Absolute Zero Results
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Transcript of Absolute Zero Results
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Results
After conducting the investigation in a fashion described in the experimental methods section of the reports a certain amount of data could be obtained. The most useful of all the data to be exported from the DataStudio interface was the pressure and the temperature of each trial. As stated before, a trail consists of submerging the testing apparatus into a certain amount of water that was being heat up by a hot plate. The apparatus would compute the data regarding the change in pressure in the sphere as well as the change in temperature. Data points were to be recorded once every 60 seconds. The trail ended once the temperature reached 50oC from its initial temperature.
After successfully collecting the pressure and temperature data points a scatter chart could be made in Microsoft Excel to produce a Temperature vs Pressure Curve. Since the purpose of the lab was to find the absolute zero of matter, the y intercept of the curves provides the intended value. The following Temperature/Pressure curves represent the results of the data collected.
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f(x) = 2.76776633279691 x − 257.655377235065R² = 0.999772299818476
Temperature vs Pressure Trial 1
Pressure (kpa))
Tem
erat
ure(
C))
Temperature vs Pressure Trial 1
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f(x) = 2.74459844139723 x − 254.723235155575R² = 0.999666858284849
Temperature vs Pressure Trial 2
pressure(kpa)
Tem
pera
ture
(C)
Pressure vs Temperature Trial 2
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f(x) = 2.68837597707645 x − 248.184061142069R² = 0.998773277206105
Temperatur vs Pressure Trial 3
pressure (kpa)
Tem
pera
ture
(C)
Pressure vs Temperature Trial 3
From the Temperature/Pressure curves created the following absolute temperatures can be recorded in to a table since as stated before the y intercepts reflect said values.
Trial (Iteration) Absolute Zero oC(Y Intercept)1 -257.662 -254.723 -248.18
Table I. Absolute Zero Results
In order to standardize the results, an average of the absolute temperatures can be taken and applied as follows
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AbsolouteTemperatureavg=Trial1+Trial2+Trial3
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AbsolouteTemperatureavg=(−257.66−254.72−248.18)C
3=−253.52C