Probl_6.12_sol - Thermodynamics of Materials HW 3 Problem...

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Unformatted text preview: Thermodynamics of Materials HW 3 Problem 6.12 a) Calculate the heat capacity based on the Debye model for aluminum at all temperatures between 0 K and 600 K spaced by 10 degrees intervals. Use a numerical integration procedure to calculate these data. Present the results in graphical and in tabulated form. b) Calculate the enthalpy change in Al when heated between 100 K and 300 K. c) For bonus points: determine the Einstein temperature for which the Einstein model provides the best match to the data calculated in (a) using the Debye model. Please provide printouts of any code that you wrote for solving these questions and explain the numerical procedures that you used. © John Kieffer, University of Michigan (2009) 1 Thermodynamics of Materials Answer a) We numerically integrate the expression ⎛T⎞ cV = 9nkB ⎜ ⎟ ⎝ θD ⎠ 3x D ∫ 0 x = θ D T , e.g., using the trapezoid rule. The code is posted on the Ctools site. The results are T (K) 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 230 240 250 260 270 280 290 300 cV (J/molK) 0.0340 0.2723 0.9156 2.1039 3.7907 5.7675 7.8093 9.7590 11.5345 13.1066 14.4770 15.6619 16.6829 17.5626 18.3217 18.9786 19.5490 20.0461 20.4812 20.8636 21.2011 21.5002 21.7664 22.0037 22.2163 22.4074 22.5799 22.7361 22.8779 23.0071 T (K) 310 320 330 340 350 360 370 380 390 400 410 420 430 440 450 460 470 480 490 500 510 520 530 540 550 560 570 580 590 600 cV (J/molK) 23.1251 23.2334 23.3330 23.4229 23.5076 23.5863 23.6569 23.7250 23.7856 23.8453 23.8977 23.9507 23.9964 24.0391 24.0840 24.1215 24.1627 24.1959 24.2271 24.2639 24.2917 24.3179 24.3427 24.3755 24.3977 24.4188 24.4388 24.4697 24.4878 24.5050 ( x4ex ex − 1 ) 2 dx , where 300 b) We numerically integrate U = 100 ∫c V dT , again using the trapezoid rule. cV data was obtained under (a), and serve as the numerical values of the function to be integrated. The temperature increment ∆T would be 10K. Code is included with the previous matlab routine. The result is ∆U = 3999.7 J/mol. c) Once we have the numerical values for cV according to the Debye model at 10 K temperature increments, we can compare these to the values that the Einstein function for cV © John Kieffer, University of Michigan (2009) 2 Thermodynamics of Materials yields for the same temperatures assuming a certain Einstein temperature. The best choice for θE is that which minimizes the difference ∑ (c 60 i =1 V , Debye (T ) − c i V ,Einstein (T )) i 2 2⎞ ⎛ ⎛ θE ⎞ eθE Ti ⎟ = min . = ∑ ⎜ cV ,Debye (Ti ) − 3nkB 2⎜ ⎟ i =1 ⎜ eθE Ti − 1 ⎝ Ti ⎠ ⎟ ⎝ ⎠ 60 2 ( ) One possible algorithm to find the optimal value for θE is to start with a value, say 1000 K that is safely larger than the expected optimal value. We calculate the error, i.e., the sum of the square of the differences between Debye and Einstein heat capacities. Then we subtract, say 100 K from the initial guess for θE and re-evaluate the error. If the new result is smaller than the old one we keep advancing with new guesses for θE in the same direction and the same increment. When the error increases, i.e., we passed the minimum, we reverse the course and use increments of half the size. We continue doing this until we drop below a specified temperature tolerance. This routine is also coded in to the posted matlab program. The Einstein temperature that gives the best agreement between the two heat capacity models is θE = 285.15 K. The agreement between the two models is shown in the figure below. Note that to achieve such agreement, the Debye and Einstein temperatures are about 100 K apart. Fig. 1 Best fit of the calculated Debye heat capacity as a function of temperature (circles) using the Einstein model (line), assuming a Debye temperature of θD = 385 K. © John Kieffer, University of Michigan (2009) 3 ...
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This note was uploaded on 10/26/2009 for the course MSE 330 taught by Professor Kiffer during the Fall '09 term at University of Michigan-Dearborn.

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