Thermal Expansion
After heating the rings, the atoms in the steel will expand and the hole will get bigger for the ball to go through.
Professor Mason is demonstrating the activity by heating the ring.
When heating a metal bar, the length will depend on change in temperature, material, time, and initial length. And we learned the formula α = (ΔL/Lo)/ ΔT. Where α is the coefficient of linear expansion.
When heating on the invar side of the bimetallic strip, the strip will curve toward invar because the α of brass is bigger than invar is.
Professor Mason showed us what actually will happen when heating on the invar side.
When heating on the brass side, the strip will still curve toward invar.
Professor Mason again showed us what will happen.
Professor showed us what would happen when the heated bimetallic strip is put in the ice box.
The result was the strip curved toward the brass side.
We are about to do the aluminum rod linear thermal expansion demo.
There is a wheel on the stand, which we will measure how much it rotates to measure how much the aluminum rod expands.
The aluminum rod is heated by the steam
This photo shows how the temperature heating on aluminum rod affect the angle rotated by the wheel.
The graph shows the minimum angel is 0.192 radians.
We used the data of the ΔL, Lo, r, ΔT and θ to find the α of the aluminum rod.
The α value we got was 19x10^-6 (/°C) and the true α value for aluminum is 24x10^-6 (/°C). This value is valid and within the uncertainty because there is a lot of uncertainties, which the diameter of the wheel was not measured as 1.5 centimeter and the initial length of the rod was measured as 1 meter. They all observed by eyes. My uncertainty of α value is 19x10^-6 ± 5x10^-6 (/°C).
Latent Heat (Heating Water)
Ice was put in the Styrofoam cup containing with heated water and we predicted and actually measure how the time vs. temperature graph would be.
This is the predicted time vs. temperature graph
This the actually graph and the slope is 0.3495 °C/s.
The heater had 293 Watts of power. We used this information and the formula Q = mcΔT to calculate the mass of water in the Styrofoam cup while heating.
We got the mass of water as 200 g, however, there might be a systematic error or random error. The graph which the ice phase in the beginning is not flat because some water was heated up by the heater in this situation.
Latent Heat Problems
We used the formula Q = mcΔT and latent heat formula Q = mL to calculate the final temperature of mixing ice and water which has phases change .
Another latent heat problem to calculate the mass of ice.
Activity: Pressure of a Column of Liquid
We put some water in the bottom of the manometer tube and blew into one end and observed the changes.
We used the formula ΔP = ρay to calculate the pressure we blew into the tube.
We measured the change of water in height was 0.045 m, thus we calculated the pressure as 441.45 Pa.
Conclusion
In today's class session, we learned about the thermal expansion and latent heat. We did activities such as heating the steel ring to letting the ball go through, heating the aluminum rod and calculating its expansion in length, heating water mixing with ice and to calculate the mass of water in the cup, doing some latent heat problems, and finally doing the activity to measure the pressure we blew into a tube containing water.
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