Linear Thermal Expansion Lab

Linear Thermal Expansion Lab:

The first picture the students were asked t predict what could happen when a metal hole is heated to see if it expands or contracts. Since the iron hole is made with the same material, it will expand allowing that a small iron ball can pass through. This indicates that when heat is applied to a metal made by the same material, it will expand at the same ratio and keeping always the distance between two points on the metal. This is called the Linear Thermal Expansion.

The second picture the material was made with different metals, such as, invar and brass. Here students were asked to predict the behavior of the metal bar.  All depends on the coefficient of linear expansion. Bigger it is, faster it will expand because it will allow more heat to excite its molecules, and by doing so, it will create a vibration and an expansion between them. And when cooled, happens the reverse of its expansion. The material will contract. In these experiment, the Brass was the one the expanded when heated, and contracted when cooled. Its Coefficient of Linear Thermal Expansion is 2.0x10^(-5)/K, and the invar is 0.09x10^(-5)/K. This means that the brass will allow more heat to cause vibration on its molecules, and causing the molecules to expand and increase its size. The same will happen when cooled, but this time the brass will bend and its molecules are been contracted, degreasing its vibration.

   
This picture represents the calculation of the Linear Thermal Expansion of a steel rod. Here the rod was heated with steam. Both the steam and the rod are at different temperature. Steam will be blow until the steam and the rod reach the same temperature. After that, students will have to calculate the angular displacement and linear displacement that the rod had; after a 300 second blowing steam through the rod, they finally reached the same temperature. The angular displacement of the rod was 0.28 rad, and the linear displacement was 12.2x10^(-5)/K. The rod is made by steel, since the calculated value is the same as the tabled linear thermal expansion's value.

 Heating Water

This picture represents a few measures that can be taken to made the mix Ice/Water going from 0 degrees Kelvin to 100 Kelvin. It also presents the formulas that can be used to perform the necessary calculations that can be asked, and the graph that describes of what happened.

The photo represents the how from the conservation of energy theorem (Heatgain=Heatlost) students were able to find the mass of water after all ice was melted. To begin, the mass of Ice was 255 g @ -5 degrees Celsius. And using the conservation of energy theorem showing on the picture, the mass of the water after melt the ice was just 14.5 g. That means that for each gram of water there will be 17.6 grams of ice.

This graph represents all the process from where the water and ice are at 0.30 degrees Celsius to 99.0 degrees Celsius, which was the boiling point of the water. The two abrupt increases in temperature showing on the graph was due to the contact between the immersion and the thermal sensor.

The picture shows the mass of the ice and the water + beaker. The starting temperature is also displayed as to be 0.30 degrees Celsius, and it took 130 s, or 2.17 m at 14 degrees Celsius to melt all the ice. The total amount of energy added until the ice melted was 37628.344 J. Students had to consider that there was a phase change, and the heat of fusion (Lf) had to be added to the formula Q=mici(Tf-Ti)+miLf. 

In this picture, the students calculated the latent heat of fusion (Lf) and the the specific heat (cw) of water. The result shown for the latent heat of fusion was not the experimental one, which is 336 J/g. And the specific heat was found to be 4.944 J/g-K. The water boiled for 86 s, and the mass of water that was turned into steam during that time was 17.5 g at 34.969 J. The Latent Heat of Vaporization (Lv) found was 1420.2 J/g has shown on the picture.

The values for the latent heat of fusion were very close to the real one, with just an .67% diference between them. Nevertheless, the specific heat and latent heat of vaporization were higher than the accepted one, where the cw was of by 18.2% and Lv by 37.05%. One possible source that can cause such discrepancy can be the simple fact the students had not considered the amount of heat that was lost, or the readings that the equipment does. Another source of error can be the room where the experiment is been held. Since the AC was on, he temperature inside the room was not at the 26 degrees Celsius, which affects the results. 

  

The last two pictures are the calculations of the uncertainty. For the Lf the uncertainty was +/- 0.085 J/g. For the Lv the uncertainty was +/- 0.503 J/g. Both Lf and Lv are calculated on the first picture. For the cw the uncertainty was +/- 0.1 J/g-k (which I believe that there was a mistake during the calculations we did. I just notice know).