Electric Flux

     Not surprisingly, we were asked to answer some questions about electric flux from the ActivPhysics website. The questions concerned how electric flux is affected by the magnitude and the sign of the electric charge that is contained in a closed surface, and the answers to those questions can be seen in the picture above. Overall, it was determined that the electric flux through a surface is directly proportional to the magnitude of the electric charge contained in the surface. Also, the sign of the electric flux is the same as the total charge contained in the surface.

     On Monday, we were also asked to construct a graph of electric flux vs. the angle of the normal vector. We used a wire square and a small bed of  nails to come up with a series of data points. First we simply laid the wire square on the bed of nails so that it contained 49 nails. The angle of the normal vector in this case was obviously 0 degrees. We then angled the wire so that it contained 42 nails and determined the angle by taking the inverse sine of the height of the square wire at that particular instance and its length. We continued to do this in decreasing increments of 7 nails until we reached zero and then used symmetry to determine the angles of up to -49 nails. These points were plotted in LoggerPro and a graph was constructed. There is a shift in the second portion of our graph, but the graph should be that of the cosine function. This graph makes sense because the electric flux through a surface can be expressed as EAcosTheta.

Electric Fields and Hockey

     On Wednesday, we calculated the electric field on a point caused by a charged 10cm uniform bar which was a certain distance away from the point. The above picture is an example of such a calculation.

     We then set up a spreadsheet of values consisting of the vertical component of the charge on the point from each segment of the bar. The values were then summed and the total charge on the point was determined to be 1.27*10^5N/C.

     We were also asked to play a couple of rounds of field hockey as a homework assignment. This is a picture of the second difficulty which I completed with one positive charge and one negative charge.

     The third difficulty level required a bit more time to complete and more charges. I ended up using one positive charge and two negative charges.

Electrical Interactions

     The first experiment we conducted on Monday involved the electrostatic forces between two pieces of scotch tape. We first had to place two 10cm strips of scotch tape onto the table with the sticky side down. We then peeled the tape off of the table and brought the non-sticky sides of the tape toward each other and observed what happened. This observation is the statement labeled A in the picture. Then, we placed two strips of tape on the table with the sticky side down and placed another strip of tape onto each of the strips on the table. They were labeled "B" for bottom and "T" for top. We then pulled the tape off of the table, separated the tape that was joined together, and answered some questions about the interactions between each of the strips. The answers to these questions are labeled 1, 2, 3, and C.

     The next experiment we engaged in was an Electric Force Law Video Analysis Activity. The first thing we did was derive an expression for an electric force acting on a hanging ball. This derivation can be seen above. We then had to do the actual video analysis and plot a graph of electric force vs. separation distance.

     This is a picture of the graph we obtained from Logger Pro. In order to construct this graph we had to set the origin at the middle of the hanging ball, click the Add Point button, and click on the middle of the ball on the stick as each frame progressed until the video ended. We then had to select the Set Active Point button, select Add Point series, and perform the same process with the hanging ball until the video ended. Then we had to create two calculated columns called the Separation Distance (distance between the two balls) and Electric Force, which was done by entering our previously derived equation. The result was the graph seen above.

     Finally, we had to answer some questions about the results we got from the experiment. The first question asked if we were able to show that electric force was inversely proportional to the square of the distance between the two charged balls in the video. The second question asked for our percent difference for our experimental value for the exponent of 2.194 with the actual value of 2 for this law. This question also asked us to determine the charge on each ball if we assumed the charge on each ball was the same and if the charge on one ball was half of that of the other. We had to use one of our data points to answer these two questions. The third question asked if we were able to determine the sign of the charge on each ball from the experiment. Finally, we were asked about sources of uncertainty in our data. We cannot tell how accurately the mass of the ball and the length of the string were measured because they were given, but how accurately we clicked on the center of each ball at each frame certainly led to some source of uncertainty.

The Diesel Cycle



This is a picture of the Diesel Cycle assignment we were given to complete as homework. It also turned out to be the quiz we were given on Wednesday. The values obtained from the calculations seemed reasonable, and the efficiency of the engine turned out to be 63.7%. I believe the ideal efficiency of a diesel engine is somewhere around 60%, so I was satisfied with my answer.

These are the calculations that were performed during the diesel cycle quiz. I did not have very much trouble performing these calculations until I tried to calculate the work for the adiabatic expansion from point 3 to point 4. I was using an equation that we derived in class(I think), and it worked well enough for the adiabatic compression. However, it did not work so well for the adiabatic expansion. In fact, the engine would have an efficiency of 398% with that value. I do not fully understand why that equation works for compression and not for expansion, but I used the formula for work in an adiabatic process provided in Professor Mason's class notes, and the calculated value made much more sense.

     This Saturday was the STEM Presentation and Poster Competition! Unfortunately, I was unable to stay for the entire event, but I was there for the first few hours or so. It was a great experience.

 



Heat Engines and Molar Heat Capacity

     On Monday, we were asked to give a theoretical analysis of a heat engine cycle. This is a picture of the graph we completed with calculated values, given values and other information such as which paths included work being done by the gas or work being done on the gas. The paths in which heat was transferred to the gas and in which heat was transferred from the gas are also noted.

     This is the page in our lab book in which I filled in all the necessary values to complete the exercise. The energy at each point on the graph was calculated using the formula E=(3/2)PV since the values of P and V were known at each point. The change in E was found by simply subtracting the internal energy of the initial point of a path from the internal energy of the terminal point of that path. All of the work done either by the gas or on the gas was done at constant pressure, so work could be calculated by W=P(V2-V1). With work and the change in internal energy known for each path, we could determine the heat energy transferred to or from the gas using the first law of thermodynamics. After all this information was found, we calculated the total work to be 1882 J, which is the same value obtained from finding the area underneath the curve.

     On Monday, we were also asked to complete some exercises using ActivPhysics. This is a screenshot of Question#6. We noted the initial values of Q and T, ran the simulation until T equaled about 400K, and recorded those final values. We determined the molar heat capacity of a monatomic gas by using the relationship dQ=nCdT and solving for C. We simply inserted our dQ and dT into the equation and found a value very close to that displayed on the screen.

     This exercise was very similar to the first, but the simulation was allowed to run for a longer time in hopes of getting a more accurate calculation of molar heat capacity at a constant pressure. Again, we obtained a value of C very close to that displayed in the picture.

     In the eighth exercise, we were asked to derive the equation we had been using to calculate the value of C. This is a picture of the derivation, and the final answer of (5/2)R can be seen toward the left. If one plugs in the value of R, 8.314 J/mol*K, into the equation the answer does indeed come out to be 20.8 J/mol*K. I can't say why Andrew decided to draw a flower.

     The last thing we did on Monday was to do an analysis of a Carnot engine cycle. However, it was a little different in that the processes of the paths had change. The paths from A to B and from C to D were both isothermal expansions. The path from B to C was an adiabatic expansion, and the path from D to A was an adiabatic compression. Ultimately, we were to determine the thermal efficiency of the Carnot engine by dividing the work done by the engine by the heat that was put in.
     We did not finish this in class, so I completed the calculations at home, and they can be seen below. I did not have to use the equation in the upper right hand portion of the paper because I knew that the heat transfer was zero in an adiabatic process. I simply used the first law of thermodynamics to compute the work done since the change in internal energy was easy to compute. The efficiency of the engine came out to be 33.8%, which is about what it should be.

The Fire Syringe Experiment and Active Physics

     This is the setup and calculations performed for the Fire Syringe experiment that was conducted on Wednesday. The experiment itself was a demonstration of an adiabatic thermodynamic process. In an adiabatic process, no heat is transferred into or out of the system. We were asked to calculate the final temperature of the gas after the plunger was pushed down very rapidly, and this calculation was made using the formula at the upper right hand corner of the picture. We first had to calculate the initial and final volumes of the cylinder using a ruler and a micrometer. The radius of the cylinder was determined by measuring the diameter of an o-ring which was attached to the plunger with a micrometer. The other measurements were made using a ruler, with the most uncertainty coming from the measurement of the final volume. All of the measurements and the calculated final temperature can be seen in the picture above. This final temperature is enough to ignite paper, and so it was determined that if a small piece of cotton was placed at the bottom of the syringe, it would ignite with a rapid decrease in volume of the trapped air. We carried out the experiment  and saw that our hypothesis was correct.  

     

     This is an actual video of the experiment. I seem to have trouble uploading videos to this blog, so if you are unable to view it you can find it on a blog belonging to either Andrew or Dennis.

 

     This is a screenshot of the first of six exercises we completed during class on Wednesday. We were asked which of the three graphs correctly displayed the relationship between volume and temperature in an isobaric thermodynamic process in which the pressure and the number of atoms remained constant. The correct graph is the topmost one in which the volume of a gas increases with increasing temperature. This shows that volume and temperature are directly proportional.



     In the second exercise, we were asked to identify the graph that correctly displays the relationship between pressure and temperature in an isochoric process in which the number of atoms and the volume of a gas remain constant. The correct answer is the graph that is displayed at the upper left hand corner of the picture. Much like the previous graph, this shows a linear, directly proportional relationship in which the pressure of the gas increases with increasing temperature.

     In the third exercise, we were asked to identify the graph which correctly displays the relationship between pressure and volume in an isothermal process in which the number of atoms and the temperature of the gas remain constant. Again, the correct answer is displayed toward the upper left hand portion of the picture. If the temperature and number of atoms remain constant, the pressure of an ideal gas decreases with increasing volume. This constitutes an inversely proportional relationship between the pressure and volume of an ideal gas.

     In the fourth exercise we were asked to calculate the final volume of a gas in an isobaric process if its initial temperature is decreased. The actual calculation can be seen below as well as the calculations for the fourth and fifth exercises.

     As you can see from the calculation, the quotient of the initial volume of gas and its initial temperature is equal to the quotient of its final volume and final temperature. The initial volume of gas was 41.6*10^-3 m^3 at 500 K. When the temperature was decreased to 301.8 K, the volume decreased to 25.1*10^-3 m^3. This answer makes sense because of the direct relationship between volume and temperature in an isobaric process.

     In the fifth exercise, we were asked to calculate the final pressure of a gas after it has experienced an increase in temperature during an isochoric process. The initial pressure of the gas was 42*10^3 Pa at a temperature of 100 K, and its pressure increased to 126*10^3 Pa after the temperature was increased to 300 K. This is another example of a direct relationship between two state variables in a thermodynamic process.



     In the last exercise we completed on Wednesday, we were asked to determine the final pressure of gas as the volume of that gas decreased in an isothermal process. One can see from the whiteboard that the product of the initial pressure and volume equals the product of the final pressure and final volume. In the first part, the volume decreased from 40*10^-3 m^3 to 20*10^-3 m^3 resulting in an increase of pressure from 62 kPa to 125 kPa. The same process was used to calculate the final pressure of the gas as its volume decreased even further to 10*10^-3 m^3.



 

 

Charles' Law I

     This is a picture of the setup for the lab we did on Monday. It is somewhat difficult to see from the picture, but the glass syringe is attached to a 25mL flask with the aid of a rubber stopper. The purpose of this experiment was to determine how temperature affects the volume of a trapped gas that is kept at a constant pressure.

     This is of course the raw data that was collected during the experiment. The procedure itself was rather simple. We took a flask of known volume and attached the glass syringe and a release valve to it through a rubber stopper. The volume of the flask was determined by subtracting the mass of the "empty" flask and stopper from the mass of the flask filled with water and stopper. These measurements were taken using an electric balance. Once this was done, we allowed a certain amount of air into the syringe and closed the valve so that the syringe read an equilibrium value of 5cc. We then began the experiment by placing a flask into a beaker of hot water and recorded the final total volume of the trapped air. This procedure was then repeated in cold water and water at room temperature. These values were entered into a spreadsheet and graphed.

     This is a picture of the measurements from the previous picture inserted into a spreadsheet and the graph of Volume vs. Temperature that was constructed using those values. As one can see, the graph of the three data points(the total volume and the temperature measured using a probe) is linear, and the volume of a trapped gas is directly proportional to its temperature. The slope of the line came out to be roughly .106 and its units are cm^3/K.

Linear Expansion and Gas Laws

     This is a picture of the setup for the first experiment that was performed on Wednesday. The setup consists of an aluminum rod with one end connected to a steam generator and the other laid on top of a small pulley. As the steam heated up the rod, the length of the rod would expand and move the pulley through a certain angular displacement. Ultimately, our goal was to find a value for alpha (in the linear expansion equation) of the aluminum rod.

     These are the calculations and relationships used to determine the value for alpha of the aluminum rod. We first had to come up with a relationship between the angular displacement of the pulley and the change in length of the rod. We then used that relationship to solve for delta L and substituted it into a previously solved equation for alpha. That final equation can be seen in the upper right hand corner of the boxed off portion of the whiteboard. We then propagated our uncertainty using only the uncertainty of the length of the rod, which was given to us. Unfortunately, the true value of 2.4*10^-5 lied outside of our calculated range of uncertainty. If we had used more values of uncertainty in our propagation, I believe the true value would have fell into our calculated range.

     The above picture simply shows the two graphs that were constructed using LoggerPro during this experiment. The graph toward the top of the picture shows the relationship of Temperature vs. Time, and the graph on the bottom shows the relationship of Angular Displacement vs. Time.

     The second experiment we conducted on Wednesday was performed in order to calculate the Heat of Vaporization of water. Unfortunately, I was so excited about starting the experiment that I forgot to take a picture of the setup. However, a picture of a similar setup consisting of a Styrofoam cup, thermistor probe, and an immersion heater can be seen on the previous post. We used the immersion heater to boil water initially at room temperature and the thermistor probe to measure the temperature as it increased. The above graph shows the data collected from the experiment. The graph should look a lot smoother than it does in our picture, but there was a bit of technical difficulty that time would not allow us to correct.



     In order to calculate the Heat of Vaporization for water, we simply needed to know the power that the immersion heater supplied, the time it took for the water to boil, and the mass of water that was actually boiled off during the experiment. The average value of power for the entire class was 290W, which we used along with an uncertainty of plus or minus 10W. We then multiplied that by the time it took for the water to boil, 101.5 seconds, in order to find the actual energy that was transferred. After that we divided the amount of energy by the total amount of liquid water lost to get a value of about 1.1*10^6. Finally, we propagated the uncertainty for our calculation. 



 

 

     After the experiment, the values calculated at each table were put into a spreadsheet and the average value was taken. Since the uncertainty we propagated was actually rather small (about 3.9*10^4), we figured that standard deviation would give us a greater value of uncertainty that would hopefully allow the true value of Heat of Vaporization to lie within our calculated range. The process of standard deviation produced a value of uncertainty of 8.58*10^5. However, even taking this massive value of uncertainty into consideration, our percent error was slightly more than 12%. Without the uncertainty, our percent error was 50% given that the true value is 2.256*10^6.

     At the top of the whiteboard are our predictions of what a graph of Pressure vs. Volume and Pressure vs. Temperature would look like. We predicted that pressure would be inversely related to volume and directly related to temperature.

     This is a graph of Pressure vs. Volume that was produced with LoggerPro from a demonstration performed by Professor Mason. As you can see from the graph, our prediction was correct and pressure is indeed inversely proportional to volume. As the pressure in the demonstration was increased, the volume decreased accordingly. The relationship should be P=1/V, but according the LoggerPro it was P=3288/V. I was never able to figure out why this was.



     This is a graph of Pressure vs. Temperature that was produced from another one of Professor Mason's demonstrations. Again, our predictions were correct and the pressure of a gas is directly related to its temperature. As the temperature increases, so does its pressure and vice versa. The relationship is clearly linear as can be seen from the graph.