3.012 Problem Set 2 Solutions

3.012 PS 2 Issued: 09.20.05 Due: 09.30.05 3.012 Fall 05 THERMODYNAMICS 1. A sheet of manganese (2 moles) at room temperature (298 K) is placed in thermal contact with a heat supply that slowly transfers 100,698 J of heat into the sample at constant pressure. Use the following thermodynamic data for Mn to answer the questions below: Mn has four solid phases, α, β, γ, and δ: C p" = 21.6 + 15.9 #10$3 T C p" = 34.9 + 2.8 #10$3 T ! C p" = 44.8 ! C p" = 47.3 ! ! J K % mole J K % mole! J K # mole ! J K # mole ! " #$ Ttrans = 993K # $% "H trans = 2,010 J mole " #$ Ttrans = 1373K ! # $% "H trans = 2,300 J mole " #$ Ttrans = 1409K # $% "H trans = 1,800 ! J mole ! a. Calculate the final temperature of the sample. b. Calculate the total enthalpy change for this process. c. Calculate the total entropy change for this process. d. What phase (or phases) are present at equilibrium at the end of this process? a. Calculation of the final temperature is made by using the heat capacity data to determine how much heat is required to heat up and through the several phase transitions present. The calculation is readily carried out by simply determining how much heat is needed for heating to each phase transition: Starting out at 298 K, we first heat the sample (α phase at room temperature) to the first phase transition α -> β, at 993K: 993 q298K "993K = # nC p dT = 298 3.012 PS 2 ! 993 # (2 moles)(21.6 + 15.9 $10 298 1 of 6 %3 T J )dT = 44,290J mole & K 10/4/05 …clearly, we have more heat to use, so we continue. The phase transition consumes a small amount of heat: q298K "993K = n#H$ "% = (2)(2,010) = 4,020J We continue summing up the heat for the next few transitions in a similar manner: 1373 ! q993K "1373K = 1373 $ nC # p dT = 993 J $ (2 moles)(34.9 + 0.0028T mole % K )dT = 29,041J 993 q" #$ = n%H " #$ = (2)(2,300) = 4,600J Our running total for heat transferred to reach the gamma phase ast 1373K is now 81,951 J. This ! 18,747 J of heat yet to be consumed. Continuing on to the next transition: leaves ! 1409 1409 $ nC p# dT = q1373K "2300K = 1373 J $ (2 moles)(44.8 mole % K )dT = 3,226J 1373 q" #$ = n%H" #$ = (2)(1,800) = 3,600J We have 11,701 J of heat yet to use up. We determine the final temperature by integrating to use up ! heat in the delta phase: the ‘left over’ ! Tf q1409K "T f = $ nC 1409 Tf # p dT = J $ (2 moles)(47.3 mole % K )dT = 11,701J 1409 94.6Tf &133,291J = 11,701J 'Tf = 1535K b. Because this process occurs at constant pressure, we immediately know: ! " dq % " (H % C p = $ rev ' = $ ' # dT & p # (T & p dqrev = dH qrev = )H *)H = 100,698J c. We determine the total entropy change in a manner similar to the integrating procedure used in part (a): ! First, the appropriate relation to get us started is: 3.012 PS 2 2 of 6 10/4/05 " dq % " (S % C p = $ rev ' = T$ ' # dT & p # (T & p Cp dT T Tf C )S = * p dT Ti T dS = Note that the last equality and integral hold only for regions where no phase transition occurs. Using this expression, we integrate over the temperature change that occurs when the 100,698 J of heat are absorbed by the system: "S = # p ! n"H# %& 1373 nC nC p& n"H & %' 1409 nC p' n"H' %( 1533 nC p( dT + + dT + + dT + + $ dT $ $ $ 993K 1373K 1373 T 1409K 1409 T 298 T 993 T 993 We account for the entropy of transition for each phase transformation crossed in the process with the terms containing the enthalpy and temperatures of transition. Plugging in and carrying out ! integrations: the "S = 119.1 J K d. Our calculation in part (a) tells us that the delta phase is the final equilibrium structure of the system. ! 2. Twenty kg of liquid bismuth at 600 K is introduced into a 10 kg alumina (Al2O3) crucible (initial temperature 298K), filling the crucible to the top; the crucible and bismuth are then surrounded by adiabatic walls (illustrated below). At equilibrium, according to the zeroth law, the temperatures of the bismuth and alumina crucible must be equal. Use the following thermodynamic data to answer the questions below: C palu min a,solid = 106.6 + 0.0178T C pBi,solid = 18.8 + 0.023T ! J K " mole C pBi,liquid = 20 + 0.00615T ! J K " mole J K " mole ! 3.012 PS 2 Tmalu min a = 2327K ! ! TmBi = 544K "H mBi = 10,900 ! 3 of 6 10/4/05 J mole a. What is the final temperature of the system? b. How much heat is transferred to the alumina? c. At the final equilibrium, is the bismuth liquid or solid? d. Consider now the same process, except that the adiabatic walls are removed. The closed system (can exchange heat…) of the bismuth in the alumina container is placed in a large room at T = 298 K, which behaves as a heat reservoir- it can transfer heat out of the system (or into it, depending on the temperatures of the reservoir and the system). At equilibrium, the bismuth and alumina will reach a final temperature of 298 K to match the environment of the room. How much heat will leave the bismuth to reach equilibrium? a. Because the crucible and liquid bismuth are enclosed by adiabatic walls, they can only exchange heat with one another. In other words, whatever heat leaves the bismuth must enter the alumina. This can be stated mathematically: qBi = "qAl2O3 The minus sign tells us that the direction of heat flow in the alumina is opposite that of the bismuth (e.g., heat flowing out of the bismuth is flowing into the alumina.) This equality immediately tells us how to determine the final temperature of the system- since it must be the same in the bismuth and the alumina: ! qBi = "qAl2O3 Tf 544 #n 600 Bi C Bi,l p dT " n Bi$H m + #n 544 Bi C Bi,s p %T f ( dT = "' # n Al2O3 C pAl2O3 ,sdT* '&298 *) A few comments: In writing the left-hand expression, we are making the assumption that the Bi will cool enough to pass through the melting point and solidify. Note that because we are passing from high to low ! temperature, we have a minus sign in front of the enthalpy of melting- we must remove heat from the sample to freeze it. To proceed, we need to calculate the number of moles of alumina and bismuth we have: 3.012 PS 2 4 of 6 10/4/05 " mole % Bi : (20,000g)$ ' = 95.7 moles # 208.9g & " mole % Al2O3 : (10,000g)$ ' = 98.08 moles # 101.96g & Plugging in the given data for alumina and bismuth, we can calculate the final temperature of the system: ! 1.1Tf2 + 1799Tf " 2,473,602 = ".8729Tf2 + 3,193,206 "10,455Tf The expression reduces to a quadratic equation, which we can solve for Tf: ! Tf = "12,254 ± (12,254) 2 " 4(1.97)("5,666,808) 2(1.93) = 433K b. The heat transferred to the alumina is just one side of the heat equality expression above: ! 1.1Tf2 + 1799Tf " 2,473,602 = "qAl2O3 qAl2O3 = 1,488,397J c. The final temperature falls well below the melting point of Bi, so the sample has solidified. ! the heat transfer that occurs to move Bi from liquid at 600K to solid Bi at d. We simply calculate 298K: 544 qBi = "n 298 Bi C Bi,l p dT # n Bi$H m + 600 "n Bi C pBi,sdT = #1,839kJ 544 3. Magnetic ! resonance imaging (MRI) is a common medical technique used for diagnostic imaging of tissues in patients. MRI is based on measuring the response of the weak magnetic dipoles in the atoms of tissues under a strong applied magnetic field. Typically, the magnetic induction in a clinical MRI machine may be ~2 Tesla. Consider the materials used to fabricate the MRI chamber that will be placed in the magnetic field, such as aluminum. a. What magnetic field strength is required to achieve a 2T induction in the MRI housing if the housing is fabricated from aluminum? b. Will the magnetization induced in the aluminum be significant compared to, say, an iron permanent magnet that has a maximal net magnetization of 1.39x105 A/m? Support your answer with a calculation. c. What is the work performed by the magnetic field in the volume of a section of the MRI aluminum housing 1 m x 1 m and 5 mm thick? 3.012 PS 2 5 of 6 10/4/05 a. We use the expression relating field strength H to induction B: B = 2T = µo H + µo M = µo (1+ " )H From the class notes, we have the susceptibility for Al: " = 2.07 #10$5 !B = 2T = 2 kg = ') 4 & #10$7 kg % m *, 1+ 2.07 #10$5 H ( ) s% C ( C2 + C A - H = 1.59 #10 6 = 1.59 #10 6 m% s m b. The magnetization is calculated directly: ! M = "H = 32.9 A m …this is only 0.02% of the magnetization of the example iron permanent magnet! c. The work is: ! 1.59"10 6 w= # VµHdH = Vµ (1+ $ ) o 0 H2 2 1.59"10 6 0 = 7,942J ! 3.012 PS 2 6 of 6 10/4/05 0.05 0.04 0.03 0.02 0.01 0 0 5 10 15 20