University:
University of Michigan
Course:
PHYSICS 250 | Fundamental Physics for the Life Sciences II
Academic year:

2022
Views:

227

Pages:

4
Author:

Trevon Bernard

Thermal expansion 1. Presentation and organization ∆l=αl o ΔT . a. Most materials expand when they heat up. linear expansion for the material. Length depends on l o ΔT . α is coefficient of 2. Introduction a. In this experiment, purpose is to measure α for copper, aluminum and steel. Three metal are isotropic which means that expansion is the same in all directions. Measure the temperatures. α would be independent of the absolute temperature. 3. Procedure Apparatues : expansion tube apparatus, steam generator, elevation block , digital multimeter, steel, Cu, and Al tubes, contriners for condensation. Procedures 1. Measure the initial length of the tube and change in length of the tube while heating/ cooling. Measure l o to within 0.5 mm. 2. Measure a change in length on the gauge. 3. Attach the thermistor lug to the threaded hole with using a thumbscrew 4. Plug the leads of multimeter into the banana plug in the center of the expansion. Record resistance of the thermistor 5. Use a block of wood to raise the apparatus. 6. Turn the outer casing of the dial gauge to align the zero point on the scale 7. Record the change in length for the tube. Connect the steam generator the copper tube. 8. Turn on the steam generator. When the thermistor resistance stabilizes, record the resistance. Repeat the experiment with copper, steel and aluminum tubes. ℃ Data L(mm) 0.8 Aluminu m steel Copper R L(mm) ℃ 10.7 82 0.75 11.74 80 0.7 13.03 78 0.65 14.68 74 0.6 16.54 70 0.5 18.85 68 0.5 21.1 65 0.45 25 61 0.4 29.1 57 0.35 34 53 0.3 40.2 49 0.25 47.2 45 0.2 56.1 42 0.15 67.3 38 0.1 84.2 33 0.05 84.4 29 R ℃ L(mm) R ℃ 60 9.59 84 119 9.4 84 55 10.9 82 110 9.86 83 50 11 80 105 9.5 82 45 12.5 76 100 10.08 80 40 14 73 95 10.67 78 35 16.5 68 90 11.31 76 30 19 64 85 12.12 74 25 23 59 80 13.06 72 20 28.5 54 75 14.05 70 15 35.9 48 70 13.18 68 65 16.68 65 60 18.296 62 55 20.3 60 50 23.2 58 45 24.7 55 40 27.5 52 Copper 0.9 0.8 0.7 f(x) = 0.01 x − 0.37 L 0.6 0.5 0.4 0.3 0.2 0.1 0 20 30 40 50 60 70 80 90 Temperatures steel 0.7 0.6 0.5 f(x) = 0.01 x − 0.46 L 0.4 0.3 0.2 0.1 0 45 50 55 60 65 70 75 80 85 90 Temperatures Aluminum 1.4 1.2 1 f(x) = 0.02 x − 0.83 L 0.8 0.6 0.4 0.2 0 50 55 60 65 70 Temperatures 75 80 85 90 Formula is ∆l=αl o ΔT . Y = Kx∗bex In this experiment, constant value of lo For copper, k was 0.0138 K = αl o so For steel K was 1.2139 so α= = | α= 0.0138 710❑ α= = α= 12∗10 experimental −theoretical Theoredicalvalue = | −3 | −6 = | −5 −5 ( K−1 ) * 100 = 40.84% α= 24∗10−6 = * 100 = 14% ( K−1 ) 1.7∗10 −12∗10 12∗10−6 0.0229 For aluminum K was = 0.0229 so α = 710 ❑ experimental −theoretical Theoredicalvalue | 1.69∗10 −6 ( K−1 ) 1.94∗10−5 K−1 So) 1.94∗10−5−17∗10−6 −6 17∗10 0.012 710❑ K lo is 710mm 17∗10−6 ( experimental −theoretical Theoredicalvalue α= K = αlo −6 3.22∗10 −24∗10 24∗10−6 −5 3.22∗10 ( K−1 ) ( K−1 ) | * 100 = 34.39% Part 2 question 3. To improve experiment accuracy, make the material dry before experiment started. If material is not dry, it might affect experiment data so that data would not be accurate. 4. B = 3 α 5. Bimetrallic strip contracts or expands the pointer which moves up and down.