3.012 PS Issued: Fall 05 Due:

Transcription

1 3.012 PS Issued: Fall 05 Due: THERMODYNAMICS 1. A sheet of manganese (2 moles) at room temerature (298 K) is laced in thermal contact with a heat suly that slowly transfers 100,698 of heat into the samle at constant ressure. Use the following thermodynamic data for Mn to answer the questions below: Mn has four solid hases, α, β, γ, and δ: C = #10 $3 T C = #10 $3 T C = 44.8 K # mole K % mole K % mole T #$ # trans = K H $% trans = 2,010 mole T #$ # trans =1373K H $% trans = 2,300 mole T #$ # trans =K H $% trans =1,800 mole C = 47.3 K # mole a. Calculate the final temerature of the samle. b. Calculate the total enthaly change for this rocess. c. Calculate the total entroy change for this rocess. d. What hase (or hases) are resent at equilibrium at the end of this rocess? a. Calculation of the final temerature is made by using the heat caacity data to determine how much heat is required to heat u and through the several hase transitions resent. The calculation is readily carried out by simly determining how much heat is needed for heating to each hase transition: Starting out at 298 K, we first heat the samle (α hase at room temerature) to the first hase transition α -> β, at K: q 298K K = # nc dt = # (2 moles)( $10 %3 T mole & K )dt = 44, PS 2 1 of 6 10/4/05

2 clearly, we have more heat to use, so we continue. The hase transition consumes a small amount of heat: q 298K K = n#h $ % = (2)(2,010) = 4,020 We continue summing u the heat for the next few transitions in a similar manner: q K 1373K = 1373 $ nc # dt = $ (2 moles)( T mole % K )dt = 29,041 q #$ = n%h #$ = (2)(2,300) = 4, Our running total for heat transferred to reach the gamma hase ast 1373K is now 81,951. This leaves 18,747 of heat yet to be consumed. Continuing on to the next transition: q 1373K 2300K = $ nc # dt = $ (2 moles)(44.8 mole % K )dt = 3, q #$ = n%h #$ = (2)(1,800) = 3, We have 11,701 of heat yet to use u. We determine the final temerature by integrating to use u the left over heat in the delta hase: T f q K T f = $ nc # dt = $ (2 moles)(47.3 mole % K )dt =11,701 T f 94.6T f &133,291 =11,701 'T f =1535K b. Because this rocess occurs at constant ressure, we immediately know: C = d q % rev $ ' # dt & d q rev = dh q rev = )H *)H =100,698 = (H % $ ' # (T & c. We determine the total entroy change in a manner similar to the integrating rocedure used in art (a): First, the aroriate relation to get us started is: PS 2 2 of 6 10/4/05

3 C = d q % rev $ ' # dt & ds = C T dt )S = T f * T i C T dt = T (S % $ ' #(T & Note that the last equality and integral hold only for regions where no hase transition occurs. Using this exression, we integrate over the temerature change that occurs when the 100,698 of heat are absorbed by the system: S = # nc $ T dt + nh 1373 & # %& K + nc $ T dt + nh & %' 1373K ' nc T dt + nh ' %( $ + K 1373 ( nc T dt We account for the entroy of transition for each hase transformation crossed in the rocess with the terms containing the enthaly and temeratures of transition. Plugging in and carrying out the integrations: S =119.1 K 1533 $ d. Our calculation in art (a) tells us that the delta hase is the final equilibrium structure of the system. 2. Twenty kg of liquid bismuth at 600 K is introduced into a 10 kg alumina (Al 2 O 3 ) crucible (initial temerature 298K), filling the crucible to the to; the crucible and bismuth are then surrounded by adiabatic walls (illustrated below). At equilibrium, according to the zeroth law, the temeratures of the bismuth and alumina crucible must be equal. Use the following thermodynamic data to answer the questions below: C alu min a,solid = T C Bi,solid = T C Bi,liquid = T K mole K mole K mole T m alu min a = 2327K T m Bi = 544K H m Bi =10,900 mole PS 2 3 of 6 10/4/05

4 a. What is the final temerature 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 rocess, excet that the adiabatic walls are removed. The closed system (can exchange heat ) of the bismuth in the alumina container is laced 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, deending on the temeratures of the reservoir and the system). At equilibrium, the bismuth and alumina will reach a final temerature 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: q Bi = q Al2 O 3 The minus sign tells us that the direction of heat flow in the alumina is oosite 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 temerature of the system- since it must be the same in the bismuth and the alumina: q Bi = q Al2 O T f % T f ( # n Bi C Bi,l dt n Bi $H m + # n Bi C Bi,s Al dt = n Al2 O 3 C 2 O ' #,s 3 dt* &' 298 )* A few comments: In writing the left-hand exression, we are making the assumtion that the Bi will cool enough to ass through the melting oint and solidify. Note that because we are assing from high to low temerature, we have a minus sign in front of the enthaly of melting- we must remove heat from the samle to freeze it. To roceed, we need to calculate the number of moles of alumina and bismuth we have: PS 2 4 of 6 10/4/05

5 mole % Bi : ( 20,000g) $ ' = 95.7 moles # 208.9g& mole % Al 2 O 3 : ( 10,000g) $ ' = moles # g& Plugging in the given data for alumina and bismuth, we can calculate the final temerature of the system: 1.1T f T f 2,473,602 =.8729T f 2 + 3,193,206 10,455T f The exression reduces to a quadratic equation, which we can solve for T f : T f = 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 exression above: 1.1T f T f 2,473,602 = q Al2 O 3 q Al2 O 3 =1,488,397 c. The final temerature falls well below the melting oint of Bi, so the samle has solidified. d. We simly calculate the heat transfer that occurs to move Bi from liquid at 600K to solid Bi at 298K: q Bi = 544 n Bi C Bi,l dt # n Bi $H m + n Bi C Bi,s dt = #1,839k Magnetic resonance imaging (MRI) is a common medical technique used for diagnostic imaging of tissues in atients. MRI is based on measuring the resonse of the weak magnetic dioles in the atoms of tissues under a strong alied magnetic field. Tyically, the magnetic induction in a clinical MRI machine may be ~2 Tesla. Consider the materials used to fabricate the MRI chamber that will be laced 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 comared to, say, an iron ermanent magnet that has a maximal net magnetization of 1.39x10 5 A/m? Suort your answer with a calculation. c. What is the work erformed by the magnetic field in the volume of a section of the MRI aluminum housing 1 m x 1 m and 5 mm thick? PS 2 5 of 6 10/4/05

6 a. We use the exression relating field strength H to induction B: B = 2T = µ o H + µ o M = µ o (1+ )H From the class notes, we have the suscetibility for Al: = 2.07 #10 $5 B = 2T = 2 kg s% C = ' 4& kg% m* ) ( #10$7,( #10 $5 )H C H =1.59 #10 6 C m % s =1.59 #106 A m b. The magnetization is calculated directly: M = H = 32.9 A m this is only 0.02% of the magnetization of the examle iron ermanent magnet! c. The work is: w = # VµHdH = Vµ o (1+ $) H = 7, PS 2 6 of 6 10/4/05

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