to the port of importation to $31,366. 80. Style B lamp shades cost $5 per lamp shade for a total cost of $27,000.

The packaging of the style B lampshade is $1,800 ($2 x 900). It will take two loads to the Port of Oakland for a total of $2,000. Insurance for style B lampshade will cost $616 because the total cost thus far is $30,800 time 2% of the value of the shipment. The ocean freight rate cost is $1,960; for a total cost of delivery of $33,376. For style C lampshades, the cost per shade for manufacturing is $6 for a total of $32,400 and the total packaging cost is $1,620. Again, two loads will be needed to deliver to the Port of Oakland for a total of $2,000. The insurance for style C lamps shades is $720 ($36,020 x . 02).

The ocean freight rate cost is $1,238. The total cost of delivering the style C shades to the port of importation is $37,978. Thus, style A lampshades would be preferred because it the least expensive out of all three styles. In conclusion, Let There Be Light Lamp Shade Company will be placing a bid for large public buildings in Asia. After analyzing how many lampshades can be loaded into the intermodal container and the total cost of delivering, the company will be placing a bid with style A lamp shades.

References

1. Murphy, P. R., & Wood, D. F. (2011). Contemporary Logistics. Upper Saddle River: Prentice-Hall.

A the centripetal force which is equal to the Tension on the Holon string, which is pulling against the spring force which is our centripetal force when our radius is constant. The normal force is always pulling up, and the force of gravity always pulling down, they are cancel each other out in this situation.

So even though we are adding more mass it does not have any effect on the centripetal force. Equation relating the net force (FCC) to the speed (v) of an object moving in uniform circular motion. F c = This equation shows that the net force required for an object to move in a circle is directly proportional to the square of the speed of the object. For a constant mass ND radius, the FCC is proportional to the speeds. VA 2 The factor by which the net force is altered is the square of the factor by which the speed is altered.

Subsequently, if the speed of the object is doubled, the net force required for that object’s circular motion is quadrupled. And if the speed of the object is halved (decreased by a factor of 2), the net force required is decreased by a factor of 4. We find the result of our experiment dose agrees with above. In our experiment we keep radius constant so our force stays the same, and as we increased mass on each trail we see our velocity keeps decreasing gradually.

Introduction:

The purpose of this lab report is to differentiate between of Newton’s Third Law and Newton’s Second Law. Newton’s Third Law states that all forces come in pairs and that the two forces in a pair act on different objects and are equal in strength and opposite in direction. Newton’s Second Law states that the acceleration of an object is proportional to the net force and inversely proportional to the mass of the object being accelerated. Using calculation equations for acceleration, force, and percent error one will be able to distinguish and evaluate the relationship between the two laws.

Procedures of experiment:

All groups had to complete two types of labs. The first lab our group had to complete was completed like so: using a car one had to tie a piece of string approximately 80 centimeters in length to a toy car on a ramp. This string was then feed one top of a super frictionless wheel. The end of the rope that had the end nearest to the ground was tied off with a 0. 2kilogram weight. The car was then pulled back by a participant until it reached 0. 8 meters and was let go. A second participant would then record the time it took for the car to reach the end of the ramp.

This was repeated a few times. One would add 0. 5 kilogram weight to the car and the time was then recorded in a similar fashion. These times would be recorded into a table and would be used as raw data. The second part of the lab was different in that: a group had to tie a small scale to opposite end of the rope where the weight was located. Immediately following that scale, one would then tie a toy car. After the toy car another scale would be attached; likewise another car would follow. One student would then pull the whole system back before the back end of the second car would touch the wall provided by the ramp.

Immediately after that student would release the system he and a second student would read the scales. Essentially, one student would have to read one scale and the other releasing would have to read one too. These force readings would be recorded in a table and would be used as raw data too.

Summary of data:

Each group had to determine the mass of each car before beginning each lab. Our group had found that both of our cars were 0. 261 kilograms. Additionally, group had to record the trail times for the first lab report like the following.  Each group had to read and record the force measurements in Newtons with the two car pulley system. Additionally, our group had used the same mass for the cars from the previous lab: 0. 261 kilograms. The magnitudes of the cars are relatively constant. IV. Analysis of Data: Before calculating many equations, one had to understand Newton’s Third Law.

The calculations for the second part of the lab would have been nearly impossible unless one understood that FT= -Fg=F1+F2. With this in mind, it was possible to ascertain that understand thatF2=-F1+Fg. Knowing this, one would be able to understand that the acceleration on the system is the same throughout. Also, it is extremely important that one must correctly change units into for the needed equations. Otherwise, almost all of the equations will be void. mass of the car | acceleration of car (m/s2) This is the first table representing the acceleration of the car, the force of the car as well as the percent error. The percent error for this particular section of the lab seems relatively low. One should see that the acceleration of the second car is significantly slower than the first car; this is because car one weighted les s than the second. This is the second table representing the acceleration, force, and percent error of both of the cars tied together. The percent error for this particular section of the lab seems relatively high. One should see that the acceleration and forces are exactly indistinguishable. This is because they are tied together forcing the cars and scales to have the same acceleration and force etc; moreover, it is because of Newton’s Third Law.

Additionally here are some sample equations I utilized during the lab report and calculations. Manipulating some of the equations was tricky especially for percent error. Substituting the accepted value with an equation is very smart thinking. percent error=(m2(9. 80m/s2))-experimental value(m2(9. 80m/s2)) (100) 56. 2934944%=1. 96-(0. 856647509)(1. 96) (100) ForceMass=acceleration 0. 485N0. 261kg=1. 858237548 m/s2 Distance= 12(acceleration)(time in seconds)2 2(Distance)(t)2= acceleration 2(. 8)(0. 93)2= 1. 849924847m/s2 V. Conclusion: There are several errors that could have occurred during this lab.

One of them might have been that I could have corrupted my calculations. I am not to entirely sure with how I calculated my percent error and therefore I might have a lower percent error yet I would not even know it or vise versa. A way to correct this problem for the future is by asking for more assistance from others in the class. I am more than sure that others who grasp the concept easier than me would be more than willing to aid me. Another error that could have occurred was that of miss reading the scales for the force in the second lab.

Although important to read the scales as accurately and as quickly as possible, one could only do so much. There was only less than a split second to read the correct or desired reading from the scale. One way to correct this error in the future is to have a larger group work on the same lab. Therefore all the students in the group could work together to figure out a solution as a team rather than an individual effort. The group would learn better as a result because there would be hands to complete the work and more brains to understand that work that is being recorded.

All thought the margin of error was higher for this lab, it is understandable because it was nearly impossible to record the force of Newtons desired in the split seconds one had. With all the information above, it is clear that there is a tie between Newton’s Third Law and his Second Law. It is possible to understand that Newton’s Second Law deals with changes in state of motion while Newton’s Third Law deals with the relation between forces On my honor, I have neither given nor received unauthorized aid on this assignment.

Experiment 1: Calorimetry Nadya Patrica E. Sauza, Jelica D. Estacio Institute of Chemistry, University of the Philippines, Diliman, Quezon City 1101 Philippines Results and Discussion Eight Styrofoam ball calorimeters were calibrated. Five milliliters of 1M hydrochloric acid (HCl) was reacted with 10 ml of 1M sodium hydroxide (NaOH) in each calorimeter. The temperature before and after the reaction were recorded; the change in temperature (? T) was calculated by subtracting the initial temperature from the final temperature. The reaction was performed twice for every calorimeter.

The heat capacity (Ccal) of each calorimeter was calculated using the formula, C_cal=(-?? H? _rxn^o n_LR)/? T[1] where ? Horxn is the total heat absorbed or evolved for every mole of reaction and nLR is the number of moles of the limiting reactant. The ? Horxn used was -55. 8kJ per mole of water while the nLR was 0. 005 mole. Table 1. Average Ccal from recorded ? T values. Trial? T, (oC)Ccal, (J)Ave Ccal, (J) 112. 2126. 82202. 91 21. 0279. 00 213. 093. 00108. 50 22. 3124. 00 310. 5558. 00558. 00 20. 5558. 00 412. 0139. 50244. 13 20. 8348. 75 513. 093. 0081. 38 24. 069. 75 612. 0139. 50209. 25 21. 0279. 00 712. 111. 60111. 60 22. 5111. 60 813. 093. 00116. 25 22. 0139. 50 Different heat capacities were calculated for each calorimeter (Table 1). After calibration, a reaction was performed in a calorimeter by each pair. A total of eight reactions were observed by the whole class. The temperature before and after the reaction were recorded. Then the change in temperature was calculated. Each reaction was performed twice to produce two trials. The experimental ? Horxn for each reaction was solved using the formula, ?? H? _rxn^o=(-C_cal ? T)/n_LR [2] where Ccal is the heat capacity previously calculated for each calorimeter.

The percent error for each reaction was computed by comparing the computed experimental ? Horxn to the theoretical ? Horxn using the formula, % error=|(computed-theoretical)/theoretical|? 100% [3] Table 2. Comparison of calculated ? Horxn and theoretical ? Horxn. RxnLRTrial? T, (oC)? Horxn, (kJ/mol)Ave ? Horxn, (kJ/mol)Theo ? Horxn, (kJ/mol)% Error 1HCl13. 5-142. 04-131. 89-132. 510. 47 23. 0-121. 75 2HOAc11. 3-26. 34-41. 61-56. 0924. 65 22. 7-56. 89 3HOAc11. 8-189. 61-203. 16-52. 47287. 18 22. 0-216. 70 4HNO311. 5-73. 24-70. 80-55. 8426. 78 21. 4-68. 36 5Mg13. 0-118. 67-138. 45-466. 8570. 34 24. 0-158. 23 6Mg15. 5-559. 4-635. 72-953. 1133. 30 27. 0-712. 01 7Zn13. 0-43. 80-43. 80-218. 6679. 97 23. 0-43. 80 8CaCl210. 00. 00-5. 8113. 07144. 47 20. 5-11. 63 There were differences in experimental and theoretical values of ? Horxn as shown by the percent error for each reaction (table 2). The discrepancies were caused by many factors. One factor was the loss of heat. The heat may have been released when the thermometer was pushed or pulled during the reaction. The heat may also have been lost because the calorimeter is not totally isolated. Another factor was the dilution of the solution. The pipette or test tube may still have been wet when used.

However, the concentration used in solving for values was the concentration of the undiluted solution. Another factor that may have contributed to the difference in the experimental and theoretical values was human error. It was manifested when reading the thermometer or measuring chemicals with different instruments. The factors aforementioned are the limitations of this experiment. References Petrucci, R. H. ; Herring, F. G. ; Madura, J. D. ; Bissonnette, C. General Chemistry, 10th ed. ; Pearson Education: Canada, 2011; Chapter 7. Appendices Appendix A Comparison of Observed and Theoretical Heats of Reactions RxnLRTrial? TnLRqrxn?

HorxnAve ? HorxnTheo ? Horxn% Error 1HCl13. 500. 00500-710. 19-142. 04-131. 89-132. 510. 47 23. 000. 00500-608. 73-121. 75 2HOAc11. 250. 00515-135. 63-26. 34-41. 61-56. 0924. 65 22. 700. 00515-292. 95-56. 89 3HOAc11. 750. 00515-976. 50-189. 61-203. 16-52. 47287. 18 22. 000. 00515-1116. 00-216. 70 4HNO311. 500. 00500-366. 19-73. 24-70. 80-55. 8426. 78 21. 400. 00500-341. 78-68. 36 5Mg13. 000. 00206-244. 13-118. 67-138. 45-466. 8570. 34 24. 000. 00206-325. 50-158. 23 6Mg15. 500. 00206-1150. 88-559. 44-635. 72-953. 1133. 30 27. 000. 00206-1464. 75-712. 01 7Zn13. 000. 00764-334. 80-43. 80-43. 80-218. 6679. 97 23. 000. 00764-334. 80-43. 0 8Na2CO3/ CaCl210. 000. 005000. 000. 00-5. 8113. 07144. 47 20. 500. 00500-58. 13-11. 63 Appendix B Sample Calculations Calibration of Calorimeter 10ml 1M NaOH + 5ml 1M HCl n. i. e. : OH-(aq) + H+(aq) ? H2O(l)? Horxn= -55. 8kJ LR: HCLnLR= 0. 005mol Grp 1 Trial 1 ?T= 2. 2oC Sol’n: C_cal=(-?? H? _rxn^o n_LR)/? T C_cal=(-(-55. 8kJ)(0. 005mol))/(? 2. 2? ^o C)? 1000J/1kJ ?(C_cal=126. 82 J) Determination of Heats of Reaction Neutralization Reaction Rxn 4 Trial 1: 10ml 1M NaOH + 5ml 1M HNO3 n. i. e. : OH-(aq) + H+(aq) ? H2O(l) LR: HNO3nLR= 0. 005mol ?T= 1. 5oCCcal= 244. 125 J Sol’n ?? H? _rxn^o=(-C_cal ? T)/n_LR ?? H? _rxn^o=(-(244. 25J)(? 1. 5? ^o C))/0. 005mol? 1kJ/1000J ? (?? H? _rxn^o=-73. 24kJ) Reaction between an Active Metal and an Acid Rxn 5 Trial 1: 15ml 1M HCl+ 0. 05g Mg n. i. e. : 2H+(aq) + Mg(s) ? Mg+2(aq) + H2(g) LR: MgnLR= 0. 00206mol ?T= 3oCCcal= 81. 375 J Sol’n ?? H? _rxn^o=(-C_cal ? T)/n_LR ?? H? _rxn^o=(-(81. 375J)(3^o C))/0. 00206mol? 1kJ/1000J ?(?? H? _rxn^o=-118. 67kJ) Displacement of One Metal by Another Rxn 7 Trial 1: 15ml 1M CuSO4 + 0. 5g Zn n. i. e. : Cu+2(aq) + Zn(s) ? Zn+2(aq) + Cu(s) LR: ZnnLR= 0. 00764mol ?T= 3oCCcal= 111. 6 J Sol’n ?? H? _rxn^o=(-C_cal ? T)/n_LR ?? H? _rxn^o=(-(111. 6J)(3^o C))/0. 00764mol? 1kJ/1000J ?(?? H? rxn^o=-43. 80kJ) Precipitation Reaction Rxn 8 Trial 1: 10ml 0. 5M Na2CO3 + 5ml 1M CaCl2 n. i. e. : CO3-2(aq) + Ca+2(aq) ? CaCO3(s) LR: Na2CO3/ CaCl2nLR= 0. 005mol ?T= 0. 5oCCcal= 116. 25 J Sol’n ?? H? _rxn^o=(-C_cal ? T)/n_LR ?? H? _rxn^o=(-(116. 25J)(? 0. 5? ^o C))/0. 005mol? 1kJ/1000J ? (?? H? _rxn^o=-11. 63kJ) Appendix C Answers to the Questions in the Lab Manual There are many possibilities that explain the discrepancy of the experimental and theoretical values of ? Horxn. First, heat might have been lost to the surroundings. This is possible whenever the thermometer is pulled out or pushed in the calorimeter during the reaction.

Also, the calorimeter might not have been thoroughly isolated. Second, the solution might have been diluted in the test tube or pipette. They might have been wet when used with the solution. Lastly, the discrepancies might have occurred due to human error. The students might have misread the thermometer when taking the temperature or the pipette when measuring the solutions. a. It is important to keep the total volume of the resulting solution to 15ml because any more or any less than that of the volume can contribute to the absorption or release of additional heat therefore affecting the ? Horxn. b.

It is important to know the exact concentrations of the reactants to solve for their number of moles and to find out the limiting reactant. c. It is important to know the exact weight of the metal solids used to solve for their number of moles and to find out whether one of them is a limiting reactant. Also, the weight is needed to solve for the heat capacity of the solid when the specific heat is given. 200ml 0. 5M HA + NaOH ? -6. 0kJ LR: HAnLR= 0. 1mole ?? H? _(rxn,mol)^o= (-6. 0 kJ)/(0. 1 mol) ?(?? H? _(rxn,mol)^o= -60 kJ) HA is a strong acid. OH-(aq) + H+(aq) ? H2O(l)? Horxn= -60 kJ/mole Calibration:15ml 2. M HCl + 5ml 2. 0M NaOH? T=5. 60oC LR: NaOHnLR= 0. 01mole Reaction:20ml 0. 450M CuSO4 + 0. 264g Zn? T=8. 83oC LR: ZnnLR= 0. 00404mole n. i. e. : OH-(aq) + H+(aq) ? H2O(l) n. i. e. : Cu+2(aq) + Zn(s) ? Zn+2(aq) + Cu(s) C_cal=(-?? H? _rxn^o n_LR)/? T C_cal=(-(-55. 8kJ)(0. 01mol))/(? 5. 60? ^o C)? 1000J/1kJ ?(C_cal=99. 6 J) ?? H? _rxn^o=(-C_cal ? T)/n_LR ?? H? _rxn^o=(-(99. 6J)(? 8. 83? ^o C))/0. 00404mol? 1kJ/1000J ? (?? H? _rxn^o=-218. 0 kJ) OH-(aq) + H+(aq) ? H2O(l)? Horxn= -55. 8kJ ?Hof,H2O= -285 kJ ?Hof,OH-= ? ?Horxn= ? Hof,product – ? Hof,reactant -55. 8 kJ = ? Hof,OH- – (-285 kJ) ?(?? H? _(f,? OH? ^-)^o=-218. 0 kJ)

Katherine Unman Introduction/Purpose: Pendulums serve a huge purpose that are often overseen by many due to technological advancements being made In the everyday world. A simple pendulum consists of a small object (the “bob”) suspended by a lightweight cord. The mass oft he pendulum is actually only the mass of the bob; the mass of the string is not included. The period of a pendulum is the amount of time for the bob to complete exactly one cycle or oscillation back and forth.

The length of the pendulum extends from the attached end of the string to the center of mass of the bob. The original aim for this investigation was to “Investigate the simple pendulum”. There are many variables on could look Into, such as displacement, angle, damping, mass of the bob etc. The most Interesting variable, however, Is the length of the swinging pendulum. The relations p between the length and the time for one swing (the period) has been researched for many centuries, and has allowed famous physicists like Isaac Newton and Galileo

Galilee to obtain an accurate value for the gravitational force acting on it, “g”. Len this simple investigation, we performed two activities to visually observe what affects the period off pendulum, mass or the length of the string. Hypothesis: With our previous knowledge of pendulums and the forces acting on a pendulum, we hypothesized that the length of the string along with gravity would affect the period a ND the mass of the bob would not. Materials: In order to complete a successful Investigation, numerous supplies were needed.

Without these materials, our observations would not have been as accurate. The mat aerials we used are: 1. Meter stick 2. Stopwatch 3. Pieces of string, 3 of the same length, and one off different length 4. Washers 5. A partner 6. Pen/pencil Procedure: When effectively Investigating what affects the period of a pendulum, some simple ye vital steps are necessary to follow. In this experiment, two activities were performed t hat share a set of Instructions. These were: 1. Gather all the materials

INTRODUCTION MATERIALS

PROCEDURE

Calibration of the Spectroscope: Using the spectroscope the four most visible lines on the scale were measured. Violet, blue, green, and yellow were all visible. With the ink pen, the measurements were recorded. A known wavelength (nm) vs. measured lines (cm) graph was then drawn from the measurements.

Observation and Measurements of the Hydrogen Spectrum: Using the calibrated spectroscope the scale position of the observable lines of the hydrogen emission spectrum was measured. Red, turquoise, violet, and purple were all visible. Using the measurements and the calibration graph the wavelength of the lines was determined.

The relative error was calculated using: Accepted Value Values of wavelength for the hydrogen atom spectrum were converted to kJ/mol. Using a form of the Rydberg equation, the Rydberg constant was calculated for each of the lines measured. This constant was used to then calculate the percentage error. Data Calibration of the Spectroscope Observations and Measurements of the Hydrogen Spectrum.

CALCULATIONS

(Convert wavelength values to corresponding energy in kJ/mol) 680 x 10^-9 2. 92 x 10^-19 J x (6. 022 x 10^23) / (1000 J) = 176 kJ/mol (Calculate the value of the Rydberg constant) (1/680)/(. 25-. 30) = . 00147059/(. 25000-. 11111)= 0. 0105042 x 10^-7 = Rh= 105,040 cm ^-1 (Calculate Percentage Error) 105040 – 109678 X 100 = 4. 23%