Lab 1 Newton’s Law of Cooling and Thermal Expansion
The Purpose
The purpose of this experiment is to demonstrate the physical effect of thermal expansion and to determine the linear expansion coefficient (alpha). Also, we want to explore Newton’s law of cooling and see if this cooling object obeys Newton's law of cooling.
The Experiment
This experiment is a fictional experiment where we were told that a metal rod was heated to near 100C and allowed to cool. The length of the rood and the temperature were recorded as functions of time. We were told that the temperature measurement has an error of +- 5% and that the length measurement has an error of +-.05mm
The Theory
Newton's Law of Cooling:
Newton's Law of cooling says that the rate of cooling is proportional to the temperature difference between the rod and the room.
dT/dt = − k(T − TR ) where Tr = 20C is the room temperature and k is a constant that depends on the specifics of the rod.
From this, we expect the temperature of the rod to be given by the following formula: T = (T0 − TR )e(−kt) + TR
Thermal Expansion:
The length of an object expands when heated. The length is given by L = L0 (1 + alpha(T − TR )
From the data we will determine the linear expansion coefficient, alpha, and demonstrate that the heated rod obeys Newton's Law of Cooling.
The Data
We were given the following data:
|
Time (s)
|
Temperature(C )
|
Length (mm)
|
|
16.31
|
85.137
|
701.322
|
|
40.22
|
76.059
|
701.177
|
|
66.98
|
59.288
|
700.747
|
|
104.42
|
48.053
|
700.513
|
|
124.01
|
48.534
|
700.63
|
|
169.17
|
36.378
|
700.293
|
|
188.86
|
32.539
|
700.181
|
|
230.33
|
32.103
|
700.294
|
|
253.11
|
25.053
|
699.974
|
|
280.09
|
27.047
|
700.146
|
The Analysis
For the analysis we need to calculate Delta T and Delta L from this data. These values are given in the following table:
|
Time (s)
|
Delta T(C)
|
Delta L (mm)
|
|
16.31
|
65.137
|
1.322
|
|
40.22
|
56.059
|
1.177
|
|
66.98
|
39.288
|
0.747
|
|
104.42
|
28.053
|
0.513
|
|
124.01
|
28.534
|
0.63
|
|
169.17
|
16.378
|
0.293
|
|
188.86
|
12.539
|
0.181
|
|
230.33
|
12.103
|
0.294
|
|
253.11
|
5.053
|
-0.026
|
|
280.09
|
7.047
|
0.146
|
Then from this data we constructed a graph of temperature vs time:
The data on this graph were fit with an exponential function. The R^2 value of the fit is near 1 which indicates the data is well described by this function. The 5% error bars were added to the graph.
Next , the graph of Delta L vs Delta T was constructed
The error bars are 0.05 mm. The data was fit with a linear function. The R^2 value is near 1 which indicates that the data is well described by this function.
From the fit the slope of the graph is
Delta L/(Delta T = 0.0213 mm/C
From the linear expansion equation
Delta L/(Delta T) = alpha *L0 = 0.0213 mm/C
Thus the linear expansion coefficient is
alpha = 0.0213mm/C/700mm = 3.0429x10−5C−1
The Conclusion
From this experiment we were able to demonstrate that the cooling rod satisfied Newton’s law of cooling. Also, the experiment demonstrated that the rod followed the relationship of linear expansion and the expansion coefficient was determined to be
alpha = 3.0429x10−5C−1