Electric Charges

We assumed that the balloon would stick to the glass after being rubbed on hair and then later after being rubbed with silk. Both assumptions were correct. We tried explaining charges to children but it was a bit more difficult than we expected.

The picture above is of an experiment we did in class with tape to show that there are multiple charges. When a piece of tape was stuck to the table then removed and placed near skin there was an attraction. When two pieces of tape were each stuck to the table, then another piece was stuck to the top of each. The tops would repel tops and the bottoms would repel bottoms but when top and bottom were placed near each other they would attract. This was consistent with what the hypothesis stated, that there were two kinds of attractions and more than one charge.

In the above picture, Mario shows us his expertise in free body diagrams and the equations of forces in the x and y directions. With this we were able to derive a formula for the force.

The relation between the electric force and the distance between the objects is inversely proportional.

We use Coulomb's law to find the electric force between two objects.

This is the graph from the experiment in which we mapped out a video of one metal ball on a stick pushing another ball hanging from a string through the use of charge. The graph on the right is the position graph of both balls and the graph on the left is of electrical force vs. separation distance. 

The above picture shows my free body diagram and my Newton equations and the angle and the Electric force. This number may not be accurate because I went with an assumed value at the end of the position graph rather than an actual value.

The two pictures above show the questions from the conclusion section from the logger pro experiment we conducted above. experiment.

The above picture shows and object that transfers electric energy from the base to the silver top through the use of belts that act as insulators. The paper on the silver top is being repelled as a result.

Professor Mason performed different experiments with the device, first with a pie tin which, at first, prevented the paper from lifting, right before it was ejected into the air. Then he turned on the device again with this fan on it, and the fan spun and also prevented the paper from being repelled.

Our assumptions were that the pie tin would fly off and that the propeller would spin clockwise, both of which were true..

The ratio between the force of electricity and the force of gravity.

Entropy

Formula for entropy and 2 examples of entropy. We also found out a new way to describe adiabatic processes, isentropic, since there is no change in entropy.

This is a Stirling engine. A temperature difference from the top and bottom causes it to run.

This picture shows what happens when you place a stirling motor above heated water and place an ice cube on top. The fan is spinning as a result.

The fan spins the opposite direction when hot water is placed on top and ice is used to cool the bottom.

The above picture shows the formula for efficiency of a Carnot Engine and also the efficiency of a Stirling engine. The graph is that of Temperature vs. Entropy.

The above picture shows how to solve for the Coefficient of Performance. We used that to find the heat required to operate a heat pump to warm the inside of a home.

The above picture illustrates how to find the effectiveness of heat engines. You take the actual output of the energy desired and divide it by the output of the reversible process. We took the efficiency of the theoretical actual process and divided it by the efficiency of the theoretical reversible process to find the effectiveness. 

In the picture above we were able to show the the final temperature is equal to the root of the two provided temperatures when entropy is equal to zero.

We used what we found in the previous picture to come up with this equation.

In the picture above we found the efficiency of a process then multiplied it by the max Coefficient of Performance so we could find the possible Coefficient of Performance. Then we found the Q of heat with that number since Q of heat is equal to the Work plus the Q of cold.

In the above picture we found the time it would take to freeze a beverage in a fridge that was providing a certain amount of cool energy to its interior. We did this by finding the the energy it takes to freeze the beverage and divided it by the energy output by the fridge.

In the picture above, Professor Mason showed us that bubbles would fall to the ground after being created. This was not the case with bubbles created with methane gas though. This was because bubbles he created had air on the inside while the methane gas bubbles were less dense than the air. allowing them to float.

The video above shows what would happen if methane gas bubbles were ignited. As expected the fire rose since it consisted of methane gas, which is less dense than air.

Carnot and Otto Engines

This shows a thermoelectric device which takes heat from one cup and powers the spinning disk, and the other side acts as a cold reservoir. The disc spins in whichever direction the flow goes. If the hot cup is on the left and the cold on the right the disk will go clockwise, and visa versa.

When the sensors are plugged into a power supply one side heats up and one side cools down.

We assumed that the direction of the spinning disk would change when the cups were swapped. We also assumed the heat would flow to one side of the device while cooling the other side.

We used the formulas Q=n*Cv*dT and E=3/2*n*R*T and Cv = 3/2*R to come up with the equation of 3/2*R = Cp - R, meaning that 5/2*R = Cp.

We show that n*dT = (p*dV + V*dp)/(Cp-Cv) by first finding that Cp-Cv = R and subbing that into the ideal gas formula where there is both a change in volume and pressure.

The picture above shows the simplification of the of the previous equation to show that dp/p + (dV/V)*(Cp/Cv) = 0. With that we were able to derive another equation by moving the V's to the other side, taking the integral of both sides, then take the exponential function of both sides and we then find the equation relating the pressures and the volumes.

We derive a formula for temperatures and volumes, by subbing for p values determined from the ideal gas law(nR cancel), on the left of the board and we derive a formula for work in an adiabatic process on the right.

We used the derived formula for work in an adiabatic process to find the work in this situation.

This is a table of a Carnot cycle The top and bottom are isothermal and the left and right sides are adiabatic. We found work on the adiabatic with the formula W=PiVi^5/3[(Vf^-2/3 - Vi^-2/3)/(-2/3)] and we found work in the isothermal stages with W=nRT*ln(Vf/Vi). 5/3 is gamma and -2/3 is 1 - gamma. We were able to fill out the rest of the table since change of energy is zero in isothermal processes and Q is zero in adiabatic processes.

This device is an example of how an Otto engine works. Air is brought into the cylindrical shell when the piston is drawn down. Then the piston goes up quickly to increase pressure and reduce volume to cause the temp to rise. The spark plug then ignites causing and explosion where the piston is forced down. As that happens the exhaust valve opens and releases the burnt air and fuel. The process is then repeated.

The left side of the board shows what we could do to improve the work output of an engine and the right side of the board shows the extra degrees of freedom in a diatonic gas.

Engines and the Four Thermodynamic Processes

At the start of class we did a computer assignment which showed the connections between Volume, Pressure, and Temperature. We also used the formula P1V1/T1=P2V2/T2 to solve for certain variables.

This graph shows the 4 processes. The isothermal explanation in the picture is not correct though, it should state that there is a gradual inverse change in both Pressure and Volume that prevents a change in temperature.

Our assumption for heating up a rubberband was that it would expand. In reality it shrinks when heated because it is a polymer.

This is our engine composed of a rubber band, ramp and hairdryer. We would move the can to the lift, heat the band, place the can on the next conveyor, cool the band, then repeat.

This picture shows where heat is added and where work is done. Lifting the can and placing the can on the other conveyor are both examples of work. Heat is added when we heat the band to shrink it.

This is the formula for Work and for Efficiency of a heat engine.

This picture shows a cycle that is isobaric on the top and bottom and isochoric on the left and right. We found internal energy at each stage with the formula E = 3/2 * PV. We then subtracted to find the change in internal energy. Net work is zero in this situation because the change in volume is the same, one being positive and one being negative.

This is a more clear picture of the same problem. We also solved for Q in this picture. To solve for Q we put the work and the change in energy together (Q = E+W). Q simply equaled the change in energy when volume was constant.

The above 3 pictures are of an experiment Professor Mason did in which a beaker was submerged first in cold water, left at room temperature, then placed in hot water. This is similar to an engine cycle.

The above picture shows what the graph should have looked like if we could have completed the experiment properly.

The above two pictures show how we found change in energy, Q, and subsequently W from the experiment. All net energies were 0. The white board picture shows our energy calculations in green and black, the mass calculation in blue, and the Q calculations in red.