26th of February's Class
Thermal Expansion
We began the class with a theoretical guess about thermal expansion of a ring rod and a ball rod. The circumference of the ring is less than the ball, therefore in the beginning the ball rod could not pass through the ring rod. The question was,"If the ring rod was heated, would the ball rod be able to pass through the ring rod?"
The image above is the prediction that our group made. When the experiment was carried out, we found out that the ball rod could pass through the ring rod after the ring rod was heated.
The image above is an experiment that was done on a metal rod. It's right end is clamped tight, so that when the thermal expansion occurs, it would only occur in one direction. It's right end is heated up with steam, while on its left, a rotary meter that measures distance by calculating its arc length.
The image above shows the formula for calculating the alpha which formula is alpha = Change in Length / (Initial Length * Change in Temperature).
Since we are using the rotary meter, which uses angular attributes, we need to convert them into their linear counterparts.
Arc length = theta * radius, hence we can obtain the length that the rod expanded by using this formula.
Calculation Steps:
-The theta's reading was 0.192 rad, and the professor eyeballed the diameter of the rotary meter to be about 1.5 cm.
-The arc length is therefore (0.192 * 0.0075) = 1.44*10^-3 m
-The initial length of the rod is 1.00 m
-The change in temperature is 100 - 23.3 = 76.7 degree Celsius
-Putting in all the values into the formula for alpha, we obtained the result, approximated to 19 * 10^-6
Thermal Expansion 2
The professor demonstrated to us that each material has different rate of thermal expansion by heating up a bimetallic strip. It is a strip which sides are made of 2 different metals, one of them is Brass, the other is Invar. He heated up its sides with a blow torch, and it bent to the side that has a lower rate of thermal expansion, in this case, Invar.
After heating it up, the professor then cooled it to room temperature and straightened it out. He then place it into a container full of ice and water, which temperature we assume to be close to 0 degree Celsius. The bimetallic strip bent to the Brass side.
This is constant with our prediction as a material that has a high rate of expansion would also possess a high rate of contraction. Since the Brass has a higher rate of expansion, it would contract much faster than Invar, causing the bimetallic strip to bend to its Brass side.
In the image above, the item that is circled red is the bimetallic strip.
Specific Heat and Latent Heat
Professor Mason conducted another experiment by putting in a stick, that heats up when plugged into the wall, into a cup filled with ice and water. The temperature is assumed to be 0 degree Celsius initially. Below is the graph of Temperature vs. Time obtained from it.
Both the beginning and end of the graph is not linear, as we can see that there are curves there. Thus we can conclude that the temperature change at those two points are not constant like its middle section.
The professor then explained that this is caused by the latent heat involved when changing water from its different states. At the beginning, there were some ice, therefore some of the heat energy is used for the conversion of water's solid state to liquid state. This latent heat is termed latent heat for fusion.
Towards the end, when water reaches its boiling point, there will be some water molecules that are converted from its liquid state to its gaseous state. This latent heat is termed latent heat for vaporization.
The image above shows how the formula for latent heat can be derived; it is shown on the left side of the whiteboard.
We calculated the Latent Heat of Fusion for water to be 219 J/g. Its true value should have been 334 J/g. Thus our error value is 334 - 219 = 115 J/g.
Complex Latent Heat and Specific Heat Question
Professor Mason gave us an example to work on, the question goes:
There is an ice cube with a mass of 255g. Its temperature is -12 degree Celsius.
Water at a temperature of 22 degree Celsius is poured onto it.
The system comes into an equilibrium when the final temperature is 0 degree Celsius and all the water poured into the ice cube freezes.
How much water was poured into the system?
Calculation Steps:
-We put all the energy equation that is involved the system and equate them to zero by the Energy Conservation Theorem
-We identified 3 different energy involved in the system. First, energy that would be gained by the ice to raise its temperature from -12 to 0 degree Celsius. Second, energy lost needed for water to drop its temperature from 22 to 0 degree Celsius. Third, the energy needed to convert water from its liquid from to its solid form using latent heat of fusion.
-The first and second energy mentioned uses Q = mC delta T. The third energy uses the formula Q = mL. m is mass, C is specific heat, delta T is the change in temperature, L is the latent heat of fusion.
-We calculated each energy, and singled out the mass of water poured onto the ice, and we got 14.6g.
Pressure
We learnt about pressure through an experiment done with manometer. We fill it with water, and then blew air into it. The higher the rate at which we blew air into it, the further the distance that the water travels in the tube.
The video above shows how pressure would affect the distance traveled by the water. And the video below would be the experiment we did to calculate the amount of pressure we apply into the manometer system.
First we mark the initial level of the water at the other end of the tube (the end that is not being blown). Then, a person would attempt to blow air at a constant rate long enough so that the final level of water could be marked.
Pressure's formula is defined by Force divided by the Area that the force is acting on.
Force's formula can be defined by Mass times Acceleration.
Mass can be defined by the Density of a material multiplied by its Volume.
We take the Acceleration to be Gravity's downward acceleration on objects at sea level.
Therefore, Pressure's formula can be written as Density x Volume x Gravity / Area.
Volume can be defined as Area x Height.
This leaves us with Pressure = Density x Height x Gravity.
Calculating all the relevant parameters give us a pressure of 0.2 x 10^-3 g / (ms^2)
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