Thermal Expansion
The top diagram shows that when a round metal with a circle cut out is heated, the metal expands and the hole gets bigger.
The bottom diagram shows a
bimetallic strip. When heated, one metal expands moroe than the other metals
and causes the metal to curve. This is because different metals have different
expansion rates. This concept was confirmed in a lab experiment conducted by Prof. Mason.
Thermal Expansion of a Metal Rod
A metal rod has one end fixed in position while the other is free to move with expansion. A pulley supports the free end of the rod. As the rod expands, the pulley rotates and the change in length can be found by measuring the angle that the pulley moves through. With the help of the diameter, circumference, and angle of motion of the pulley, we can then use the original length of the rod to find the coefficient of thermal expansion of the metal (alpha). We found this coefficient to be 2.93 X 10^-5.
However, within every measured value there is a certain degree of uncertainty. To account for this, we propagated for the uncertainty of alpha considering the uncertainties in the measurements of temperature, diameter, and angle movement (theta). We propagated for the uncertainty in alpha and found that uncertainty was +/- 4.26 X 10^-6
This now means that alpha is 2.93 X 10^-5 +/- 4.26 X 10^-6.
We checked a thermal expansion table of metals and compared it with our results. We found several metals that were within the value and uncertainty range we came up with. However, we can make an educated guess and be fairly confident that the rod material was aluminum since it is cheap and it does not look like lead or glass, etc.
Latent Heat of Vaporization
In this experiment, we set up logger pro to track a temperature change in water as heat is added to it by a 308 W immersion heater. The initial mass of the water and cup was 150 g. We started the data collection and we can see the temperature rising in the graph.
At this point, the water is boiling and we can see from the graph below that the temperature keeps constant at 100*C.
We allowed the water to boil long enough for sufficient mass to be lost by steam. The final mass measured was 133.0 g, which means there was 17.7 g mass loss due to steam. However, the amount actually lost is greater because steam mass was being lost between the end of the data collection to the time when we measured the mass.
We measured the change in time when water was boiling by selecting the region where the line of the graph was horizontal at 100*C. This came out to be 116.5 s.
Now that we have the change in time, power, and mass we can solve for the latent heat of vaporization by doing the following.
First we took that the amount of energy is proportional to the power and time (W=Pt). This gave us that 35882 J of energy were used in the 116.5 s.
Next we took the amount of energy used and divided it by the amount of mass lost due to steam. This gave us the latent heat of vaporization for water to be 2027 J/g or 2.027 x 10^6 J/kg.
We took the results of experimental latent heat of vaporization for water and put them on a spreadsheet. With this we were able to calculate the standard deviation for the experiments. The standard deviation was 8.58x10^5 J/kg. We can take this to be the uncertainty in the experiment. This means that our value for latent heat of vaporization was 2.027 x 10^6 J/kg +/- 8.58x10^5 J/kg. This range of uncertainty covers the true value therefore we conclude that the our experimental value is accurate.
Latent Heats
In this problem, we will determine the final temperature of a 790 g block of ice that has 215 kJ of heat added to it. The total heat is equal to the heat required to take the temperature up from -5 *C to 0 *C, then the latent heat of fusion for water, and then the heat required to take the temperature of liquid water up to its final temperature.
We calculated that it takes 8255.5 J to rise the temperature 5 C* to 0 *C. Then we calculated that the latent heat of fusion was 1777500 J. This means that to change the temperature of solid water from -5 *C to liquid water at 0 *C, it takes 1785.76 kJ (82.555 kJ + 1777.5 kJ) of energy. This means 215 kJ is not enough to melt the block ice, therefore, the final temperature is 0 *C.
We also summed the energies and came up with a negative result. The negative implies that there needs to be more energy applied to the ice.
Relationships Between Pressure, Volume, and Temperature
This shows the relationship between pressure (P) and volume (V). We can see that the pressure and volume are inversely proportional. It is somewhat intuitive since if volume gets closer and closer to zero, the pressure must be approaching infinity. So P=A/V where A is a constant.
A demonstration with hot and cold water was conducted with a pressure sensor. The graph below shows the relationships between temperature and pressure.
This shows the relationship between pressure (P) and temperature (T). The pressure is proportional to the temperature. The graph of P vs. T is linear as seen from the demonstration.
