Help me do the Lab Thermal Expansion

Science

Academy College

Question Description

Help me do Lab 3: Thermal Expansion.

4 Data Analysis 1. Derive a symbolic expression for the uncertainty in the expansion coefficient (δα) using the partial derivative method (Equation 2). • Your expression should be in terms of the following quantities: L, ∆L, ∆T, δL, δ(∆T), δ(∆L). 2. To determine the temperature values that correspond with the measured resistance values, assume the relationship between between R and T is linear (this is a good assumption only over small temperature ranges), and interpolate between the adjacent data points. • The uncertainty stated by Pasco for the thermistor values is ±0.2 ◦C. • Assume 3 sigfigs for given temperature values. For example, assume the temperature value listed as 25◦C is actually 25.0 ◦C 3. Calculate ∆T ± δ(∆T). Recall that the uncertainty in a sum or difference is given by Equation 3 in the introduction. 4. Use Equation 1 to find α. 5. Use the expression derived in step 1 of the data analysis instructions to find δα. 6. Compare with the theoretical α value.

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Lab 3: Thermal Expansion Phys. 4C 1 Objective • To find the coefficient of thermal expansion for copper and aluminum from given data, and perform an error analysis to understand the precision of the experiment. 2 Introduction Most materials expand when heated through a temperature range that does not produce a phase change. The added heat increases the average amplitude of vibration of the atoms in the material which increases the average separation between the atoms. If an object of length L at some initial temperature Ti is heated to temperature Tf , the increase in length ∆L = Lf − Li is characteristic of the composition of the object. The coefficient of linear expansion α is given by: ∆L (1) L∆T In this lab we will use data gathered by students in a previous term to calculate α and do an error analysis to investigate the precision of the lab and see whether the data agrees with the accepted values of α for copper and aluminum: α= αCu = 17.6 × 10−6 (◦ C)−1 αAl = 23.4 × 10−6 (◦ C)−1 Recall the uncertainty (δf ) of a function of x, y, and z is given by s   2  2 2 ∂f ∂f ∂f 2 2 2 δf = (δx) + (δy) + (δz) ∂x ∂y ∂z (2) Where δx, δy, and δz (the uncertainties in x, y, and z, respectively) are independent and random. The uncertainty in a sum or product f = x ± y is determined by adding the uncertainties of the inputs in quadrature: q 2 2 δf = (δx) + (δy) (3) 3 Experimental Procedure This is an explanation of the experimental procedure followed in previous terms. You will not actually perform these steps. However, you need to understand what was done so that you can properly analyze the data and draw conclusions. The basic idea of the experiment is to heat a metal tube by running steam into it, indirectly measure the change in its temperature, and also measure the change in its length. 1. The length of the copper tube at room temperature, L, was measured with a meterstick. Care was taken to measure just the length of the part of the rod whose expansion would be measured. The L was taken to be the distance from the pin in one end of the rod, which was held fixed in place by the Steam generator Ohmmeter Metal tube Dial gauge Insulation Bracket 30 0 15 60 50 40 90 70 80 0 10 30 20 Thermistor Expansion base Pin on tube held fixed in slot Figure 1: Schematic of the apparatus for the experiment. © 1990 PASCO scientific $2.00 apparatus, to the bracket attached at the far end of the rod. The end of the rod with the bracket could move horizontally to allow for the expansion, which was measured with a dial gauge. 2. Water was added to the steam generator and heated to a boil. 3. The copper rod was fit into in expansion base as shown in Figure 1. The stainless steel pin on the tube fit into the slot that fixed it in place, and the bracket at the other end fit against the spring arm of the dial gauge. 4. A device called a thermistor, which is a resister whose resistance depends on temperature more sensitively than a normal resistor, was attached to the side of the tube, making good contact with it and a foam insulator is put over it. 5. An ohmmeter was used to measure the resistance of the thermistor at room temperature, Ri . 6. The vent of the steam generator was connected to the inside of the tube using black flexible rubber tubing. The end of the expansion base that held the tube where it was connected to the steam generator was propped up so that condensation inside the tube ran down and out the other end, where it was collected in a small cup. 7. On the dial gauge, each increment is equivalent to 0.01 mm of tube expansion. 8. The value on the ohmmeter, R, and the value on the dial gauge were observed as the tube warmed. When the thermistor resistance (ohmmeter reading) stabilized, Rhot and the reading on the dial gauge, ∆L, were recorded. 9. This procedure was repeated for aluminum. 4 Data Analysis 1. Derive a symbolic expression for the uncertainty in the expansion coefficient (δα) using the partial derivative method (Equation 2). • Your expression should be in terms of the following quantities: L, ∆L, ∆T, δL, δ(∆T ), δ(∆L). 012-04394C Thermal Expansion Apparatus ➁ Calculate ΔT = Thot – Trm. Record the result in the table. ➂ Using the equation ΔL = αLvalues ΔT, calculate a for copper, with steel, and 2. To determine the temperature that correspond the aluminum. measured resistance values, assume the relationship between between R and T is linear (this is a good assumption only over small temperature αCu = and__________________ ranges), interpolate between the adjacent data points. = α __________________ steel • The uncertainty stated by Pasco for the thermistor values is ±0.2◦ C. α = __________________ Al • Assume 3 sigfigs for given temperature values. For example, assume the temperature value listed as 25◦ C is actually 25.0◦ C Questions ➀ Look up the accepted the linear expansion coefficient foror copper, steel, and 3. Calculate ∆T ± δ(∆T ). values Recallforthat the uncertainty in a sum difference is given by Equation 3 in aluminum. Compare these values with your experimental values. What is the percentage the introduction. difference in each case? Is your experimental error consistently high or low? 4. Use ➁ Equation 1 tooffind On the basis yourα. answers in question 1, speculate on the possible sources of error in your experiment. How might you improve the accuracy of the experiment? 5. Use the expression derived in step 1 of the data analysis instructions to find δα. ➂ From your result, can you calculate the coefficients of volume expansion for copper, aluminum, andtheoretical steel? (i.e. ΔV = αvolV ΔT) 6. Compare with the α value. THERMISTOR CONVERSION TABLE: Temperature versus Resistance Res. (Ω) 351,020 332,640 315,320 298,990 283,600 269,080 255,380 242,460 230,260 218,730 207,850 197,560 187,840 178,650 169,950 161,730 153,950 146,580 139,610 133,000 126,740 120,810 115,190 109,850 104,800 100,000 Temp. (°C) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 Res. (Ω) 95,447 91,126 87,022 83,124 79,422 75,903 72,560 69,380 66,356 63,480 60,743 58,138 55,658 53,297 51,048 48,905 46,863 44,917 43,062 41,292 39,605 37,995 36,458 34,991 33,591 32,253 Temp. (°C) 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 5 Figure 2 Res. (Ω) Temp. (°C) 30,976 29,756 28,590 27,475 26,409 25,390 24,415 23,483 22,590 21,736 20,919 20,136 19,386 18,668 17,980 17,321 16,689 16,083 15,502 14,945 14,410 13,897 13,405 12,932 12,479 12,043 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 Res. (Ω) 11,625 11,223 10,837 10,467 10,110 9,767.2 9,437.7 9,120.8 8,816.0 8,522.7 8,240.6 7,969.1 7,707.7 7,456.2 7,214.0 6,980.6 6,755.9 6,539.4 6,330.8 6,129.8 5,936.1 5,749.3 5,569.3 Temp. (°C) 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 5 Tables Copy and complete the following tables. Write the symbolic δα expression on the same page as the tables, they will be uploaded together in question 1 on the Canvas quiz. 5.1 5.2 5.3 Data L (mm) ∆L (mm) Ri (kΩ) Rhot (kΩ) Copper 705.0 0.830 96.3 7.04 Aluminum 701.0 1.165 107.0 7.44 Ti (◦ C) Thot (◦ C) ∆T (◦ C) Uncertainties δL (mm) δ∆L (mm) 0.5 0.005 δRi (kΩ) δRhot (kΩ) δTi (◦ C) δThot (◦ C) 0.2 0.2 Results Accepted α, 10−6 (◦ C)−1 Copper 17.6 Aluminum 23.4 α ± δα, 10−6 (◦ C)−1 δ∆T (◦ C) Thermal Expansion Lab Experimental Setup April 25, 2020 1/5 Thermistor Connection April 25, 2020 2/5 Dial Gauge April 25, 2020 3/5 ➂ From your result, can you calculate the coefficients of volume expansion for copper, aluminum, and steel? (i.e. ΔV = αvolV ΔT) THERMISTOR CONVERSION TABLE: Temperature versus Resistance Res. (Ω) 351,020 332,640 315,320 298,990 283,600 269,080 255,380 242,460 230,260 218,730 207,850 197,560 187,840 178,650 169,950 161,730 153,950 146,580 139,610 133,000 126,740 120,810 115,190 109,850 104,800 100,000 Temp. (°C) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 Res. (Ω) 95,447 91,126 87,022 83,124 79,422 75,903 72,560 69,380 66,356 63,480 60,743 58,138 55,658 53,297 51,048 48,905 46,863 44,917 43,062 41,292 39,605 37,995 36,458 34,991 33,591 32,253 Temp. (°C) 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 Res. (Ω) Temp. (°C) 30,976 29,756 28,590 27,475 26,409 25,390 24,415 23,483 22,590 21,736 20,919 20,136 19,386 18,668 17,980 17,321 16,689 16,083 15,502 14,945 14,410 13,897 13,405 12,932 12,479 12,043 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 Res. (Ω) 11,625 11,223 10,837 10,467 10,110 9,767.2 9,437.7 9,120.8 8,816.0 8,522.7 8,240.6 7,969.1 7,707.7 7,456.2 7,214.0 6,980.6 6,755.9 6,539.4 6,330.8 6,129.8 5,936.1 5,749.3 5,569.3 Temp. (°C) 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 5 April 25, 2020 4/5 Temperature vs. Resistance Thermistor Data April 25, 2020 5/5 ...
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