材料热力学与动力学复习题答案
热力学与动力学往年考试整理
判断题：1.由亚稳相向稳定相转变不需要推动力。
X2.压力可以改变材料的结构，导致材料发生相变。
V3.对于凝聚态材料，随着压力升高，熔点提高。
V4.热力学第三定律指出：在0K时任何纯物质的熵值等于零。
X5.在高温下各种物质显示相同的比热。
V6.溶体的性质主要取决于组元间的相互作用参数。
V7.金属和合金在平衡态下都存在一定数量的空位，因此空位是热力学稳定的缺陷。
V8.固溶体中原子定向迁移的驱动力是浓度梯度。
X9.溶体中析出第二相初期，第二相一般与母相保持非共格以降低应变能。
X10.相变过程中如果稳定相的相变驱动力大于亚稳相，一定优先析出。
X1.根据理查德规则，所有纯固体物质具有大致相同的熔化熵。
2.合金的任何结构转变都可以通过应力驱动来实现。
3.在马氏体相变中，界面能和应变能构成正相变的阻力，但也是逆相变的驱动力。
4.在高温下各种纯单质固体显示相同的等容热容。
5.二元溶体的混合熵只和溶体的成分有关，与组元的种类无关。
6.材料相变形核时，过冷度越大，临界核心尺寸越大。
7.二元合金在扩散时，两组元的扩散系数总是相同。
8.焓具有能量单位，但它不是能量，也不遵守能量守恒定律；但是系统的焓变可由能量表达。
9.对于凝聚态材料，随着压力升高，熔点提高,BCC—FCC转变温度也升高。
10.由于马氏体相变属于无扩散切变过程，因此应力可以促发形核和相变。
简答题：1．一般具有同素异构转变的金属从高温冷却至低温时，其转变具有怎样的体积特征？试根据高温和低温下自由能与温度的关系解释此现象。
有一种具有同素异构转变的常用金属和一般金属所具有的普遍规律不同，请指出是那种金属？简要解释其原因？（8分）答：在一定温度下元素的焓和熵随着体积的增加而增大，因此疏排结构的焓和熵大于密排结构。
G=H-TS,低温下，TS项贡献很小，G主要取决于H。
而疏排结构的H大于密排结构，疏排结构的自由能G也大于密排结构。
所以低温下密排结构是稳定相。
高温下，G主要取决于TS项，而疏排结构的熵大于密排结构，其自由能G则小于密排结构。
材料热力学与动力学复习资料+课后习题
材料热力学与动力学（复习资料）一、 概念•热力学基本概念和基本定律1. 热0：一切互为热平衡的物体，具有相同的温度。
2. 热1： - 焓：恒压体系→吸收的热量=焓的增加→焓变等于等压热效应 - 变化的可能性→过程的方向；限度→平衡3. 热2：任何不受外界影响体系总是单向地趋向平衡状态→熵+自发过程+可逆过程→隔绝体系的熵值在平衡时为最大→熵增原理（隔离体系）→Gibbs 自由能：dG<0，自发进行（同T ，p ： ）4. 热3：- (H.W.Nernst ，1906)： - (M ．Plank ，1912)：假定在绝对零度时，任何纯物质凝聚态的熵值为零S*(0K)=0 - (Lewis ，Gibson ，1920)：对于过冷溶体或内部运动未达平衡的纯物质，即使在0K 时，其熵值也不等于零，而是存在所谓的“残余熵” - Final ：在OK 时任何纯物质的完美晶体的熵值等于零• 单组元材料热力学1. 纯金属固态相变的体积效应- 除非特殊理由，所有纯金属加热固态相变都是由密排结构（fcc ）向疏排结构（bcc ）的转变→加热过程发生的相变要引起体积的膨胀→BCC 结构相在高温将变得比其他典型金属结构（如FCC 和HCP 结构）更稳定（除了Fe ）- 热力学解释1→G ：温度相同时，疏排结构的熵大于密排结构；疏排结构的焓大于密排结构→低温：H ；高温：TS - 热力学解释2→ Maxwell 方程： - α-Fe →γ-Fe ：磁性转变自由能- Richard 规则：熔化熵-Trouton 规则：蒸发熵 （估算熔沸点）2. 晶体中平衡状态下的热空位- 实际金属晶体中空位随着温度升高浓度增加，大多数常用金属（Cu 、Al 、Pb 、W 、Ag …）在接近熔点时，其空位平衡浓度约为10-4；把高温时金属中存在的平衡空位通过淬火固定下来，形成过饱和空位状态，对金属中的许多物理过程（例如扩散、时效、回复、位错攀移等）产生重要影响3. 晶体的热容- Dulong-Petit ：线性谐振动子+能量均分定律→适应于较高温度及室温附近，低温时与实验不符U Q W∆=-dH PV U d Q =+=)(δRd Q S Tδ=()d dH TdS G H d TS =--=00lim()lim()0p T T T GS T→→∂∆-=∆=∂()()V T T P V V S ∂∂=∂∂//()()()T T T V P V V S T V H ∂∂+∂∂=∂∂///RK mol J T H S mm m ≈⋅≈∆=∆/3.8/K mol J T H S b v v ⋅≈∆=∆/9.87/3V V VQ dU C RdT dT δ⎛⎫⎛⎫=== ⎪ ⎪⎝⎭⎝⎭-Einstein(固体振动热容理论)：晶体总共吸收了n 个声子，被分配到3N 个谐振子中；不适用于极低温度，无法说明在极低温度时定容热容的实验值与绝对温度的3次方成比例。
材料热力学与动力学：热力学定律习题
p2 p1
vapHm R
(1 T2
1) T1
（定积分式）
ln p vapH m / RT C （不定积分式）
第一章 热力学定律 习题课
13. Gibbs-Helmholtz方程
( A / T )
T
V
U T2
(G /Tபைடு நூலகம்)
T
p
H T2
第一章 热力学定律 习题课
1．判断下列说法是否正确： ① 状态给定后，状态函数就有一定的值，反之亦然。
② 状态函数改变后，状态一定改变。
③ 状态改变后，状态函数一定都改变。
解释：若外界条件不变，即状态给定后，所有的状态函数都有定 值。当某一个或某几个状态函数发生变化时，状态一定改变；反 之，当状态发生改变时，状态函数中，一定有某一个或几个发生 变化，而不一定全部的状态函数都发生变化。 如：气体节流膨胀过程，焓不变。
第一章 热力学定律 习题课
10.热力学基本关系式
dU TdS pdV dH TdS Vdp
dA SdT pdV dG SdT Vdp
11.克拉佩龙方程
dT dp
TVm Hm
第一章 热力学定律 习题课
12.克劳修斯—克拉佩龙方程
d ln p dT
Vap H m RT 2
（微分式）
ln
第一章 热力学定律 习题课
⑤ 恒温过程一定是可逆过程。
解释：恒温过程是指系统与环境的温度始终保持相 等且恒定，是一个自始至终保持热平衡的过程，即 为可逆过程。数学表达式dT = 0； ✓若对于一个ΔT = 0的过程，则说明该过程始终态温 度相等，并不一定是恒温过程。 ✓等温可逆过程一定是恒温过程。
可以这么看，一个思想封闭的人遇到烦恼时就 会焦躁不按或者抑郁忧愁，也就是内心混乱度增 加。但是当这个人向他人敞开心扉获得安慰之后 ，他就会趋于平静。所以，同学们都要学会向他 人倾诉，同他人交流，不要把自己封闭在一个狭 小的世界里。
材料热力学与动力学复习题答案
一、常压时纯Al 的密度为ρ=2.7g/cm 3，熔点T m =660.28℃，熔化时体积增加5%。
用理查得规则和克-克方程估计一下，当压力增加1Gpa 时其熔点大约是多少？ 解：由理查德规则RTm Hm R Tm Hm Sm ≈∆⇒≈∆=∆ …①由克-克方程VT H dT dP ∆∆=…② 温度变化对ΔH m 影响较小，可以忽略，①代入②得V T H dT dP ∆∆=dT T 1V Tm R dp V T Tm R ∆≈⇒∆≈…③ 对③积分 dT T1V T Tm R p d T Tm Tm pp p ⎰⎰∆+∆+∆= 整理 ⎪⎭⎫ ⎝⎛∆+∆=∆Tm T 1ln V Tm R p V T R V Tm R Tm T ∆∆=∆⨯∆≈ Al 的摩尔体积 V m =m/ρ=10cm 3=1×10-5m 3Al 体积增加 ΔV=5%V m =0.05×10-5m 3K 14.60314.810510R V p T 79=⨯⨯=∆∆=∆- Tm ’=Tm+T ∆=660.28+273.15+60.14=993.57K二、热力学平衡包含哪些内容，如何判断热力学平衡。
内容：（1）热平衡，体系的各部分温度相等；（2）质平衡：体系与环境所含有的质量不变；（3）力平衡：体系各部分所受的力平衡，即在不考虑重力的前提下，体系内部各处所受的压力相等；（4）化学平衡：体系的组成不随时间而改变。
热力学平衡的判据：（1）熵判据：由熵的定义知dS Q T δ≥不可逆可逆对于孤立体系，有0Q =δ，因此有dS 可逆不可逆0≥，由于可逆过程由无限多个平衡态组成，因此对于孤立体系有dS 可逆不可逆0≥，对于封闭体系，可将体系和环境一并作为整个孤立体系来考虑熵的变化，即平衡自发环境体系总0S S S ≥∆+∆=∆（2）自由能判据 若当体系不作非体积功时，在等温等容下，有()0d ,≤V T F 平衡状态自发过程上式表明，体系在等温等容不作非体积功时，任其自然，自发变化总是向自由能减小的方向进行，直至自由能减小到最低值，体系达到平衡为止。
材料热力学习题答案1
The problems of the first law1. a lead bullet is fired at a frigid surface. At what speed must it travel to melt on impact, if its initial temperature is 25℃ and heating of the rigid surface of the rigid surface is neglected? The melting point of lead is 327℃. The molar heat of fusion of the lead is 4.8kJ/mol. The molar heat capacity C P of lead may be taken as 29.3J/(mol K) (1.1)Solution: )/(5.112.20721]108.4)25327(3.29[2121)(2322s m V v n n WQ nMv mv W H T C n Q Q Q absorb melting p melt increase absorb ==⨯+-⨯===∆+∆=+=2. what is the average power production in watts of a person who burns 2500 kcal of food in a day? Estimate the average additional powder production of 75Kg man who is climbing a mountain at eh rate of 20 m/min (1.2)Solution )/(24560208.975)/(12160602410467000//)(104670001868.4102500sin 3S J t h mg P S J t Q t W P J Q gincrea Burning Burning =⨯⨯=∆==⨯⨯====⨯⨯=3 One cubic decimeter (1 dm 3) of water is broken into droplets having a diameter of one micrometer (1 um) at 20℃. (1.3)(a) what is the total area of the droplets?(b) Calculate the minimum work required to produce the droplets. Assume that the droplets arerest (have zero velocity)Water have a surface tension of 72.75 dyn/cm at 20℃ (NOTES: the term surface energy (ene/cm 2) is also used for surface tension dyn/cm)Solution)(25.218)106103(1075.72)(103)101(4)101(34)101(232523263631J S W m nS S Single total =⨯-⨯⨯⨯=∆=⨯=⨯⨯⨯⨯⨯⨯⨯⨯==-+----σππ4.Gaseous helium is to be used to quench a hot piece of metal. The helium is in storage in aninsulated tank with a volume of 50 L and a temperature of 25℃, the pressure is 10 atm. Assume that helium is an ideal gas.(a) when the valve is opened and the gas escapes into the quench chamber (pressure=1 atm), whatwill be the temperature of the first gas to hit the specimen?(b) As the helium flows, the pressure in the tank drops. What will be the temperature of thehelium entering the quench chamber when the pressure in the tank has fallen to 1 atm? (1.4)Solution: )(180118298)(1185.229810101325501010101325)5500(1)()(118)101(298)()(0334.0/00K T T T K RR nC W T b K T P PT T Adiabatic a p C R P=-=∆-==⨯⨯⨯⨯⨯⨯⨯-⨯==∆=⨯==--5 An evacuated (P=0), insulted tank is surrounded by a very large volume (assume infinite volume) of an ideal gas at a temperature T 0. The valve on the tank is opened and the surrounding gas is allowed to flow quickly into the tank until the pressure inside the tank is equals the pressure outside. Assume that no heat flow takes place. What is the final tempeture of the gas in the tank? The heat capacity of the gas, C p and C v each may be assumed to be constant over the temperature rang spanned by the experiment. You answer may be left in terms of C p and C vhint: one way to approach the problem is to define the system as the gas ends up in the tank. (1.5)solution 0/000/00)0()(T P P T T P PT T Adiabatic PPC R C R ≈-==6. Calculate the heat of reaction of methane with oxygen at 298K, assuming that the products of reaction are CO 2 and CH 4 (gas)[This heat of reaction is also called the low calorific power of methane] convert the answer into unites of Btu/1000 SCF of methane. SCF means standard cubic feet, taken at 298 and 1atmNOTE: this value is a good approximation for the low calorific powder of natural gas (1.6)DA TA:)()()(224g O H g CO g CH FOR80.5705.9489.17]/[0298---•∆mol g Kcal Hsolution)1000/(9.2610252103048.01101076.191)/(76.191)89.1780.57205.94()2(22333332982982224422SCF Btu mol g Kcal H H H H H OH CO O CH CH O H CO =⨯⨯⨯⨯⨯=•=∆+⨯---=∆-∆+∆-=∆+=+-7. Methane is delivered at 298 K to a glass factory, which operates a melting furnace at 1600 K. The fuel is mixed with a quantity of air, also at 298 K, which is 10% in excess of the amount theoretically needed for complete combustion (air is approximately 21% O 2 and 79% N 2) (1.7) (a) Assuming complete combustion, what is the composition of the flue gas (the gas followingcombustion)?(b) What is the temperature of the gas, assuming no heat loss?(c) The furnace processes 2000kg of glass hourly, and its heat losses to the surroundings average400000 kJ/h. calculate the fuel consumption at STP (in m 3/h) assuming that for gas H 1600-H 298=1200KJ/KG(d) A heat exchanger is installed to transfer some of the sensible heat of the flue gas to thecombustion air. Calculate the decrease in fuel consumption if the combustion air is heated to 800KDA TA STP means T=298K, P=1atm22224O N O H CO CH for 2.82.89.117.1316)/(C mol cal C P •Solution)(210448.1125.9100076.191298)/(25.9)]87.012.72(2.843.179.1171.87.13[01.0)(%87.0%%12.72%%43.17%2%%71.8)11.1(221791.1231%22)(0,,222222224K T T T C mol cal X C C b O N CO O H CO O H CO O CH a i i p p p =⨯⨯+=∆+=•=+⨯+⨯+⨯=======-⨯+⨯⨯+=+=+∑)/(1644)0224.011868.448.11)8001600(48.1125.9189570(102800000)/(189570)298800)](48.1187.8)48.1125.9[(100076.191)()/(87.848.11/]211002.22.816[)()/(3214)0224.011868.448.11)2981600(48.1125.9100076.191(102800000)/(280000040000020001200)(33min ,,,,298,,33min h m V mol g cal dTn C n C H H C mol cal X C C d h m V h KJ P C gConsu i i r p i i p p i i p r p g Consu =⨯⨯-⨯-⨯=•=-⨯-⨯-⨯=--∆=∆•=⨯⨯+===⨯⨯-⨯-⨯⨯==+⨯=⎰∑∑∑8.In an investigation of the thermodynamic properties of a-manganese, the following heat contents were determined:H 700-H 298=12113 J/(g atom) H 1000-H 298=22803 J/(g atom)Find a suitable equation for H T -H 298 and also for C P as a function of temperature in the form (a+bT) Assume that no structure transformation takes place in the given tempeture rang. (1.8)Solution )298(0055.0)298(62.35011.062.35011.062.3522803)2981000(2)2981000(12113)298700(2)298700(]2[2229822222982---=∆-=-===-+-=-+-+=+==∆⎰⎰T T H TC b a ba ba T baT bTdT a dT C H TP T P9.A fuel gas containing 40% CO, 10% CO 2, and the rest N 2 (by volume) is burnt completely with air in a furnace. The incoming and ongoing temperatures of the gases in the furnace are 773K and 1250K,respectively. Calculate (a) the maximum flame temperature and (b) heat supplied to the furnace per cu. ft of exhaust gas (1.9)molJ Hmol J H CO f CO f /393296/1104580,298,0,298,2-=∆-=∆)/(10184.403.29)/(1067.11010.492.19)/(1037.81020.935.44)/(1042.01097.345.283,253,253,253,222molK J T C molK J T T C molK J T T C molK J T T C N P O P CO P CO P -------⨯+=⨯-⨯+=⨯-⨯+=⨯-⨯+=Solution?0)499.0321.018.1()1067.01019.277.28(28.282831067.01038.477.289.0)1019.01058.528.33(2.0282838)()/(1019.01058.528.33722.0278.0)/(1067.01038.477.281.065.005.02.0)()/(282838110458393296%2.72%8.27%10%65%5%20)4/(1122298127332981523733253253298,,,,298,253,,,,,253,,,,,,,0,298,0,298,298,22222222222222==+--⨯+⨯++⨯=⨯-⨯++⨯⨯-⨯+-⨯=--∆=∆⨯-⨯+=+==⨯-⨯+=+++===-=∆-∆=∆========+-----------⎰⎰⎰∑∑⎰∑∑∑∑T T T T T T T dTT T dTT T dT n C n C n H H molK J T T C C n C C molK J T T C C C C n C C a mol J n H n H H N CO production O N CO CO reation then O N air mole need fuel mole when CO O CO T TT i i r p i i p p i i N P CO P i i p p r p O P N P CO P CO P i i p p r p i p f i r f idTT T Q dT T T Q b T T T T T T T dT T T dTT T dT n C n C n H H T TT i i r p i i p p i i 9.0)1019.01058.528.33(2.02828389.0)1019.01058.528.33(2.0282838)(0)499.0321.018.1()1067.01019.277.28(28.282831067.01038.477.289.0)1019.01058.528.33(2.0282838)(253125029812502982531250298125029829812125029815231250253253298,,,,298,⨯⨯-⨯++⨯-=⨯⨯-⨯++⨯-===+--⨯+⨯++⨯=⨯-⨯++⨯⨯-⨯+-⨯=--∆=∆-----------⎰⎰⎰⎰⎰∑∑⎰10. (a) for the reaction 2221CO O CO →+,what is the enthalpy of reaction (0H ∆) at 298 K ? (b) a fuel gas, with composition 50% CO, 50% N 2 is burned using the stoichiometric amount of air. What is the composition of the flue gas?(c) If the fuel gas and the air enter there burner at 298 K, what is the highest temperature theflame may attain (adiabatic flame temperature)? DA TA :standard heats of formation f H ∆ at 298 K (1.10))/(393000)/(1100002mol J CO mol J CO -=-=Heat capacities [J/(mol K)] to be used for this problem N 2=33, O 2=33, CO=34, CO 2=57 Solution)(21100)298)(39889.0(222.02830000)/(3975.03325.057)/(33111.034222.033666.033)(%,75%%,251.111002.22%%1.11%%,6.66%%,2.222.0/25.015.0%)()/(283000393000110000)(,0,,,,,,22220,298,0,298,0K T T dT C n H H K mol J X C C K mol J X C C C N CO product O N CO fuel b mol J n H n H H a P p p i P r i P r i P p i P p i P f i r f ==-⨯-⨯=-∆=∆•=⨯+⨯==•=⨯+⨯+⨯====-====+==+-=∆-∆=∆⎰∑∑∑∑11.a particular blast furnace gas has the following composition by (volume): N 2=60%, H 2=4, CO=12%, CO 2=24%(a) if the gas at 298K is burned with the stochiometric amount of dry air at 298 K, what is the composition of the flue gas? What is the adiabatic flame temperature? (b) repeat the calculation for 30% excess combustion air at 298K(C)what is the adiabatic flame temperature when the blast furnace gas is preheated to 700K (the dry air is at 298K)(d) suppose the combustion air is not dry ( has partial pressure of water 15 mm Hg and a total pressure of 760 mm Hg) how will the flame temperature be affected? DA TA(k J/mol) （1.11）2CO CO FOR513.393523.110)/(--∆mol kJ H f 2222,)(O N g O H CO CO FOR34505733]/[K mol J C P •Solution)(1052)(75438286370])295.03450(241604[026.0])335.03457(110523393513[079.0])([%8.66%%,8.6%%,6.2%%,8.15%%,9.72.0/83.110012%)()(1122)(82538313430])295.03450(241604[029.0])335.03457(110523393513[086.0])([%7.65%%,7.5%%,9.2%%,1.17%%,6.82.0/810012%2121)(,,,,,,,02222,,,,,,,0222222222K T K T T n C T T X C dT n C n C H x H N O H CO CO b K T K T T n C T T X C dT n C n C H x H N O H CO CO OH O H CO O CO a i i r P ii P i i r P i i p P i i i i r P ii P i i r P i i p P i i ===∆=∆-∆-⨯--+∆-⨯---=+--∆=∆=====⨯+====∆=∆-∆-⨯--+∆-⨯---=+--∆=∆=====+=→+→+∑∑∑⎰∑∑∑∑∑⎰∑∑)(1419),(11213842594034286.0)402(2.39714.0])295.03450(241604[029.0])335.03457(110523393513[086.0)3(K T K T T T T T H ===∆=∆⨯--∆⨯-∆-⨯--+∆-⨯---=∆12．A bath of molten copper is super cooled to 5℃ below its true melting point. Nucleation of solid copper then takes place, and the solidification proceeds under adiabatic conditions. What percentage of the bath solidifies?DA TA: Heat of fusion for copper is 3100 cal/mol at 1803℃(the melting point of copper) C P,L =7.5(cal/mol ℃), C P,S =5.41+(1.5*10-3T )(cal/mol ℃) （1.12） Solution)/(310355.75.0)17981803(105.1541.5310002231798,1798,17981803,18031798,1803,mol cal H H dT C dT C H L S SL L P S P L S =⨯-⨯-⨯+⨯+==+++-⎰⎰13．Cuprous oxide (Cu 2O) is being reduced by hydrogen in a furnace at 1000K, (a)write the chemical reaction for the reduced one mole of Cu 2O(b)how much heat is release or absorbed per mole reacted? Given the quantity of heat and state whether heat is evolved (exothermic reaction) or absorbed (endothermic reaction)DA TA: heat of formation of 1000K in cal/mol Cu 2O=-41900 H 2O=-59210 （1.13） solution)/(173104190059210222mol cal H OH Cu H O Cu =-=∆+=+,exothermic reaction14. (a) what is the enthalpy of pure, liquid aluminum at 1000K?(b) an electric resistance furnace is used to melt pure aluminum at the rate of 100kg/h. the furnace is fed with solid aluminum at 298K. The liquid aluminum leaves the furnace at 1000K. what is the minimum electric powder rating (kW) of furnace.DA TA : For aluminum : atomic weight=27g/mol, C p,s =26(J/molK), C p,L =29(J/molK), Melting point=932K, Heat of fusion=10700J/mol (1.14)Solution )(28.0)(7.2793600110002727184)/(2718410700)9321000(29)298932(261000932,932298,1000,kW W P mol J H dT C dT C H SLL P S P l ==⨯⨯==+-⨯+-⨯=++=⎰⎰15 A waste material (dross from the melting of aluminum) is found to contain 1 wt% metallic aluminum. The rest may be assumed to aluminum oxide. The aluminum is finely divided and dispersed in the aluminum oxide; that is the two material are thermally connected.If the waster material is stored at 298K. what is the maximum temperature to which it may rise if all the metallic aluminum is oxidized by air/ the entire mass may be assumed to rise to the same temperature. Data : atomic weight Al=27g/mol, O=16g/mol, C p,s,Al =26(J/molK), C p,s,Al2O3=104J/mol, heat formation of Al 2O 3=-1676000J/mol (1.15)Solution;)(600)(3021041029927275.116122711676000K T K T T ==∆∆⨯⨯++⨯⨯=⨯⨯16 Metals exhibit some interesting properties when they are rapidly solidified from the liquid state. An apparatus for the rapid solidification of copper is cooled by water. In the apparatus, liquid copper at its melting point (1356K) is sprayed on a cooling surface, where it solidified and cools to 400K. The copper is supplied to the apparatus at the rate of one kilogram per minute. Cooling water is available at 20℃, and is not allowed to raise above 80℃. What is the minimum flow rate of water in the apparatus, in cubic meters per minute?DA TA; for water: C p =4.184J/g k, Density=1g/cm 3; for copper: molecular weight=63.54g/mol C p =7cal/mol k, heat of fusion=3120 cal/mol (1.16)Solution:min)/(10573.2)2080(1min /min54.631000)]4001356(73120[min /33m V VQ Q Water Copper -⨯=-=⨯⨯-⨯+=17 water flowing through an insulated pipe at the rate of 5L/min is to be heated from 20℃ to 60℃ b an electrical resistance heater. Calculate the minimum power rating of the resistance heater in watts. Specify the system and basis for you calculation. DA TA; For water C p =4.184J/g k, Density=1g/cm 3 (1.17)Solution: )(139476010005)2060(184.4W W =⨯⨯-⨯=18 The heat of evaporation of water at 100℃ and 1 atm is 2261J/mol (a) what percentage of that energy is used as work done by the vapor?(b)if the density of water vapor at 100℃ and 1 atm is 0.597kg/m 3 what is the internal energy change for the evaporation of water? (1.18)Solution: )/(375971822613101%6.71822613101%)/(31010224.0273373101325mol J Q W U mol J V P =⨯+-=+=∆=⨯==⨯⨯=∆19 water is the minimum amount of steam (at 100℃ and 1 atm pressure) required to melt a kilogram of ice (at 0℃)? Use data for problem 1.20 (1.19) Solution )(125,3341000)10018.42261(g m m =⨯=⨯+20 in certain parts of the world pressurized water from beneath the surface of the earth is available as a source of thermal energy. To make steam, the geothermal water at 180℃is passed through a flash evaporator that operates at 1atm pressure. Two streams come out of the evaporator, liquid water and water vapor. How much water vapor is formed per kilogram of geothermal water? Is the process reversible? Assume that water is incompressible. The vapor pressure of water at 180℃is1.0021 Mpa( about 10 atm) Data: C P,L=4.18J/(g k), C P,v=2.00J/(g k), △H V=2261J/g, △H m=334 J/g (1.20)Solution:leirreversibgxxx)(138),1000(8018.4)8018.48022261(=-⨯⨯=⨯-⨯+。
[工学]南昌大学材料热力学答案
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The problems of the first law1.1 a lead bullet is fired at a frigid surface. At what speed must it travel to melt on impact, if its initial temperature is 25℃ and heating of the rigid surface of the rigid surface is neglected? The melting point of lead is 327℃. The molar heat of fusion of the lead is 4.8kJ/mol. The molar heat capacity C P of lead may be taken as 29.3J/(mol K)Solution: )/(363102.20721]108.4)25327(3.29[2121)(23322s m V v n n WQ nMv mv W H T C n Q Q Q absorb melting p melt increase absorb =⨯=⨯+-⨯===∆+∆=+=-1.2 what is the average power production in watts of a person who burns 2500 kcal of food in a day? Estimate the average additional powder production of 75Kg man who is climbing a mountain at the rate of 20 m/minSolution )/(24560208.975)/(12160602410467000//)(104670001868.4102500sin 3S J t h mg P S J t Q t W PJ Q gincrea Burning Burning =⨯⨯=∆==⨯⨯====⨯⨯=1.3 One cubic decimeter (1 dm 3) of water is broken into droplets having a diameter of one micrometer (1 um) at 20℃.(a) what is the total area of the droplets?(b) Calculate the minimum work required to produce the droplets. Assume that the droplets are rest (have zerovelocity)Water have a surface tension of 72.75 dyn/cm at 20℃ (NOTES: the term surface energy (ene/cm 2) is also used for surface tension dyn/cm)Solution)(6.436)106103(1075.72)(106)105.0(4)105.0(34)101(232523263631J S W m nS S Single total =⨯-⨯⨯⨯=∆=⨯=⨯⨯⨯⨯⨯⨯⨯⨯==-+----σππ1.4 Gaseous helium is to be used to quench a hot piece of metal. The helium is in storage in an insulated tank with a volume of 50 L and a temperature of 25℃, the pressure is 10 atm. Assume that helium is an ideal gas.(a) when the valve is opened and the gas escapes into the quench chamber (pressure=1 atm), what will be thetemperature of the first gas to hit the specimen?(b) As the helium flows, the pressure in the tank drops. What will be the temperature of the helium entering thequench chamber when the pressure in the tank has fallen to 1 atm?Solution: )(180118298)(1185.229810101325501010101325)5500(1)()(118)101(298)()(0334.0/00K T T T K RR nC W T b K T P P T T Adiabatica p CR P=-=∆-==⨯⨯⨯⨯⨯⨯⨯-⨯==∆=⨯==--1.5 An evacuated (P=0), insulted tank is surrounded by a very large volume (assume infinite volume) of an ideal gas at a temperature T 0. The valve on the tank is oðened and the surrounding gas is allowed to flow suickly into t(e tank until the pressure insi`e the tank is equals the pressure outside. Assume that no heat flow takes place. What is the0final tempeture kf t èe gaS in the tank? The heat cap!city mf the gas, C p and C v each íay be(assumed to be c/nsuant over thå temperature rang!spanNed by the døperiment. You answer may be meft in terms of C p and S v hint: one way to approach the xroblem is to define the system as the gas ends up in the tank.solution 0/000/00)0()(T P P T T P PT T Adiabatic PPC R C R ≈-==1.6 Calculate the heat of reaction of methane with oxygen at 298K, assuming that the products of reaction are CO 2 and CH 4 (gas)[This heat of reaction is also called the low calorific power of methane] convert the answer into unites of Btu/1000 SCF of methane. SCF means standard cubic feet, taken at 298 and 1atm NOTE: this value is a good approximation for the low calorific powder of natural gasDATA: )()()(224g O H g CO g CH FOR80.5705.9489.17]/[0298---•∆mol g Kcal Hsolution)1000/(9.2610252103048.01101076.191)/(76.191)89.1780.57205.94()2(22333332982982224422SCF Btu mol g Kcal H H H H H OH CO O CH CH O H CO =⨯⨯⨯⨯⨯=•=∆+⨯---=∆-∆+∆-=∆+=+- 1.7Methane is delivered at 298 K to a glass factory, which operates a melting furnace at 1600 K. The fuel is mixed with a quantity of air, also at 298 K, which is 10% in excess of the amount theoretically needed for complete combustion (air is approximately 21% O 2 and 79% N 2)(a) Assuming complete combustion, what is the composition of the flue gas (the gas following combustion)? (b) What is the temperature of the gas, assuming no heat loss?(c) The furnace processes 2000kg of glass hourly, and its heat losses to the surroundings average 400000 kJ/h.calculate the fuel consumption at STP (in m 3/h) assuming that for gas H 1600-H 298=1200KJ/KG(d) A heat exchanger is installed to transfer some of the sensible heat of the flue gas to the combustion air.Calculate the decrease in fuel consumption if the combustion air is heated to 800K DATA STP means T=298K, P=1atm22224O N O H CO CH for2.82.89.117.1316)/(C mol cal C P •Solution)(210448.1125.9100076.191298)/(25.9)]87.012.72(2.843.179.1171.87.13[01.0)(%87.0%%12.72%%43.17%2%%71.8)11.1(221791.1231%22)(0,,222222224K T T T C mol cal X C C b O N CO O H CO O H CO O CH a i i p p p =⨯⨯+=∆+=•=+⨯+⨯+⨯=======-⨯+⨯⨯+=+=+∑)/(1644)0224.011868.448.11)8001600(48.1125.9189570(102800000)/(189570)298800)](48.1187.8)48.1125.9[(100076.191)()/(87.848.11/]211002.22.816[)()/(3214)0224.011868.448.11)2981600(48.1125.9100076.191(102800000)/(280000040000020001200)(33min ,,,,298,,33min h m V mol g cal dTn C n C H H C mol cal X C C d h m V h KJ P C gConsu i i r p i i p p i i p r p g Consu =⨯⨯-⨯-⨯=•=-⨯-⨯-⨯=--∆=∆•=⨯⨯+===⨯⨯-⨯-⨯⨯==+⨯=⎰∑∑∑1.8 In an investigation of the thermodynamic properties of a-manganese, the following heat contents were determined:H 700-H 298=12113 J/(g atom) H 1000-H 298=22803 J/(g atom)Find a suitable equation for H T -H 298 and also for C P as a function of temperature in the form (a+bT) Assume that no structure transformation takes place in the given tempeture rang.Solution )298(0055.0)298(62.35011.062.35011.062.3522803)2981000(2)2981000(12113)298700(2)298700(]2[2229822222982---=∆-=-===-+-=-+-+=+==∆⎰⎰T T H TC b a ba ba T baT bTdT a dT C H T P T P1.9 A fuel gas containing 40% CO, 10% CO 2, and the rest N 2 (by volume) is burnt completely with air in a furnace. The incoming and ongoing temperatures of the gases in the furnace are 773K and 1250K,respectively. Calculate (a) the maximum flame temperature and (b) heat supplied to the furnace per cu. ft of exhaust gasmolJ Hmol J H CO f CO f /393296/1104580,298,0,298,2-=∆-=∆)/(10184.403.29)/(1067.11010.492.19)/(1037.81020.935.44)/(1042.01097.345.283,253,253,253,222molK J T C molK J T T C molK J T T C molK J T T C N P O P CO P CO P -------⨯+=⨯-⨯+=⨯-⨯+=⨯-⨯+= Solution?0)499.0321.018.1()1067.01019.277.28(28.282831067.01038.477.289.0)1019.01058.528.33(2.0282838)()/(1019.01058.528.33722.0278.0)/(1067.01038.477.281.065.005.02.0)()/(282838110458393296%2.72%8.27%10%65%5%20)4/(1122298127332981523733253253298,,,,298,253,,,,,253,,,,,,,0,298,0,298,298,22222222222222==+--⨯+⨯++⨯=⨯-⨯++⨯⨯-⨯+-⨯=--∆=∆⨯-⨯+=+==⨯-⨯+=+++===-=∆-∆=∆========+-----------⎰⎰⎰∑∑⎰∑∑∑∑T T T T T T T dTT T dTT T dTn C n C n H H molK J T T C C n C C molK J T T C C C C n C C a mol J n Hn H H N CO production O N CO CO reation then O N air mole need fuel mole when CO O CO T TT i i r p i i p p i i N P CO P i i p p r p O P N P CO P CO P i i p p r p i pf i rf idTT T Q dT T T Q b T T T T T T T dTT T dTT T dT n C n C n H H T TT i i r p i i p p i i 9.0)1019.01058.528.33(2.02828389.0)1019.01058.528.33(2.0282838)(0)499.0321.018.1()1067.01019.277.28(28.282831067.01038.477.289.0)1019.01058.528.33(2.0282838)(253125029812502982531250298125029829812125029815231250253253298,,,,298,⨯⨯-⨯++⨯-=⨯⨯-⨯++⨯-===+--⨯+⨯++⨯=⨯-⨯++⨯⨯-⨯+-⨯=--∆=∆-----------⎰⎰⎰⎰⎰∑∑⎰1.10 (a) for the reaction2221CO O CO →+,what is the enthalpy of reaction (0H ∆) at 298 K ?(b) a fuel gas, with composition 50% CO, 50% N 2 is burned using the stoichiometric amount of air. What is the composition of the flue gas?(c) If the fuel gas and the air enter there burner at 298 K, what is the highest temperature the flame may attain(adiabatic flame temperature)? DATA :standard heats of formationfH ∆ at 298 K)/(393000)/(1100002mol J CO mol J CO -=-=Heat capacities [J/(mol K)] to be used for this problem N 2=33, O 2=33, CO=34, CO 2=57Solution )(21100)298)(39889.0(222.02830000)/(3975.03325.057)/(33111.034222.033666.033)(%,75%%,251.111002.22%%1.11%%,6.66%%,2.222.0/25.015.0%)()/(283000393000110000)(,0,,,,,,22220,298,0,298,0K T T dT C n H H K mol J X C C K mol J X C C C N CO product O N CO fuel b mol J n H n H H a P p p i P r i P r i P p i P p i P f i r f ==-⨯-⨯=-∆=∆•=⨯+⨯==•=⨯+⨯+⨯====-====+==+-=∆-∆=∆⎰∑∑∑∑1.11 a particular blast furnace gas has the following composition by (volume): N 2=60%, H 2=4, CO=12%, CO 2=24%(a) if the gas at 298K is burned with the stochiometric amount of dry air at 298 K, what is the composition of the flue gas? What is the adiabatic flame temperature?(b) repeat the calculation for 30% excess combustion air at 298K(C)what is the adiabatic flame temperature when the blast furnace gas is preheated to 700K (the dry air is at 298K) (d) suppose the combustion air is not dry ( has partial pressure of water 15 mm Hg and a total pressure of 760 mm Hg) how will thE dlaMe temperature be affected?DaTA(k J?mol)2CO CO FOR513.393523.110)/(--∆mol kJ H f2222,)(O N g O H CO CO FOR ?? 34505733]/[K mol J C P •SolutionOH O H CO O CO a 222222121)(→+→+ 416.0)(104.0)(:22==N n O n Air6.0)(04.0)(24.0)(12.0)(:222====N n H n CO n CO n Fuel 32.0)(08.0)(:22==N n O n Air 92.0)(04.0)(36.0)(:222===N n O H n CO n Flue)(98.1108)(8108.53106308.43)/(8.533492.05004.05736.092.004.036.06308.43)08.241(04.0)523.11051.393(12.03,,222222K T K T K J C C C n CKJH H H N O H CO ii r P O H H CO CO ==⨯=∆=⨯+⨯+⨯=++==-⨯+-⨯=∆+∆=∆∑--(b)repeat the calculation for 30% excess0combustion air at 298K6.0)(04.0)(24.0)(12.0)(:222====N n H n CO n CO n Fuel024.0)(016.1)(04.0)(36.0)(:2222====O n N n O H n CO n Flue December 13, 2020(C)what is the adiabatic flame temperature when the blasp furnace gas is preheated to 700K (the dry air is at 298K)6.0)(04.0)(24.0)(12.0)(:222====N n H n CO n CO n Fuel 32.0)(08.0)(:22==N n O n Air 92.0)(04.0)(36.0)(:222===N n O H n CO n Flue)(6.1401)(6.11038.5310373.59)/(8.533492.05004.05736.096.004.036.0373.59)346.02804.05724.03312.0()298700()08.241(04.0)523.11051.393(12.03,,298700222222K T K T K J C C C n CKJH H H H N O H CO ii r P fuel O H H CO CO ==⨯=∆=⨯+⨯+⨯=++==⨯+⨯+⨯+⨯⨯-+-⨯+-⨯=∆+∆+∆=∆∑---(d) suppose the combustion air is not dry ( has partial pressure of water 15 mm Hg and a total pressure of 760 mm Hg) how will the flame temperature be affected?)(8.1051)(8.75388.57106308.43)/(88.5734024.034016.15004.05736.0024.0016.104.036.06308.43)08.241(04.0)523.11051.393(12.03,,2222222K T K T K J C C C C n CKJH H H O N O H CO ii r P O H H CO CO ==⨯=∆=⨯+⨯+⨯+⨯=+++==-⨯+-⨯=∆+∆=∆∑--6.0)(04.0)(24.0)(12.0)(:222====N n H n CO n CO n Fuel 008.04.01576015)(32.0)(08.0)(:222=-===O H n N n O n Air 92.0)(048.0)(36.0)(:222===N n O H n CO n Flue)(1103)(8052.54106308.43)/(2.543492.050048.05736.092.0048.036.06308.43)08.241(04.0)523.11051.393(12.03,,222222K T K T K J C C C n CKJH H H N O H CO ii r P OH H CO CO ==⨯=∆=⨯+⨯+⨯=++==-⨯+-⨯=∆+∆=∆∑--1.12 A bath of molten copper is super cooled to 5℃ below its true melting point. Nucleation of solid copper then takes place, and the solidification proceeds under adiabatic conditions. What percentage of the bath solidifies? DATA: Heat of fusion for copper is 3100 cal/mol at 1803℃(the melting point of copper) C P,L =7.5(cal/mol ℃), C P,S =5.41+(1.5*10-3T )(cal/mol ℃) Solution)/(310355.75.0)17981803(105.1541.5310002231798,1798,17981803,18031798,1803,mol cal HH dT C dT C H LS SL L P S P L S =⨯-⨯-⨯+⨯+==+++-⎰⎰1.13 Cuprous oxide (Cu 2O) is being reduced by hydrogen in a furnace at 1000K, (a)write the chemical reaction for the reduced one mole of Cu 2O(b)how much heat is release or absorbed per mole reacted? Given the quantity of heat and state whether heat is evolved (exothermic reaction) or absorbed (endothermic reaction)DATA: heat of formation of 1000K in cal/mol Cu 2O=-41900 H 2O=-59210 solution)/(173104190059210222mol cal H OH Cu H O Cu =-=∆+=+,exothermic reaction1.14(a) what is the enthalpy of pure, liquid aluminum at 1000K?(b) an electric resistance furnace is used to melt pure aluminum at the rate of 100kg/h. the furnace is fed with solid aluminum at 298K. The liquid aluminum leaves the furnace at 1000K. what is the minimum electric powder rating (kW) of furnace.DATA : For aluminum : atomic weight=27g/mol, C p,s =26(J/molK), C p,L =29(J/molK), Melting point=932K, Heat of fusion=10700J/mol Solution)(28.0)(7.2793600110002727184)/(2718410700)9321000(29)298932(261000932,932298,1000,kW W P mol J H dT C dT C H SLL P S P l ==⨯⨯==+-⨯+-⨯=++=⎰⎰1.15 A waste material (dross from the melting of aluminum) is found to contain 1 wt% metallic aluminum. The rest may be assumed to aluminum oxide. The aluminum is finely divided and dispersed in the aluminum oxide; that is the two material are thermally connected.If the waster material is stored at 298K. what is the maximum temperature to which it may rise if all the metallic aluminum is oxidized by air/ the entire mass may be assumed to rise to the same temperature. Data : atomic weight Al=27g/mol, O=16g/mol, C p,s,Al =26(J/molK), C p,s, Al2O3=104J/mol, heat formation of Al 2O 3=-1676000J/molSolution;)(600)(3021041029927275.116122711676000K T K T T ==∆∆⨯⨯++⨯⨯=⨯⨯1.16 Metals exhibit some interesting properties when they are rapidly solidified from the liquid state. An apparatus for the rapid solidification of copper is cooled by water. In the apparatus, liquid copper at its melting point (1356K) is sprayed on a cooling surface, where it solidified and cools to 400K. The copper is supplied to the apparatus at the rate of one kilogram per minute. Cooling water is available at 20℃, and is not allowed to raise above 80℃. What is the minimum flow rate of water in the apparatus, in cubic meters per minute? DATA; for water: C p =4.184J/g k, Density=1g/cm 3; for copper: molecular weight=63.54g/mol C p =7cal/mol k, heat of fusion=3120 cal/molSolution:min)/(10573.2)2080(1min /min54.631000)]4001356(73120[min /33m V VQQ WaterCopper -⨯=-=⨯⨯-⨯+=1.17 water flowing through an insulated pipe at the rate of 5L/min is to be heated from 20℃ to 60℃ b an electrical resistance heater. Calculate the minimum power rating of the resistance heater in watts. Specify the system and basis for you calculation. DA TA; For water C p =4.184J/g k, Density=1g/cm 3 Solution:)(139476010005)2060(184.4W W =⨯⨯-⨯=1.18 The heat of evaporation of water at 100℃ and 1 atm is 2261J/mol (a) what percentage of that energy is used as work done by the vapor?(b)if the density of water vapor at 100℃ and 1 atm is 0.597kg/m 3 what is the internal energy change for the evaporation of water?Solution:)/(375971822613101%6.71822613101%)/(31010224.0273373101325mol J Q W U mol J V P =⨯+-=+=∆=⨯==⨯⨯=∆1.19 water is the minimum amount of steam (at 100℃ and 1 atm pressure) required to melt a kilogram of ice (at 0℃)? Use data for problem 1.20 Solution)(125,3341000)10018.42261(g m m =⨯=⨯+1.20 in certain parts of the world pressurized water from beneath the surface of the earth is available as a source of thermal energy. To make steam, the geothermal water at 180℃ is passed through a flash evaporator that operates at 1atm pressure. Two streams come out of the evaporator, liquid water and water vapor. How much water vapor is formed per kilogram of geothermal water? Is the process reversible? Assume that water is incompressible. The vapor pressure of water at 180℃ is 1.0021 Mpa( about 10 atm) Data: C P,L =4.18J/(g k), C P ,v =2.00J/(g k), △H V =2261J/g, △H m =334 J/g Solution:leirreversib g x x x )(138),1000(8018.4)8018.48022261(=-⨯⨯=⨯-⨯+The problems of the second law2.1 The solar energy flux is about 4J cm 2/min. in no focusing collector the surface temperature can reach a value of about 900℃. If we operate a heat engine using the collector as the heat source and a low temperature reservoir at 25℃, calculate the area of collector needed if the heat engine is to produce 1 horse power. Assume the engine operates at maximum efficiency.Solution)(25.6)(7466010427390)2590(24m S W tWP StQ T T T W H H L H ===⨯⨯+-=-=2.2 A refrigerator is operated by 0.25 hp motor. If the interior of the box is to be maintained at -20℃ ganister a maximum exterior temperature of 35℃, what the maximum heat leak (in watts) into the box that can be tolerated if the motor runs continuously? Assume the coefficient of performance is 75% of the value for a reversible engine.Solution:)(64374625.02035202734375.0W P P T T T P Q T T T W L LL LH HHLH =⨯⨯+-⨯=-=-=2.3 suppose an electrical motor supplies the work to operate a Carnot refrigerator. The interior of the refrigerator is at 0℃. Liquid water is taken in at 0℃ and converted to ice at 0℃. To convert 1 g of ice to 1 g liquid. △H=334J/g is required. If the temperature outside the box is 20℃, what mass of ice can be produced in one minute by a 0.25 hp motor running continuously? Assume that the refrigerator is perfectly insulated and that the efficiencies involved have their largest possible value.Solution:)(4576033474625.020273g m M m P P T T T P L LLLH ===⨯⨯=-=2.4 under 1 atm pressure, helium boils at 4.126K. The heat of vaporization is 84 J/mol what size motor (in hp) is needed to run a refrigerator that must condense 2 mol of gaseous helium at 4.126k to liquid at the same temperature in one minute? Assume that the ambient temperature is 300K and that the coefficient of performance of the refrigerator is 50% of the maximum possible.Solution:)(52.0)(393'60284216.4216.4300'5.0%50hp W P P T T T P P Q T T T W L L L H LLLH ==⨯⨯-=-==-=2.5 if a fossil fuel power plant operating between 540 and 50℃ provides the electrical power to run a heat pump that works between 25 and 5℃, what is the amount of heat pumped into the house per unit amount of heat extracted from the power plant boiler.(a) assume that the efficiencies are equal to the theoretical maximum values(b) assume the power plant efficiency is 70% of maximum and that coefficient of performance of the heat pumpis 10% of maximum(c) if a furnace can use 80% of the energy in fossil foe to heat the house would it be more economical in terms ofoverall fissile fuel consumption to use a heat pump or a furnace ? do the calculations for cases a and b solution:1,2,2,1,212,2,2,2,21,1,1,1,198.82527352527354050540)(H H H H H H L H H H L H P P P P P P P T T T P P T T T P a =+-=+-=-=-= .,)(6286.0)(1,2,not is b ok is a c PP b H H =2.6 calculate △U and △S when 0.5 mole of liquid water at 273 K is mixed with 0.5 mol of liquid water at 373 K and the system is allowed to reach equilibrium in an adiabatic enclosure. Assume that C p is 77J /(mol K) from 273K to 373K Solution:)/(933.0)273323ln(5.0)373323ln(5.0)ln()ln()(02211K J C C T T C n T T C n S J U P P E P E P =+=+=∆=∆2.7 A modern coal burning power plant operates with a steam out let from the boiler at 540℃ and a condensate temperature of 30℃.(a) what is the maximum electrical work that can be produced by the plant per joule of heat provided to theboiler?(b) How many metric tons (1000kg) of coal per hour is required if the plant out put is to be 500MW (megawatts).Assume the maximum efficiency for the plant. The heat of combustion of coal is 29.0 MJ/k g(c) Electricity is used to heat a home at 25℃ when the out door temperature is 10℃ by passing a currentthrough resistors. What is the maximum amount of heat that can be added to the home per kilowatt-hour of electrical energy supplied?Solution:)(3.69)(6937136005000.29)()(89.013054030540)(ton kg m T T T mb J Q T T T W a LH LH H L H ==⨯=-=+-=-= )(9.191102525273)(J Q Q T T T W c H HHLH =-+=-=2.8 an electrical resistor is immersed in water at the boiling temperature of water (100℃) the electrical energy input into the resistor is at the rate of one kilowatt(a) calculate the rate of evaporation of the water in grams per second if the water container is insulated that is noheat is allowed to flow to or from the water except for that provided by the resistor(b) at what rate could water could be evaporated if electrical energy were supplied at the rate of 1 kw to a heatpump operating between 25 and 100℃data for water enthalpy of evaporation is 40000 J/mol at 100℃; molecular weight is 18g/mol; density is 1g/cm 3solution:)(23.2,2510027310010004000018)()(45.0,10004000018)(g m m b g m ma =-+===2.9 some aluminum parts are being quenched (cooled rapidly ) from 480℃ to -20℃ by immersing them in abrine , which is maintained at -20℃ by a refrigerator. The aluminum is being fed into the brine at a rate of one kilogram per minute. The refrigerator operates in an environment at 30℃; that is the refrigerator may reject heat at 30℃. what is them minus power rating in kilowatts, of motor required to operate the refrigerator?Data for aluminum heat capacity is 28J/mol K; Molecular weight 27g/mol Solution:)(5.102)(102474202732030)20480(28271000kW W P P T T T P P L L L L H W L ==---=-=--⨯=2.10 an electric power generating plant has a rated output of 100MW. The boiler of the plant operates at 300℃. The condenser operates at 40℃(a) at what rate (joules per hour) must heat be supplied to the boiler?(b) The condenser is cooled by water, which may under go a temperature rise of no more than 10℃. Whatvolume of cooling water in cubic meters per hour, is require to operate the plant?(c) The boiler tempeture is to be raised to 540℃,but the condensed temperature and electric output will remainthe same. Will the cooling water requirement be increased, decreased, or remain the same?Data heat capacity 4.184, density 1g/cm 3 Solution: )(109.7)(102.21040300273300)(1188J t P Q W P T T T P a H H L H H H ⨯==⨯=-+=-= )(1003.1184.41010)(103.4)(34611m V Q V J Q b LL ⨯==⨯⨯⨯⨯= noW P T T T P c L H H H )(10626.11040540273540)(88⨯=-+=-= 2.11 (a) Heat engines convert heat that is available at different temperature to work. They have been several proposals to generate electricity y using a heat engine that operate on the temperature differences available at different depths in the oceans. Assume that surface water is at 20℃, that water at a great depth is at 4℃, and that both may be considered to be infinite in extent. How many joules of electrical energy may be generated for each joule of energy absorbed from surface water? (b) the hydroelectric generation of electricity use the drop height of water as the energy source. in a particular region the level of river drops from 100m above sea level to 70m above the sea level . what fraction of the potential energy change between those two levels may be converted into electrical energy? how much electrical energy ,in kilowatt-hours, may be generated per cubic meter of water that undergoes such a drop? Solution: )/(1006.136001000)()(055.0127320420)(6h kW h mg P b J Q T T T W a H H L H ⨯=⨯∆==+-=-=2.12 a sports facility has both an ice rink and a swimming pool. to keep the ice frozen during the summer requires the removal form the rink of 105 KJ of thermal energy per hour. It has been suggested that this task be performed by a thermodynamic machine, which would be use the swimming pool as the high temperature reservoir. The ice in the rink is to be maintain at a temperature of –15℃, and the swimming pool operates at 20℃, (a) what is the theoretical minimum power, in kilowatts, required to run the machine? (b) how much heat , in joule per hour , would be supplied t the pool by this machine?Solution: )(1014.1101527320273)()(77.33600/10152731520)(555kJ Q b kW P T T T P a H L L L H ⨯=-+==-+=-=2.13solution:)/(81.6810ln 314.877.45277.6282.4)/(152940)()/(67.4977.45277.6282.4)()/(152940)(22)(2molK cal S mol cal H d molK cal S c mol cal H b AlNN Al a -=+-⨯-⨯=∆=∆-=-⨯-⨯=∆=∆=+ 2.14 solution:)/(2257412000)27340273ln 184.4273336263273ln1.2()(400,010,K J dT T C T H dT T C m S WATER P m m ICE P =+++=+∆+=∆⎰⎰-2.15)(70428)(2896100077773002J W J Q T T T W L L L H ==-=-= 2.16)(4.3719))2.4300(314.85.13.83(3002.4300)(7.58663.832.42.4300J Q T T T W J Q T T T W H H L H L L L H =-⨯+-=-==-=-=2.17 yes d Q c K J P P nR S b J pdV n W Q OU T a )(0)()/(1.1910ln 314.81ln)()(570410ln 298314.810)(0==⨯⨯==∆=⨯⨯=-=-==∆=∆⎰ 2.18)(1222335273020********g m m m T T T L L H =-=-=⨯ Property Relations 1. At -5︒C, the vapor pressure of ice is 3.012mmHg and that of supercooled liquid water is 3.163mmHg. The latent heat of fusion of ice is 5.85kJ/mol at -5︒C. Calculate ∆G and ∆S per mole for the transition of from water to ice at -5︒C. (3.2, 94) Solution: mol J P P RT G waterO H iceO H /9.1089523.0ln 268314.8163.3012.3ln )5273(314.8ln ,,22-=⨯⨯=-⨯==∆mol J H /1085.53⨯=∆)/(23.22268)9.108(5850K mol J T G H S S T H G ⋅=--=∆-∆=∆∴∆-∆=∆ 2. (1) A container of liquid lead is to be used as a calorimeter to determine the heat of mixing of two metals, A andB. It has been determined by experiment that the “heat capacity ” of the bath is 100cal/︒C at 300︒C. With the bath originally at 300︒C, the following experiments are performed;(2) A mechanical mixture of 1g of A and 1g of B is dropped into the calorimeter. A and B were originally at 25︒C. When the two have dissolved, the temperature of the bath is found to have increased 0.20︒C. 2. Two grams of a 50:50(wt.%) A-B alloy at 25︒C is dropped similarly into the calorimeter. The temperature decreases 0.40︒C. (a) What is the heat of mixing of the 50:50 A-B alloy (per gram of alloy)? (b) To what temperature does it apply ? (3.5, 94)Solution: mol J K cal C bath P /418/100,==(a) g cal T C Q bath P /102/2.01002/,=⨯=∆=This is the heat of mixing.(b) The heat capacity of C P, alloy : )/(072.06.27424.0100)254.0300(2,,K g cal TC C bath P alloy P ⋅=⨯⨯=--⨯∆⨯=Assuming that the calorimeter can be applied to the maximum of T ︒C, the for mixing to form 1 gram of alloy:10)'300(,1+-=T C Q bath P , )'(,2T T C Q alloy P -⋅=, 21Q Q =)'(10)'300(,,T T C T C alloy P bath P -=+-3. The equilibrium freezing point of water is 0︒C. At that temperature the latent heat of fusion of ice (the heat required to melt the ice) is 6063J/mol. (a) What is the entropy of fusion of ice at 0︒C ? (b) What is the change of Gibbs free energy for ice →water at 0︒C?(c) What is the heat of fusion of ice at -5︒C ? C P(ice) = 0.5 cal/(g. ︒C); C P(water) = 1.0 cal/(g. ︒C). (d) Repeat parts a and b at -5︒C. (3.6, p94)Solution: (a) At 0︒C, ∆G =0, ∴ T m ∆S = ∆H)./(09.222736030K mol J T H S m ==∆=∆(b) At 0︒C, ∆G =0© )./(62.37)./(1818.45.0)./(5.0,K mol J K mol J K g cal C ice P =⨯⨯==)./(24.75)./(1818.40.1)./(0.1,K mol J K mol J K g cal C water P =⨯⨯==a reversible process can be designed as follows to do the calculation:。
“材料热力学”补充习题参考答案
“材料热力学”补充习题参考答案教材各章习题参考答案 (魏)3.2 ΔG = -108.9 J/mol; ΔS = -21.42 J/(mol.K) 3.6 (a )22.09/(.)S J mol K ?=；(b) At 0?C, ?G =0;(c) ?H = 5841.9 J;(d) ?S =21.39J /(mol.K)，?G = 109.38 J/mol4.1 (a ) 2898.28J/mol; ( b ) No; ( c ) 345 J/mol; ( d ) 14939 atm; ( e )4921 J/mol 4.2 ( a ) 272.8K; ( b ) Pa P 610345?≈? ; ( c ) 249.46K4.3 1202K4.4 P=5.73?10-6 atm 4.5 0.16P 4.708.10430685ln +-=TP4.8 ( a ) 1180K; ( b ) 695.3K; ( c ) 114.4kJ/mol; ( d ) 7123 J/mol; ( e )4.2J/mol4.9 In the initial state: 4.06 mol %; in the final state:5.3 mol% 4.10 ( a )348 kJ; ( b ) 2.3×10-3Pa ;( c ) “ solution not possible ”; (d ) “solution not possible ”5.1atmp H 0005.0=5.2、atmp o 1221007.1-?=If the error in enthalpy is 500cal, the uncertainty in the pressure calculated is 28.6%, and if the error in enthalpy is -500cal, the uncertainty is -22.1%5.3、(a) T =462K; (b) T = 420K 5.4 (a)atmP O 2621014.1-?=, (b) P O2 =2.28?10-10 atm., (c) The equilibriumoxygen pressure remains the same when the total pressure increases, which means a higher purity level of N 2 .5.5 (a) 略； (b) Pa atm P H 8.181013056.1800019.0)('2=?==; (c) 21.5L Ar is needed to be bubbled into the melt.5.6（a ）0.880.900.920.940.960.981.00 1.02 1.04 1.06 1.087.27.47.67.88.08.28.48.6l n K a1/T , 10-31/K=-=?ooG kJ H1000;50- 66.6kJ(b) Ja = 3 < Ka, the reaction will proceed from left to right, and theatmosphere will not oxidize Ni. 5.7 略5.8. (a) P SiO = 8.1?10-8 (atm) (b) ?H o = 639500J; ?So =334.9J/K(c ) PO2 =10-30 atm 5.9 5.10．JH o72250=?，the reaction is an endothermic one.5.11. (a),166528J Ho=?the reaction is an endothermic one.;(b) At 1168K, the equilibrium pressure of CO2 equals one atmosphere. 5.12 (a) 略， (b) MgCO P P =; (c) T = 2037 K5.13 (a) 略； (b) 13109.2?=K ; (c) ppm 186.05.14 (a) 略； (b) kJ H 52.267=?; (c) K T 1592=5.15 (a) )(106.13atm -?≈; (b) )(1028.210)(2atm P g O H -?=5.16 (a)97.9=K ; (b) atm x 14.4=; (c) if the temperature is increased,the fraction of water reacted will increase since the equilibria constant increases with increasing temperature.6.2 （a ）1.287V;(b) When the water impure, the voltage will go higher; (c)1.219V 6.4 (a) 145.3kJ;(b) The maximum work that could be derived is 702.36kJ; (c) In this case, the maximum work that could be derived is 696.56kJ.)(106.08)(atm Pg u -?=6.5 (a) -6252J/mol; (b) 370.0)(=II Cda ; (c) )(42.3mmHg P Cd =;6.67.87?10-4 V 6.7 (a))(22g Cl Mg MgCl+=(b)Pa P Cl 21'1086.82-?=;(c) 2.485V6.8 (a) PaP O 11'2105.5-?=;(b) Anode: e Ni Ni 2+→Cathode: -→+2222/1Oe O ;(c) 0.757V; (d) 0.261V6.10 (a) )(509.3V E o=;(b) 0.074kJ;(c) 4.1?106J;(d) Y es. In this case, the open circuit voltage is 3.648V;(e) In this case, to keep the temperature constant, 3.92?106J heatshould be removed from the battery per hour. 6.11(a)TGCOAl C O Al o26.3211008.12/322/36232-?=+=+Δ(b) The minimum voltage at which the electrolysis may be carriedout at 1250K is 1.172V .7.1 0.117 atm 7.5 ( a ) ,82.52.52.5BA BA BB T PV V V x x x x x ??=+=-- ?,102.5 2.5 2.5A B A A B A T PV V V x x x x x ??=+=-- ?( b) B A Mx x V5.2=7.72)1(736.0ln Sn Snx --=γ7.8 The maximum solubility of MgF2 in liquid MgCl at 900?Cis 19mol% .7.9 ( a ) 1121K; ( b ) 1. 8 cal/K 9.69.8 Solution:(a) 90 mol%B is the composition of the first solid to form;10 mol % is the composition of the last liquid drop.(b) solid (60 mol%B is the composition) is about 77% ; liquid(15 mol%B is the composition) is 23%9.9 (a) 2900℃, α(12%) (b) 2300℃, liq(95%) (c) 8.2%α(compositionis 24% )+91.8%β(85%)补充习题参考答案（魏）1.ΔS m =19.1J/mol.K, ΔG m = -5740 J/mol, ΔF m = -5740 J/molIsothermally expan ds to a vacuum: w = 0, ΔH m =0 , ΔU m =0,ΔS m =19.1J/mol.K, ΔG m = -5740 J/mol, ΔF m = -5740 J/mol2. 3.68 × 10-3 atm3、x x Br H C Cl H C 406.0;594.05556==Pa p Pa p Br H C Cl H C 26838;744445556==4.5、JGJ GJ G a a exmix idmix mix B A R B R A 5302)5(;6912)4(;1610)3(;788.1;62.1)2(;894.0;81.0)1()()(=?-=?-=?====γγ6. a endothermic one; b. Y es; c JJ EMn EFe 704;1584==μμd;/9363mol J G m mix -=?ePap Pa p Fe Mn 4;1198==S n P b4578-=ω;418.0=Sn a8. a) Mg boils and which makes oS ?more negative, so the slope changes for larger; b) Firstly, we should avoid using metallic material for this purpose since the melting points of metals are mostly too low. Ceramic materials, usually composed of oxides and having high melting points can be chosenThe material should not be reduced by pure silicon at 1600oC. By examing Ellingham diagram, crucibles (坩埚) made of Al 2O 3 .c ) -890kJ /molO2;d ) -170kJ /molFeO; e) -30kJ; f) Pa2110－;g)721063.0/?=pco p COPure Substance as Standard Statepq（b ）I 、II 、IIIIII:AA x RT T p RT T T ln )(ln)()(**+=+=μμμk A9、① 650oC ，1220 oC and 1520 oC ; ② 1480 oC ;③ When the temperature is equal to or higher that 710 oC ;④ 2/100molO kJ G o-=?⑤ 900 oC; ⑥,102/112,1010'25'2=?=-=?=--G Pa PmolO kJ G Pa P O O ,Pap e O 10')(210-=⑦ 510-=K;⑧ 1220oC10、a) -489120+197.06TlnT J/mol; b) 2.89×10-54 ; c)JG 749429=?; Ni is stable under thiscondition, and NiO is not stable; d)p e o 58')(21046.3?= e) fromthe calculation, we found that at 1000oC,Pap e o 58')(21046.3?=. Soat 1000oC, when the oxygen pressure is less than 3.46×1058Pa, Ni is stable and can not be oxidized, and NiO will be reduced to Ni under this condition. 11. a)molkJ G o/2.23=?; b)43.42=eCO CO p p . This is theminimum CO/CO2 ratio required to reduce pure FeO to Fe at 1600oC. c)2.142=eCO CO p p . This is the minimum CO/CO2ratio required to reduce FeO in a slag( 炉渣) to Fe in a metallic iron melt under the given conditions at 1600oC.12.%10?-Ar .3 15 % 100 ) 10。
材料热力学习题解答
《材料热力学》复习思考题解答3. 在1560℃时，C 在液态铁中的活度系数和偏摩尔超额焓由下列式表示： 2l n 0.37711.7c C C X X γ=-++25.415.017.25E C C C H X X =++(K Cal) 其标准态为纯石墨，计算1560℃时液相与石墨平衡的相线的斜率。
解：以石墨为标准态时，C 在液态铁中的化学位为：l n (1)LC CC R T a μμ=+ 石墨 当液相与石墨平衡时，L C Cμμ=石墨。
即ln 0C α=。
又ln ln ln C C C X αγ=+ln ln 0(2)C C X γ∴+=由(2)式得：平衡时0.2067C X =两边取微分得：(ln )(ln )1[](1/)[]0(1/)C C C X T C C C C d T dX dX T X X γγ∂∂++=∂∂ (ln )[](1/)ln ln 1(1/)[()]1()CC X EC C C C C T C TC C CdX H X T d T R X X X X γγγ∂-∂∴==⋅∂∂-++∂∂2(5.415.017.25) 4.1810000.20678.311(723.4)278.6C C CC X X X X ++⨯⨯=-⋅++=- 2C dX T dT=-CdX 又d(1/T)5221278.68.310(1560273)C dX dT T -∴=-==⨯+C dX d(1/T) 1()K - 4. 在1000K 时，A-B 二元溶液中，当0.01B X =时，0.1B a =。
在盛有大量A 的量热计中加入少量的B 组元时，测得吸热7000Cal/mol ，假定2ln ln B A B X γγ=。
求1500K 时，当0.02B X =时，B 组元的活度。
解：在1000K 时，当0.01B X =时，0.1B a =0.1100.01B γ∴== 又022ln ln10ln 2.3490.99B B A X γγ=== 又ln [](1/)ii P H R T γ∂∆=∂15001500010001000l n (1/)BBH d d T Rγ∆∴=⎰⎰1500100011[ln ][ln ]()15001000B B B H R γγ∆∴=+-7000 4.18112.349()8.31150010001.175⨯=+-= 202l n (l n )0.981.175B A B X γγ∴==⨯ 1.128= l n 3.09B γ∴= 3.090.020.0B B B a X γ==⨯=7. 若A-B 二元合金系在液、固态两组元均能无限互溶，且均为理想溶液。
2012硕士《材料热力学与动力学》复习练习题
Question 16
1） 指出各水平线的三相平衡反应 2） w(SiO2)=0.40 的系统(图中 R 点)从 1700C 冷却到 1000C 时的冷却曲线示意图。 注明每一阶段系统有哪些相？发生哪些 变化？指出各阶段的自由度数？ 3） w(SiO2)=0.10 的系统 12 kg，冷却到 1400C 时，液相中含 MnO 多少 kg？ 4） w(SiO2)=0.60 的系统 1500C 以哪些相存在？计算其相对 量。
4
2012 研究生《材料热力学与动力学》复习练习题（10 月 8 日交，手写完成）
Question 1 进行下述过程时，系统的ΔU、ΔH、ΔS和ΔG何者为零？ 1.1 非理想气体的卡诺循环； 1.2 隔离系统中的任意过程； 1.3 在100C，1大气压下1mol水蒸发成水蒸汽； 1.4 绝热可逆过程。 Question 2 1mol 理想气体等容升温到状态 3，求 Q,W,ΔU,ΔH。 若将理想气体先等压膨胀到状态 2，然后再等温（可 逆）压缩到状态 3，求 Q,W,ΔU,ΔH，并与直接从 1 到 3 的途径相比较。
Question 11 导出液相中 Bi 的活度系数的估算公式。
H m T a Bi exp ( 1) RT Tm
其中，熔化热为 H m 纯 Bi 的熔点为 Tm，R 为气体常数。
Question 12 对下列二元相图，指出其中的错误 (用相律说明原因)
2
Question 13
Trouton's定律为表示为:
1 (V1,T1) 2 (V2,T2) V
H vap 90Tb
单位J/mol, 其中Tb为沸点(K), 汞的沸点为630 K. 计算在
298K液态汞的分压. 用Troutons定律估算汞的汽化热.
热力学习题及答案解析
热力学习题及答案解析
热力学学习题及答案解析
热力学是物理学的一个重要分支，研究能量转化和热力学系统的性质。
在学习
热力学的过程中，我们经常会遇到各种热力学学习题，通过解题可以加深对热
力学知识的理解。
下面我们就来看看一些常见的热力学学习题及答案解析。
1. 问题：一个理想气体在等温过程中，体积从V1扩大到V2，求气体对外界所
做的功。
答案解析：在等温过程中，理想气体对外界所做的功可以用以下公式表示：
W = nRTln(V2/V1)，其中n为气体的摩尔数，R为气体常数，T为温度。
根据这
个公式，我们可以计算出气体对外界所做的功。
2. 问题：一个物体从20摄氏度加热到80摄氏度，求其温度变化时吸收的热量。
答案解析：物体温度变化时吸收的热量可以用以下公式表示：Q = mcΔT，其
中m为物体的质量，c为物体的比热容，ΔT为温度变化。
根据这个公式，我们
可以计算出物体温度变化时吸收的热量。
3. 问题：一个热机从高温热源吸收了500J的热量，向低温热源放出了300J的
热量，求该热机的热效率。
答案解析：热机的热效率可以用以下公式表示：η = 1 - Q2/Q1，其中Q1为
热机从高温热源吸收的热量，Q2为热机向低温热源放出的热量。
根据这个公式，我们可以计算出该热机的热效率。
通过以上几个热力学学习题及答案解析，我们可以看到在解题的过程中，需要
灵活运用热力学知识，并且掌握一定的计算方法。
希望通过不断的练习和思考，我们能够更好地理解和掌握热力学知识，提高解题能力。
材料热力学 习题答案
The problems of the first law1. a lead bullet is fired at a frigid surface. At what speed must it travel to melt on impact, if its initial temperature is 25℃ and heating of the rigid surface of the rigid surface is neglected? The melting point of lead is 327℃. The molar heat of fusion of the lead is 4.8kJ/mol. The molar heat capacity C P of lead may be taken as 29.3J/(mol K) (1.1)Solution: )/(5.112.20721]108.4)25327(3.29[2121)(2322s m V v n n WQ nMv mv W H T C n Q Q Q absorb melting p melt increase absorb ==⨯+-⨯===∆+∆=+=2. what is the average power production in watts of a person who burns 2500 kcal of food in a day? Estimate the average additional powder production of 75Kg man who is climbing a mountain at eh rate of 20 m/min (1.2)Solution )/(24560208.975)/(12160602410467000//)(104670001868.4102500sin 3S J t h mg P S J t Q t W P J Q gincrea Burning Burning =⨯⨯=∆==⨯⨯====⨯⨯=3 One cubic decimeter (1 dm 3) of water is broken into droplets having a diameter of onemicrometer (1 um) at 20℃. (1.3)(a) what is the total area of the droplets?(b) Calculate the minimum work required to produce the droplets. Assume that the dropletsare rest (have zero velocity)Water have a surface tension of 72.75 dyn/cm at 20℃ (NOTES: the term surface energy (ene/cm 2) is also used for surface tension dyn/cm)Solution)(25.218)106103(1075.72)(103)101(4)101(34)101(232523263631J S W m nS S Single total =⨯-⨯⨯⨯=∆=⨯=⨯⨯⨯⨯⨯⨯⨯⨯==-+----σππ4.Gaseous helium is to be used to quench a hot piece of metal. The helium is in storage in an insulated tank with a volume of 50 L and a temperature of 25℃, the pressure is 10 atm. Assume that helium is an ideal gas.(a) when the valve is opened and the gas escapes into the quench chamber (pressure=1 atm),what will be the temperature of the first gas to hit the specimen?(b) As the helium flows, the pressure in the tank drops. What will be the temperature of thehelium entering the quench chamber when the pressure in the tank has fallen to 1 atm? (1.4)Solution: )(180118298)(1185.229810101325501010101325)5500(1)()(118)101(298)()(0334.0/00K T T T K RR nC W T b K T P PT T Adiabatic a p C R P=-=∆-==⨯⨯⨯⨯⨯⨯⨯-⨯==∆=⨯==--5 An evacuated (P=0), insulted tank is surrounded by a very large volume (assume infinite volume) of an ideal gas at a temperature T 0. The valve on the tank is opened and the surrounding gas is allowed to flow quickly into the tank until the pressure inside the tank is equals the pressure outside. Assume that no heat flow takes place. What is the final tempeture of the gas in the tank? The heat capacity of the gas, C p and C v each may be assumed to be constant over the temperature rang spanned by the experiment. You answer may be left in terms of C p and C vhint: one way to approach the problem is to define the system as the gas ends up in the tank. (1.5)solution 0/000/00)()(T P P T T P PT T Adiabatic PPC R C R ≈-==6. Calculate the heat of reaction of methane with oxygen at 298K, assuming that the products of reaction are CO 2 and CH 4 (gas)[This heat of reaction is also called the low calorific power of methane] convert the answer into unites of Btu/1000 SCF of methane. SCF means standard cubic feet, taken at 298 and 1atmNOTE: this value is a good approximation for the low calorific powder of natural gas (1.6)DA TA:)()()(224g O H g CO g CH FOR80.5705.9489.17]/[0298---∙∆mol g Kcal Hsolution)1000/(9.2610252103048.01101076.191)/(76.191)89.1780.57205.94()2(22333332982982224422SCF Btu mol g Kcal H H H H H OH CO O CH CH O H CO =⨯⨯⨯⨯⨯=∙=∆+⨯---=∆-∆+∆-=∆+=+-7. Methane is delivered at 298 K to a glass factory, which operates a melting furnace at 1600 K. The fuel is mixed with a quantity of air, also at 298 K, which is 10% in excess of the amount theoretically needed for complete combustion (air is approximately 21% O 2 and 79% N 2) (1.7)(a) Assuming complete combustion, what is the composition of the flue gas (the gasfollowing combustion)?(b) What is the temperature of the gas, assuming no heat loss?(c) The furnace processes 2000kg of glass hourly, and its heat losses to the surroundingsaverage 400000 kJ/h. calculate the fuel consumption at STP (in m 3/h) assuming that for gas H 1600-H 298=1200KJ/KG(d) A heat exchanger is installed to transfer some of the sensible heat of the flue gas to thecombustion air. Calculate the decrease in fuel consumption if the combustion air is heated to 800KDA TA STP means T=298K, P=1atm22224O N O H CO CH for 2.82.89.117.1316)/(C mol cal C P ∙Solution)(210448.1125.9100076.191298)/(25.9)]87.012.72(2.843.179.1171.87.13[01.0)(%87.0%%12.72%%43.17%2%%71.8)11.1(221791.1231%22)(0,,222222224K T T T C mol cal X C C b O N CO O H CO O H CO O CH a i i p p p =⨯⨯+=∆+=∙=+⨯+⨯+⨯=======-⨯+⨯⨯+=+=+∑)/(1644)0224.011868.448.11)8001600(48.1125.9189570(102800000)/(189570)298800)](48.1187.8)48.1125.9[(100076.191)()/(87.848.11/]211002.22.816[)()/(3214)0224.011868.448.11)2981600(48.1125.9100076.191(102800000)/(280000040000020001200)(33min ,,,,298,,33min h m V mol g cal dTn C n C H H C mol cal X C C d h m V h KJ P C gConsu i i r p i i p p i i p r p g Consu =⨯⨯-⨯-⨯=∙=-⨯-⨯-⨯=--∆=∆∙=⨯⨯+===⨯⨯-⨯-⨯⨯==+⨯=⎰∑∑∑8.In an investigation of the thermodynamic properties of a-manganese, the following heat contents were determined: H 700-H 298=12113 J/(g atom) H 1000-H 298=22803 J/(g atom)Find a suitable equation for H T -H 298 and also for C P as a function of temperature in the form (a+bT) Assume that no structure transformation takes place in the given tempeture rang. (1.8)Solution )298(0055.0)298(62.35011.062.35011.062.3522803)2981000(2)2981000(12113)298700(2)298700(]2[2229822222982---=∆-=-===-+-=-+-+=+==∆⎰⎰T T H TC b a ba ba T baT bTdT a dT C H TP T P9.A fuel gas containing 40% CO, 10% CO 2, and the rest N 2 (by volume) is burnt completely with air in a furnace. The incoming and ongoing temperatures of the gases in the furnace are 773K and 1250K,respectively. Calculate (a) the maximum flame temperature and (b) heat supplied to the furnace per cu. ft of exhaust gas (1.9)molJ Hmol J H CO f CO f /393296/1104580,298,0,298,2-=∆-=∆)/(10184.403.29)/(1067.11010.492.19)/(1037.81020.935.44)/(1042.01097.345.283,253,253,253,222molK J T C molK J T T C molK J T T C molK J T T C N P O P CO P CO P -------⨯+=⨯-⨯+=⨯-⨯+=⨯-⨯+=Solution?0)499.0321.018.1()1067.01019.277.28(28.282831067.01038.477.289.0)1019.01058.528.33(2.0282838)()/(1019.01058.528.33722.0278.0)/(1067.01038.477.281.065.005.02.0)()/(282838110458393296%2.72%8.27%10%65%5%20)4/(1122298127332981523733253253298,,,,298,253,,,,,253,,,,,,,0,298,0,298,298,22222222222222==+--⨯+⨯++⨯=⨯-⨯++⨯⨯-⨯+-⨯=--∆=∆⨯-⨯+=+==⨯-⨯+=+++===-=∆-∆=∆========+-----------⎰⎰⎰∑∑⎰∑∑∑∑T T T T T T T dT T T dTT T dT n C n C n H H molK J T T C C n C C molK J T T C C C C n C C a mol J n Hn H H N CO production O N CO CO reation then O N air mole need fuel mole when CO O CO T TT i i r p i i p p i i N P CO P i i p p r p O P N P CO P CO P i i p p r p i pf i rf idTT T Q dT T T Q b T T T T T T T dT T T dTT T dT n C n C n H H T TT i i r p i i p p i i 9.0)1019.01058.528.33(2.02828389.0)1019.01058.528.33(2.0282838)(0)499.0321.018.1()1067.01019.277.28(28.282831067.01038.477.289.0)1019.01058.528.33(2.0282838)(253125029812502982531250298125029829812125029815231250253253298,,,,298,⨯⨯-⨯++⨯-=⨯⨯-⨯++⨯-===+--⨯+⨯++⨯=⨯-⨯++⨯⨯-⨯+-⨯=--∆=∆-----------⎰⎰⎰⎰⎰∑∑⎰10. (a) for the reaction 2221CO O CO →+,what is the enthalpy of reaction (0H ∆) at 298 K ?(b) a fuel gas, with composition 50% CO, 50% N 2 is burned using the stoichiometric amount of air. What is the composition of the flue gas?(c) If the fuel gas and the air enter there burner at 298 K, what is the highest temperaturethe flame may attain (adiabatic flame temperature)? DA TA :standard heats of formation f H ∆ at 298 K (1.10))/(393000)/(1100002mol J CO mol J CO -=-=Heat capacities [J/(mol K)] to be used for this problem N 2=33, O 2=33, CO=34, CO 2=57 Solution)(21100)298)(39889.0(222.02830000)/(3975.03325.057)/(33111.034222.033666.033)(%,75%%,251.111002.22%%1.11%%,6.66%%,2.222.0/25.015.0%)()/(283000393000110000)(,0,,,,,,22220,298,0,298,0K T T dT C n H H K mol J X C C K mol J X C C C N CO product O N CO fuel b mol J n H n H H a P p p i P r i P r i P p i P p i P f i r f ==-⨯-⨯=-∆=∆∙=⨯+⨯==∙=⨯+⨯+⨯====-====+==+-=∆-∆=∆⎰∑∑∑∑11.a particular blast furnace gas has the following composition by (volume): N 2=60%, H 2=4, CO=12%, CO 2=24%(a) if the gas at 298K is burned with the stochiometric amount of dry air at 298 K, what is the composition of the flue gas? What is the adiabatic flame temperature? (b) repeat the calculation for 30% excess combustion air at 298K(C)what is the adiabatic flame temperature when the blast furnace gas is preheated to 700K (the dry air is at 298K)(d) suppose the combustion air is not dry ( has partial pressure of water 15 mm Hg and a total pressure of 760 mm Hg) how will the flame temperature be affected? DA TA(k J/mol) （1.11）2CO CO FOR513.393523.110)/(--∆m o lkJ H f 2222,)(O N g O H CO CO FOR34505733]/[K mol J C P ∙Solution)(1052)(75438286370])295.03450(241604[026.0])335.03457(110523393513[079.0])([%8.66%%,8.6%%,6.2%%,8.15%%,9.72.0/83.110012%)()(1122)(82538313430])295.03450(241604[029.0])335.03457(110523393513[086.0])([%7.65%%,7.5%%,9.2%%,1.17%%,6.82.0/810012%2121)(,,,,,,,02222,,,,,,,0222222222K T K T T n C T T X C dT n C n C H x H N O H CO CO b K T K T T n C T T X C dT n C n C H x H N O H CO CO OH O H CO O CO a i i r P ii P i i r P i i p P i i i i r P ii P i i r P i i p P i i ===∆=∆-∆-⨯--+∆-⨯---=+--∆=∆=====⨯+====∆=∆-∆-⨯--+∆-⨯---=+--∆=∆=====+=→+→+∑∑∑⎰∑∑∑∑∑⎰∑∑)(1419),(11213842594034286.0)402(2.39714.0])295.03450(241604[029.0])335.03457(110523393513[086.0)3(K T K T T T T T H ===∆=∆⨯--∆⨯-∆-⨯--+∆-⨯---=∆12．A bath of molten copper is super cooled to 5℃ below its true melting point. Nucleation of solid copper then takes place, and the solidification proceeds under adiabatic conditions. What percentage of the bath solidifies?DATA: Heat of fusion for copper is 3100 cal/mol at 1803℃(the melting point of copper) C P,L =7.5(cal/mol ℃), C P,S =5.41+(1.5*10-3T )(cal/mol ℃) （1.12） Solution)/(310355.75.0)17981803(105.1541.5310002231798,1798,17981803,18031798,1803,mol cal H H dT C dT C HL S SL L P S P LS =⨯-⨯-⨯+⨯+==+++-⎰⎰13．Cuprous oxide (Cu 2O) is being reduced by hydrogen in a furnace at 1000K, (a)write the chemical reaction for the reduced one mole of Cu 2O(b)how much heat is release or absorbed per mole reacted? Given the quantity of heat and state whether heat is evolved (exothermic reaction) or absorbed (endothermic reaction) DATA: heat of formation of 1000K in cal/mol Cu 2O=-41900 H 2O=-59210 （1.13） solution)/(173104190059210222mol cal H OH Cu H O Cu =-=∆+=+,exothermic reaction14. (a) what is the enthalpy of pure, liquid aluminum at 1000K?(b) an electric resistance furnace is used to melt pure aluminum at the rate of 100kg/h. the furnace is fed with solid aluminum at 298K. The liquid aluminum leaves the furnace at 1000K. what is the minimum electric powder rating (kW) of furnace.DATA : For aluminum : atomic weight=27g/mol, C p,s =26(J/molK), C p,L =29(J/molK), Melting point=932K, Heat of fusion=10700J/mol (1.14)Solution )(28.0)(7.2793600110002727184)/(2718410700)9321000(29)298932(261000932,932298,1000,kW W P mol J H dT C dT C H SLL P S P l ==⨯⨯==+-⨯+-⨯=++=⎰⎰15 A waste material (dross from the melting of aluminum) is found to contain 1 wt% metallic aluminum. The rest may be assumed to aluminum oxide. The aluminum is finely divided and dispersed in the aluminum oxide; that is the two material are thermally connected.If the waster material is stored at 298K. what is the maximum temperature to which it may rise if all the metallic aluminum is oxidized by air/ the entire mass may be assumed to rise to the same temperature. Data : atomic weight Al=27g/mol, O=16g/mol, C p,s,Al =26(J/molK), C p,s,Al2O3=104J/mol, heat formation of Al 2O 3=-1676000J/mol(1.15)Solution;)(600)(3021041029927275.116122711676000K T K T T ==∆∆⨯⨯++⨯⨯=⨯⨯16 Metals exhibit some interesting properties when they are rapidly solidified from the liquid state. An apparatus for the rapid solidification of copper is cooled by water. In the apparatus, liquid copper at its melting point (1356K) is sprayed on a cooling surface, where it solidified and cools to 400K. The copper is supplied to the apparatus at the rate of one kilogram per minute. Cooling water is available at 20℃, and is not allowed to raise above 80℃. What is the minimum flow rate of water in the apparatus, in cubic meters per minute? DATA; for water: C p =4.184J/g k, Density=1g/cm 3; for copper: molecular weight=63.54g/mol C p =7cal/mol k, heat of fusion=3120 cal/mol (1.16)Solution:min)/(10573.2)2080(1min /min54.631000)]4001356(73120[min /33m V VQ Q Water Copper -⨯=-=⨯⨯-⨯+=17 water flowing through an insulated pipe at the rate of 5L/min is to be heated from 20℃ to 60℃ b an electrical resistance heater. Calculate the minimum power rating of the resistance heater in watts. Specify the system and basis for you calculation. DATA; For water C p =4.184J/g k, Density=1g/cm 3 (1.17) Solution: )(139476010005)2060(184.4W W =⨯⨯-⨯=18 The heat of evaporation of water at 100℃ and 1 atm is 2261J/mol (a) what percentage of that energy is used as work done by the vapor?(b)if the density of water vapor at 100℃ and 1 atm is 0.597kg/m 3 what is the internal energy change for the evaporation of water? (1.18)Solution: )/(375971822613101%6.71822613101%)/(31010224.0273373101325mol J Q W U mol J V P =⨯+-=+=∆=⨯==⨯⨯=∆19 water is the minimum amount of steam (at 100℃ and 1 atm pressure) required to melt a kilogram of ice (at 0℃)? Use data for problem 1.20 (1.19) Solution )(125,3341000)10018.42261(g m m =⨯=⨯+20 in certain parts of the world pressurized water from beneath the surface of the earth is available as a source of thermal energy. To make steam, the geothermal water at 180℃ is passed through a flash evaporator that operates at 1atm pressure. Two streams come out of the evaporator, liquid water and water vapor. How much water vapor is formed per kilogram of geothermal water? Is the process reversible? Assume that water is incompressible. The vapor pressure of water at 180℃ is 1.0021 Mpa( about 10 atm) Data: C P,L =4.18J/(g k), C P,v =2.00J/(g k), △H V =2261J/g, △H m =334 J/g (1.20) Solution:leirreversib g x x x )(138),1000(8018.4)8018.48022261(=-⨯⨯=⨯-⨯+The problems of the second law1 The solar energy flux is about 4J cm 2/min. in no focusing collector the surface temperature can reach a value of about 900℃. If we operate a heat engine using the collector as the heat source and a low temperature reservoir at 25℃, calculate the area of collector needed if the heat engine is to produce 1 horse power. Assume the engine operates at maximum efficiency. (2.1)Solution )(664.0)(74660104273900)25900(24m S W tWP StQ T T T W H H L H ===⨯⨯+-=-=2 A refrigerator is operated by 0.25 hp motor. If the interior of the box is to be maintained at -20℃ ganister a maximum exterior temperature of 35℃, what the maximum heat leak (in watts) into the box that can be tolerated if the motor runs continuously? Assume the coefficient of performance is 75% of the value for a reversible engine. (2.2)Solution:)(114474625.02035202733475.0%75W P P T T T P Q T T T W L LLLH HHLH =⨯⨯+-⨯=-=-=3 suppose an electrical motor supplies the work to operate a Carnot refrigerator. The interior of the refrigerator is at 0℃. Liquid water is taken in at 0℃ and converted to ice at 0℃. To convert 1 g of ice to 1 g liquid. △H=334J/g is required. If the temperature outside the box is 20℃, what mass of ice can be produced in one minute by a 0.25 hp motor runningcontinuously? Assume that the refrigerator is perfectly insulated and that the efficiencies involved have their largest possible value. (2.3)Solution: )(4576033474625.020273g m M m P P T T T P L LLLH ===⨯⨯=-=4 under 1 atm pressure, helium boils at 4.126K. The heat of vaporization is 84 J/mol what size motor (in hp) is needed to run a refrigerator that must condense 2 mol of gaseous helium at 4.126k to liquid at the same temperature in one minute? Assume that the ambient temperature is 300K and that the coefficient of performance of the refrigerator is 50% of the maximum possible. (2.4)Solution: )(52.0)(393'60284216.4216.4300'5.0%50hp W P P T T T P P Q T T T W L L L H LLLH ==⨯⨯-=-==-= 5 if a fossil fuel power plant operating between 540 and 50℃ provides the electrical powerto run a heat pump that works between 25 and 5℃, what is the amount of heat pumped into the house per unit amount of heat extracted from the power plant boiler. (a) assume that the efficiencies are equal to the theoretical maximum values(b) assume the power plant efficiency is 70% of maximum and that coefficient ofperformance of the heat pump is 10% of maximum(c) if a furnace can use 80% of the energy in fossil foe to heat the house would it be moreeconomical in terms of overall fissile fuel consumption to use a heat pump or a furnace ? do the calculations for cases a and b (2.5)solution:1,2,2,1,212,2,2,2,21,1,1,1,198.82527352527354050540)(H H H H H H L H H H L H P P P P P P P T T T P P T T T P a =+-=+-=-=-=.,)(6286.0)(1,2,not is b ok is a c P P b H H =6 calculate △U and △S when 0.5 mole of liquid water at 273 K is mixed with 0.5 mol of liquid water at 373 K and the system is allowed to reach equilibrium in an adiabaticenclosure. Assume that C p is 77J /(mol K) from 273K to 373K (2.6) Solution:)/(933.0)273323ln(5.0)373323ln(5.0)ln()ln()(02211K J C C T T C n T T C n S J U P P E P E P =+=+=∆=∆ 7 A modern coal burning power plant operates with a steam out let from the boiler at 540℃and a condensate temperature of 30℃.(a) what is the maximum electrical work that can be produced by the plant per joule of heatprovided to the boiler?(b) How many metric tons (1000kg) of coal per hour is required if the plant out put is to be500MW (megawatts). Assume the maximum efficiency for the plant. The heat of combustion of coal is 29.0 MJ/k g(c) Electricity is used to heat a home at 25℃ when the out door temperature is 10℃ bypassing a current through resistors. What is the maximum amount of heat that can be added to the home per kilowatt-hour of electrical energy supplied? (2.7)Solution:)(3.69)(6937136005000.29)()(89.013054030540)(ton kg m T T T mb J Q T T T W a LH LH H L H ==⨯=-=+-=-=)(9.191102525273)(J Q Q T T T W c H HHLH =-+=-=8 an electrical resistor is immersed in water at the boiling temperature of water (100℃) the electrical energy input into the resistor is at the rate of one kilowatt(a) calculate the rate of evaporation of the water in grams per second if the water containeris insulated that is no heat is allowed to flow to or from the water except for that provided by the resistor(b) at what rate could water could be evaporated if electrical energy were supplied at therate of 1 kw to a heat pump operating between 25 and 100℃data for water enthalpy of evaporation is 40000 J/mol at 100℃; molecular weight is 18g/mol; density is 1g/cm 3 (2.8)solution:)(23.2,2510027310010004000018)()(45.0,10004000018)(g m m b g m ma =-+===9 some aluminum parts are being quenched (cooled rapidly ) from 480℃ to -20℃ byimmersing them in a brine , which is maintained at -20℃ by a refrigerator. The aluminum is being fed into the brine at a rate of one kilogram per minute. The refrigerator operates in an environment at 30℃; that is the refrigerator may reject heat at 30℃. what is them minuspower rating in kilowatts, of motor required to operate the refrigerator? Data for aluminum heat capacity is 28J/mol K; Molecular weight 27g/mol (2.9)Solution:)(5.102)(102474202732030)20480(28271000kW W P P T T T P P L L L L H W L ==---=-=--⨯=10 an electric power generating plant has a rated output of 100MW. The boiler of the plantoperates at 300℃. The condenser operates at 40℃(a) at what rate (joules per hour) must heat be supplied to the boiler?(b) The condenser is cooled by water, which may under go a temperature rise of no morethan 10℃. What volume of cooling water in cubic meters per hour, is require to operate the plant?(c) The boiler tempeture is to be raised to 540℃,but the condensed temperature and electricoutput will remain the same. Will the cooling water requirement be increased, decreased, or remain the same?Data heat capacity 4.184, density 1g/cm 3 (2.10)Solution: )(109.7)(102.21040300273300)(1188J t P Q W P T T T P a H H L H H H ⨯==⨯=-+=-=)(1003.1184.41010)(103.4)(34611m V Q V J Q b L L ⨯==⨯⨯⨯⨯=noW P T T T P c L H H H )(10626.11040540273540)(88⨯=-+=-=11 (a) Heat engines convert heat that is available at different temperature to work. Theyhave been several proposals to generate electricity y using a heat engine that operate on the temperature differences available at different depths in the oceans. Assume that surface water is at 20℃, that water at a great depth is at 4℃, and that both may be considered to be infinite in extent. How many joules of electrical energy may be generated for each joule of energy absorbed from surface water? (b) the hydroelectric generation of electricity use the drop height of water as the energy source. in a particular region the level of river drops from 100m above sea level to 70m above the sea level . what fraction of the potential energy change between those two levels may be converted into electrical energy? how much electrical energy ,in kilowatt-hours, may be generated per cubic meter of water that undergoes such a drop? (2.11)Solution:)/(1006.136001000)()(055.0127320420)(6h kW hmg P b J Q T T T W a H H L H ⨯=⨯∆==+-=-=12 a sports facility has both an ice rink and a swimming pool. to keep the ice frozen during the summer requires the removal form the rink of 105 KJ of thermal energy per hour. It has been suggested that this task be performed by a thermodynamic machine, which would be use the swimming pool as the high temperature reservoir. The ice in the rink is to be maintain at a temperature of –15℃, and the swimming pool operates at 20℃, (a) what is the theoretical minimum power, in kilowatts, required to run the machine? (b) how much heat , in joule per hour , would be supplied t the pool by this machine? (2.12)Solution:)(1014.1101527320273)()(77.33600/10152731520)(555kJ Q b kW P T T T P a H L L L H ⨯=-+==-+=-=13solution:)/(81.6810ln 314.877.45277.6282.4)/(152940)()/(67.4977.45277.6282.4)()/(152940)(22)(2molK cal S mol cal H d molK cal S c mol cal H b AlNN Al a -=+-⨯-⨯=∆=∆-=-⨯-⨯=∆=∆=+14solution:)/(2257412000)27340273ln 184.4273336263273ln1.2()(40,010,K J dT T C T H dT T C m S WATER P m mICE P =+++=+∆+=∆⎰⎰- 15)(70428)(2896100077773002J W J Q T T T W L L L H ==-=-=16)(4.3719))2.4300(314.85.13.83(3002.4300)(7.58663.832.42.4300J Q T T T W J Q T T T W H H L H L L L H =-⨯+-=-==-=-=17yesd Q c K J PPnR S b J pdV n W Q OU T a )(0)()/(1.1910ln 314.81ln )()(570410ln 298314.810)(0==⨯⨯==∆=⨯⨯=-=-==∆=∆⎰18)(122233527302033560500g m m m T T T L L H =-=-=⨯教材各章习题参考答案 (魏)3.2 ΔG = -108.9 J/mol; ΔS = -21.42 J/(mol.K)3.6 (a ) 22.09/(.)S J mol K ∆=；(b) At 0︒C, ∆G =0; (c) ∆H = 5841.9 J;(d) ∆S =21.39J /(mol.K)，∆G = 109.38 J/mol4.1 (a ) 2898.28J/mol; ( b ) No; ( c ) 345 J/mol; ( d ) 14939 atm; ( e )4921 J/mol4.2 ( a ) 272.8K; ( b ) Pa P 610345⨯≈∆ ; ( c ) 249.46K 4.3 1202K4.4 P=5.73⨯10-6 atm 4.5 0.16P4.7 08.10430685ln +-=TP 4.8 ( a ) 1180K; ( b ) 695.3K; ( c ) 114.4kJ/mol; ( d ) 7123 J/mol; ( e )4.2J/mol4.9 In the initial state: 4.06 mol %; in the final state:5.3 mol% 4.10 ( a )348 kJ; ( b ) 2.3×10-3Pa ;( c ) “ solution not possible ”; (d ) “solution not possible ”5.1 atm p H 0005.0= 5.2、atmp o 1221007.1-⨯=If the error in enthalpy is 500cal, the uncertainty in the pressure calculated is 28.6%, and if the error in enthalpy is -500cal, the uncertainty is -22.1%5.3、(a) T =462K; (b) T = 420K5.4 (a) atm P O 2621014.1-⨯=, (b) P O2 =2.28⨯10-10 atm., (c) The equilibriumoxygen pressure remains the same when the total pressure increases, which means a higher purity level of N 2 .5.5 (a) 略； (b) Pa atm P H 8.181013056.1800019.0)('2=⨯==; (c) 21.5L Ar isneeded to be bubbled into the melt.5.6（a ）l n K a1/T, 10-31/K=∆-=∆o o G kJ H 1000;50- 66.6kJ(b) Ja = 3 < Ka, the reaction will proceed from left to right, and theatmosphere will not oxidize Ni. 5.7 略5.8. (a) P SiO = 8.1⨯10-8 (atm) (b) ∆H o = 639500J; ∆So =334.9J/K (c ) PO2 =10-30 atm 5.9 5.10．J H o72250=∆，the reaction is an endothermic one.5.11. (a),166528J H o =∆ the reaction is an endothermic one.; (b) At 1168K, the equilibrium pressure of CO2 equals one atmosphere.)(106.08)(atm Pg u -⨯=5.12 (a) 略 ， (b) Mg CO P P =; (c) T = 2037 K 5.13 (a) 略； (b) 13109.2⨯=K ; (c) ppm 186.0 5.14 (a) 略； (b) kJ H 52.267=∆; (c) K T 1592= 5.15 (a) )(106.13atm -⨯≈; (b) )(1028.210)(2atm P g O H -⨯=5.16 (a) 97.9=K ; (b) atm x 14.4=; (c) if the temperature is increased, the fraction of water reacted will increase since the equilibria constant increases with increasing temperature.6.2 （a ）1.287V;(b) When the water impure, the voltage will go higher; (c) 1.219V 6.4 (a) 145.3kJ;(b) The maximum work that could be derived is 702.36kJ; (c) In this case, the maximum work that could be derived is696.56kJ.6.5 (a) -6252J/mol; (b) 370.0)(=II Cd a ; (c) )(42.3mmHg P Cd =; 6.67.87⨯10-4 V 6.7 (a))(22g Cl Mg MgCl +=(b) Pa P Cl 21'1086.82-⨯=;(c) 2.485V6.8 (a) Pa P O 11'2105.5-⨯=;(b) Anode: e Ni Ni 2+→Cathode: -→+2222/1O e O ;(c) 0.757V; (d) 0.261V6.10 (a) )(509.3V E o=;(b) 0.074kJ;(c) 4.1⨯106J;(d) Yes. In this case, the open circuit voltage is 3.648V;(e) In this case, to keep the temperature constant, 3.92⨯106J heatshould be removed from the battery per hour. 6.11(a) TG CO Al C O Al o 26.3211008.12/322/36232-⨯=+=+Δ(b) The minimum voltage at which the electrolysis may be carriedout at 1250K is 1.172V .7.1 0.117 atm 7.5 ( a ) ,82.5 2.5 2.5B A BA BB T PV V V x x x x x ⎛⎫∂=+=--⎪∂⎝⎭ ,102.5 2.5 2.5A B A A B A T PV V V x x x x x ⎛⎫∂=+=-- ⎪∂⎝⎭( b) B A M x x V 5.2=7.7 2)1(736.0ln Sn Sn x --=γ7.8 The maximum solubility of MgF2 in liquid MgCl at 900︒C is 19。
热力学习题与答案（原件）讲解
材料热力学习题1、阐述焓H 、内能U 、自由能F 以及吉布斯自由能G 之间的关系，并推导麦克斯韦方程之一：T P PST V )()(∂∂-=∂∂。
答： H=U+PV F=U-TS G=H-TS U=Q+W dU=δQ+δWdS=δQ/T, δW=-PdV dU=TdS-PdVdH=dU+PdV+VdP=TdS+VdP dG=VdP-SdTdG 是全微分，因此有：TP P TP ST V ，PT G T P G ，T V P G T P T G P S T G P T P G )()()()()()(2222∂∂-=∂∂∂∂∂=∂∂∂∂∂=∂∂∂∂=∂∂∂∂∂-=∂∂∂∂=∂∂∂因此有又而2、论述： 试绘出由吉布斯自由能—成分曲线建立匀晶相图的过程示意图，并加以说明。
（假设两固相具有相同的晶体结构）。
由吉布斯自由能曲线建立匀晶相图如上所示，在高温T 1时，对于所有成分，液相的自由能都是最低；在温度T 2时，α和L 两相的自由能曲线有公切线，切点成分为x1和x2，由温度T 2线和两个切点成分在相图上可以确定一个液相线点和一个固相线点。
根据不同温度下自由能成分曲线，可以确定多个液相线点和固相线点，这些点连接起来就成为了液相线和固相线。
在低温T 3，固相α的自由能总是比液相L 的低，因此意味着此时相图上进入了固相区间。
3、论述：通过吉布斯自由能成分曲线阐述脱溶分解中由母相析出第二相的过程。
第二相析出：从过饱和固溶体α中(x0)析出另一种结构的β相(xβ)，母相的浓度变为xα. 即：α→β+ α1α→β+ α1 的相变驱动力ΔGm的计算为ΔGm=Gm(D)-Gm(C)，即图b中的CD段。
图b中EF是指在母相中出现较大为xβ的成分起伏时，由母相α析出第二相的驱动力。
4、根据Boltzman方程S=kLnW，计算高熵合金FeCoNiCuCrAl和FeCoNiCuCrAlTi0.1（即FeCoNiCuCrAl各为1mol，Ti为0.1mol）的摩尔组态熵。
材料热力学习题答案2
The problems of the second law1 The solar energy flux is about 4J cm 2/min. in no focusing collector the surface temperature can reach a value of about 900℃. If we operate a heat engine using the collector as the heat source and a low temperature reservoir at 25℃, calculate the area of collector needed if the heat engine is to produce 1 horse power. Assume the engine operates at maximum efficiency. (2.1)Solution )(664.0)(74660104273900)25900(24m S W tWP StQ T T T W H H L H ===⨯⨯+-=-=2 A refrigerator is operated by 0.25 hp motor. If the interior of the box is to be maintained at -20℃ ganister a maximum exterior temperature of 35℃, what the maximum heat leak (in watts) into the box that can be tolerated if the motor runs continuously? Assume the coefficient of performance is 75% of the value for a reversible engine. (2.2)Solution:)(114474625.02035202733475.0%75W P P T T T P Q T T T W L LLLH HHLH =⨯⨯+-⨯=-=-=3 suppose an electrical motor supplies the work to operate a Carnot refrigerator. The interior of the refrigerator is at 0℃. Liquid water is taken in at 0℃ and converted to ice at 0℃. To convert 1 g of ice to 1 g liquid. △H=334J/g is required. If the temperature outside the box is 20℃, what mass of ice can be produced in one minute by a 0.25 hp motor running continuously? Assume that the refrigerator is perfectly insulated and that the efficiencies involved have their largest possible value. (2.3)Solution: )(4576033474625.020273g m M m P P T T T P L LLLH ===⨯⨯=-=4 under 1 atm pressure, helium boils at 4.126K. The heat of vaporization is 84 J/mol what sizemotor (in hp) is needed to run a refrigerator that must condense 2 mol of gaseous helium at 4.126k to liquid at the same temperature in one minute? Assume that the ambient temperature is 300K and that the coefficient of performance of the refrigerator is 50% of the maximum possible. (2.4)Solution: )(52.0)(393'60284216.4216.4300'5.0%50hp W P P T T T P P Q T T T W L L L H LLLH ==⨯⨯-=-==-= 5 if a fossil fuel power plant operating between 540 and 50℃ provides the electrical power torun a heat pump that works between 25 and 5℃, what is the amount of heat pumped into the house per unit amount of heat extracted from the power plant boiler.(a) assume that the efficiencies are equal to the theoretical maximum values(b) assume the power plant efficiency is 70% of maximum and that coefficient of performance ofthe heat pump is 10% of maximum(c) if a furnace can use 80% of the energy in fossil foe to heat the house would it be moreeconomical in terms of overall fissile fuel consumption to use a heat pump or a furnace ? do the calculations for cases a and b (2.5)solution:1,2,2,1,212,2,2,2,21,1,1,1,198.82527352527354050540)(H H H H H H L H H H L H P P P P P P P T T T P P T T T P a =+-=+-=-=-=.,)(6286.0)(1,2,not is b ok is a c P P b H H =6 calculate △U and △S when 0.5 mole of liquid water at 273 K is mixed with 0.5 mol of liquid water at 373 K and the system is allowed to reach equilibrium in an adiabatic enclosure. Assume that C p is 77J /(mol K) from 273K to 373K (2.6) Solution:)/(933.0)273323ln(5.0)373323ln(5.0)ln()ln()(02211K J C C T T C n T T C n S J U P P E P E P =+=+=∆=∆ 7 A modern coal burning power plant operates with a steam out let from the boiler at 540℃ and acondensate temperature of 30℃.(a) what is the maximum electrical work that can be produced by the plant per joule of heatprovided to the boiler?(b) How many metric tons (1000kg) of coal per hour is required if the plant out put is to be500MW (megawatts). Assume the maximum efficiency for the plant. The heat of combustionof coal is 29.0 MJ/k g(c) Electricity is used to heat a home at 25℃ when the out door temperature is 10℃ by passinga current through resistors. What is the maximum amount of heat that can be added to the home per kilowatt-hour of electrical energy supplied? (2.7)Solution:)(3.69)(6937136005000.29)()(89.013054030540)(ton kg m T T T mb J Q T T T W a LH LH H L H ==⨯=-=+-=-=)(9.191102525273)(J Q Q T T T W c H HHLH =-+=-=8 an electrical resistor is immersed in water at the boiling temperature of water (100℃) the electrical energy input into the resistor is at the rate of one kilowatt(a) calculate the rate of evaporation of the water in grams per second if the water container isinsulated that is no heat is allowed to flow to or from the water except for that provided by the resistor(b) at what rate could water could be evaporated if electrical energy were supplied at the rate of 1kw to a heat pump operating between 25 and 100℃data for water enthalpy of evaporation is 40000 J/mol at 100℃; molecular weight is 18g/mol; density is 1g/cm 3 (2.8)solution:)(23.2,2510027310010004000018)()(45.0,10004000018)(g m m b g m ma =-+===9 some aluminum parts are being quenched (cooled rapidly ) from 480℃ to -20℃ by immersingthem in a brine , which is maintained at -20℃ by a refrigerator. The aluminum is being fed into the brine at a rate of one kilogram per minute. The refrigerator operates in an environment at 30℃; that is the refrigerator may reject heat at 30℃. what is them minus power rating in kilowatts, of motor required to operate the refrigerator?Data for aluminum heat capacity is 28J/mol K; Molecular weight 27g/mol (2.9)Solution:)(5.102)(102474202732030)20480(28271000kW W P P T T T P P L L L L H W L ==---=-=--⨯=10 an electric power generating plant has a rated output of 100MW. The boiler of the plantoperates at 300℃. The condenser operates at 40℃(a) at what rate (joules per hour) must heat be supplied to the boiler?(b) The condenser is cooled by water, which may under go a temperature rise of no more than10℃. What volume of cooling water in cubic meters per hour, is require to operate the plant?(c) The boiler tempeture is to be raised to 540℃,but the condensed temperature and electricoutput will remain the same. Will the cooling water requirement be increased, decreased, or remain the same?Data heat capacity 4.184, density 1g/cm 3 (2.10)Solution: )(109.7)(102.21040300273300)(1188J t P Q W P T T T P a H H L H H H ⨯==⨯=-+=-=)(1003.1184.41010)(103.4)(34611m V Q V J Q b L L ⨯==⨯⨯⨯⨯=noW P T T T P c L H H H )(10626.11040540273540)(88⨯=-+=-=11 (a) Heat engines convert heat that is available at different temperature to work. They have beenseveral proposals to generate electricity y using a heat engine that operate on the temperature differences available at different depths in the oceans. Assume that surface water is at 20℃, that water at a great depth is at 4℃, and that both may be considered to be infinite in extent. How many joules of electrical energy may be generated for each joule of energy absorbed from surface water? (b) the hydroelectric generation of electricity use the drop height of water as the energy source. in a particular region the level of river drops from 100m above sea level to 70m above the sea level . what fraction of the potential energy change between those two levels may be converted into electrical energy? how much electrical energy ,in kilowatt-hours, may be generated per cubic meter of water that undergoes such a drop? (2.11)Solution:)/(1006.136001000)()(055.0127320420)(6h kW hmg P b J Q T T T W a H H L H ⨯=⨯∆==+-=-=12 a sports facility has both an ice rink and a swimming pool. to keep the ice frozen during the summer requires the removal form the rink of 105 KJ of thermal energy per hour. It has been suggested that this task be performed by a thermodynamic machine, which would be use the swimming pool as the high temperature reservoir. The ice in the rink is to be maintain at a temperature of –15℃, and the swimming pool operates at 20℃, (a) what is the theoretical minimum power, in kilowatts, required to run the machine? (b) how much heat , in joule per hour , would be supplied t the pool by this machine? (2.12)Solution:)(1014.1101527320273)()(77.33600/10152731520)(555kJ Q b kW P T T T P a H L L L H ⨯=-+==-+=-=13solution:)/(81.6810ln 314.877.45277.6282.4)/(152940)()/(67.4977.45277.6282.4)()/(152940)(22)(2molK cal S mol cal H d molK cal S c mol cal H b AlNN Al a -=+-⨯-⨯=∆=∆-=-⨯-⨯=∆=∆=+14solution:)/(2257412000)27340273ln 184.4273336263273ln1.2()(40,010,K J dT T C T H dT T C m S WATER P m mICE P =+++=+∆+=∆⎰⎰- 15)(70428)(2896100077773002J W J Q T T T W L L L H ==-=-=16)(4.3719))2.4300(314.85.13.83(3002.4300)(7.58663.832.42.4300J Q T T T W J Q T T T W H H L H L L L H =-⨯+-=-==-=-=17yesd Q c K J PPnR S b J pdV n W Q OU T a )(0)()/(1.1910ln 314.81ln )()(570410ln 298314.810)(0==⨯⨯==∆=⨯⨯=-=-==∆=∆⎰)(122233527302033560500g m m m T T T L L H =-=-=⨯。
材料热力学与动力学复习题答案word版本
材料热力学与动力学复习题答案一、常压时纯Al 的密度为ρ=2.7g/cm 3，熔点T m =660.28℃，熔化时体积增加5%。
用理查得规则和克-克方程估计一下，当压力增加1Gpa 时其熔点大约是多少？ 解：由理查德规则RTm Hm R Tm Hm Sm ≈∆⇒≈∆=∆ …①由克-克方程VT H dT dP ∆∆=…② 温度变化对ΔH m 影响较小，可以忽略，①代入②得 V T H dT dP ∆∆=dT T1V Tm R dp V T Tm R ∆≈⇒∆≈…③ 对③积分 dT T1V T Tm R p d T Tm Tm pp p ⎰⎰∆+∆+∆= 整理 ⎪⎭⎫ ⎝⎛∆+∆=∆Tm T 1ln V Tm R p V T R V Tm R Tm T ∆∆=∆⨯∆≈ Al 的摩尔体积 V m =m/ρ=10cm 3=1×10-5m 3Al 体积增加 ΔV=5%V m =0.05×10-5m 3K 14.60314.810510R V p T 79=⨯⨯=∆∆=∆- Tm’=Tm+T ∆=660.28+273.15+60.14=993.57K二、热力学平衡包含哪些内容，如何判断热力学平衡。
内容：（1）热平衡，体系的各部分温度相等；（2）质平衡：体系与环境所含有的质量不变；（3）力平衡：体系各部分所受的力平衡，即在不考虑重力的前提下，体系内部各处所受的压力相等；（4）化学平衡：体系的组成不随时间而改变。
热力学平衡的判据：（1）熵判据：由熵的定义知dS Q T δ≥不可逆可逆对于孤立体系，有0Q =δ，因此有dS 可逆不可逆0≥，由于可逆过程由无限多个平衡态组成，因此对于孤立体系有dS 可逆不可逆0≥，对于封闭体系，可将体系和环境一并作为整个孤立体系来考虑熵的变化，即平衡自发环境体系总0S S S ≥∆+∆=∆ （2）自由能判据 若当体系不作非体积功时，在等温等容下，有()0d ,≤V T F 平衡状态自发过程上式表明，体系在等温等容不作非体积功时，任其自然，自发变化总是向自由能减小的方向进行，直至自由能减小到最低值，体系达到平衡为止。
材料热力学与动力学005相变热力学与动力学2
4 4 V r 3 ( t ) 3 3 3
所以新相总体积
4 dV 3t 3 I vVa dt 3
相变开始阶段ห้องสมุดไป่ตู้
V V
4 dV 3t 3 I vV dt 3
所以
将V作为１，经过t时间产生新相的 体积分数为x, dx=dVß：
4 t x I v 3t 3dt 3 0
对于曲面晶界，R可由下式求得
P
8R 4R 2
2 R
1 1 1 1 ( ) R 2 R1 R2
R1,R2为曲面晶界的最大及最小半径
上式表明，由界面能提供的作用于单位面积晶界的驱动 力与界面能成正比，与界面曲率半径成反比，力的方向 指向曲率中心。对于平直界面， R 驱动力为零。 在三个不同的晶粒交点处（如A点)，为保持界 面张力平衡，即保持三个交角均为1200，晶界必将 凸向大晶粒一方，出现曲面晶界，在驱动力作用下， 小晶粒中原子越过界面向大晶粒迁移。
2 3 DV C a ( ) r 3 r03 t kt 2 RT
r 0 为粗化开始时颗粒的平均半径；r 为经过t时间粗化后颗粒的平均半径
3
3
考虑颗粒尺寸分布，Wagner公式： 2 9 DV C a ( ) 3 3
r r0 8 RT
t
五、晶粒长大 1.晶粒正常长大：在界面曲率驱动力下，晶粒发生均匀 长大的过程。 母相全部转变成新相后，还将通过晶界的迁移发 生晶粒的粗化。推动晶界迁移的驱动力来自界面能的 降低。50nm晶粒系统晶界能为104J/m3。
根据Gibbs-Thomson方程
Ca (r ) 2VB ln Ca () RTr
材料热力学和动力学答案
Gibbs-Thomson effect:1.The Gibbs–Thomson Effect, in common physics usage, refers to variations in vaporpressure or chemical potential across a curved surface or interface. The existence of a positive interfacial energy will increase the energy required to form small particles with high curvature, and these particles will exhibit an increased vapor pressure.See Ostwald–Freundlich equation.2.More specifically, the Gibbs–Thomson effect refers to the observation that small crystalsare in equilibrium with their liquid melt at a lower temperature than large crystals. In cases of confined geometry, such as liquids contained within porous media, this leads to a depression in the freezing point / melting point that is inversely proportional to the pore size, as given by the Gibbs–Thomson equation.Why at a relatively lower temperature solute transport tends to become more effective via grain boundary than through the lattice or through dislocation? Please have an example material to clear.（为什么在一个相对较低的温度，溶质趋向于通过晶界的运输比晶格和位错运输更有效？请举一种材料作为例子详述）答：在多晶体中的扩散除了再晶粒点阵内部进行之外，还会沿表面、晶界、位错等缺陷部位进行。
材料热力学与动力学-1.4热力学第三定律
T2
T1
T2 dT = ∫ CpdLnT Cp T1 T
Cp
△S可求
T
lnT
1.4.3 熵的计算
任何温度下 S 为0K时1mol体系的熵 当T=298K时 S = S + ∫ C d lnT 当T=XK时 S = S + ∫ C d ln T T=298K，10 Pa，1mol标准熵 T=298K 10 Pa 1mol
1.4.1 能斯特热定律
G = H − TS 自由焓和焓定义 恒温下状态改变时 ∆GT = ∆HT − T∆ST 微分 ( ∂∆G ) p = ( ∂∆H ) p − T ( ∂∆S ) p − ∆S
∂T ∂T ∂T
( ∂∆G ) p = −∆S ∂T ∂∆H ∂∆S ( )p = T ) p = ∆Cp ∂T ∂T
ST = S 0 + ∫ Cpd ln T
0 298 298 0 p T
0
0
X
T
298
p
298
5
S 298 = S 0 + ∫
0
298
0
Cpd lnT
任何温度下的标准熵
ST = S 298 + ∫ Cpd ln T
298
0 0
T
1.4.2 热力学第三定律
任何均匀物质在内部完全平衡条件下， 在绝对零度时，其熵值为零。
1.4.3 熵的计算
熵和热容的定义 dS = δQ / T δQ = C dT H 等压的可逆过程 dS = (δTQ ) = (δT ) = C dT T 在一定成分的封闭体系在恒压下
(可逆 )
p
p
p
p
p
∆S = S (T 2, p ) − S (T 1, p ) = ∫
材料热力学与动力学动力学部分复习
3、速率方程的积分形式
A：一级反应 反应速率与反应物浓度一次方成正比的反应称一级反应(first order reaction)。如： AB
dc A kcA dt
dcA c A k dt
dc A kcA dt
t dc A k dt C A,0 c 0 A CA
2、化学反应的速率方程 表示反应速率与参加反应的各种物质浓度及影响反应的各种因 素之间的关系，也称为化学反应的动力学方程。 元反应：
aA bB lL mM
V ∝ caAcbB 或 V=KcaAcbB
（质量作用定律） 反应速率
K：反应速率常数(rate constant of reaction)或反应比速。 K 物理意义：各反应物质皆为单位浓度时的反应速率。 影响K值因素：反应类型、温度、溶剂、催化剂等。
（1）计算此反应的活化能？ （2）欲使A在10min内转化率达到90%，反应温 度应控制在多少？
二、相变过程动力学
1、形核率 单位时间单位体积母相中形成的新相晶核数，用 I表示 （1）均匀形核
G * Q I B exp( ) exp( ) kT kT
(2)成核速率随温度变化的关系
2 3 T 64 1 * * m G As 3 (H m ) 2 (T ) 2 3
c
A,0
dy kdt yc B,0 y
或
c A,0
dy 1 d y kdt c B,0 c B,0 y c A,0 y
由t = 0到t = t积分上式，得
c A0
c B,0 c A,0 y 1 ln kt c B,0 c A,0 c B,0 y
活化能Ea： 为了能发生化学反应，普通分子（具有平均能量的分子）必须吸 收足够能量先变成活化分子 ，在此变化过程中所要吸收的最小
材料热力学试题2009-2010-v1
2009-2010 年 秋 季学期研究生课程考试试题考 试 科 目：材料热力学与动力学 学生所在院（系）：材料学院、航天学院学生所在学科：材料学、材料加工工程 (* 题签与答题纸一起上交)一、仔细阅读下列论述，判断正误，如果错误，请说明该论述违反了哪些热力学原理，并给出正确的论述。
（16分）(1) 纯金属中不存在空位时的吉布斯自由能最低。
(2) 可采用公切线法则确定晶界偏析（晶界相成分）。
(3) 材料变形过程的热力学流是扩散通量。
(4) 低压下不可能将石墨转变为金刚石。
二、（1）固态纯组元的G-T 曲线如下图所示，请判断哪条线正确，并解释原因。
(6分)（2）A-B 二元系中，固相和液相的摩尔自由能-成分曲线如下图所示。
请在自由能-成分曲线上，图示出体系成分为X*处，固相纯A 和液相纯B 混合后的吉布斯自由能的变化量∆G mix ，并说明原因。
(6分) （此题请直接答在题签上）三、简答题：(1) 请说明化学位和活度的物理意义。
（6分）(2) 简述Calphad 的三要素及其主要功能。
（6分）(3) 请解释Onsager 倒易关系、最小熵产生原理。
（6分）四、某些在平衡条件下固态互不混溶的二元系（I AB >0），通过高能球磨或高压扭转等剧烈塑性形变（应变量非常大）实现纳米化后，可发现两个组元的相互固溶度会显著提高。
试从热力学的角度解释该现象。
（8分）学院学号 姓名五、15克金和25克银混合后形成单相理想固溶体，问（1）金和银的摩尔分数各是多少？（2）固溶体总的混合熵是多少？（3）500o C 时，摩尔自由能变化为多少？（4）令纯金和纯银的自由能为零，那么500o C 时金和银的化学势各为多少？（5）在500o C 加入一个金原子，固溶体的自由能变化为多少？已知：金的原子量为197，银的原子量为108，气体常数R 为8.314J·mol -1·K -1. (15分)六、试通过如图所示的A-B 二元相图，判断A-B 固溶体的性质、溶体组元间的相互作用能。
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一、常压时纯Al 的密度为ρ=2.7g/cm 3，熔点T m =660.28℃，熔化时体积增加5%。
用理查得规则和克-克方程估计一下，当压力增加1Gpa 时其熔点大约是多少？ 解：由理查德规则RTm Hm R Tm Hm Sm ≈∆⇒≈∆=∆ …①由克-克方程VT H dT dP ∆∆=…② 温度变化对ΔH m 影响较小，可以忽略，①代入②得V T H dT dP ∆∆=dT T 1V Tm R dp V T Tm R ∆≈⇒∆≈…③ 对③积分 dT T1V T Tm R p d T Tm Tm pp p ⎰⎰∆+∆+∆= 整理 ⎪⎭⎫ ⎝⎛∆+∆=∆T m T 1ln V T m R p V T R V Tm R Tm T ∆∆=∆⨯∆≈ Al 的摩尔体积 V m =m/ρ=10cm 3=1×10-5m3Al 体积增加 ΔV=5%V m =0.05×10-5m 3K 14.60314.810510R V p T 79=⨯⨯=∆∆=∆- Tm’=Tm+T ∆=660.28+273.15+60.14=993.57K二、热力学平衡包含哪些内容，如何判断热力学平衡。
内容：（1）热平衡，体系的各部分温度相等；（2）质平衡：体系与环境所含有的质量不变；（3）力平衡：体系各部分所受的力平衡，即在不考虑重力的前提下，体系内部各处所受的压力相等；（4）化学平衡：体系的组成不随时间而改变。
热力学平衡的判据：（1）熵判据：由熵的定义知dS Q T δ≥不可逆可逆对于孤立体系，有0Q =δ，因此有dS 可逆不可逆0≥，由于可逆过程由无限多个平衡态组成，因此对于孤立体系有dS 可逆不可逆0≥，对于封闭体系，可将体系和环境一并作为整个孤立体系来考虑熵的变化，即平衡自发环境体系总0S S S ≥∆+∆=∆ （2）自由能判据 若当体系不作非体积功时，在等温等容下，有()0d ,≤V T F 平衡状态自发过程上式表明，体系在等温等容不作非体积功时，任其自然，自发变化总是向自由能减小的方向进行，直至自由能减小到最低值，体系达到平衡为止。
（3）自由焓判据 若当体系不作非体积功时，在等温等压下，有0d ≤G 平衡状态自发过程所以体系在等温等容不作非体积功时，任其自然，自发变化总是向自由能减小的方向进 行，直至自由能减小到最低值，体系达到平衡为止。
三、试比较理想熔体模型与规则熔体模型的异同点。
（1）理想熔体模型：在整个成分范围内每个组元都符合拉乌尔定律，这样的溶体称为理想溶体，其特征为混合热为零，混合体积变化为零，混合熵不为零。
从微观上看，组元间粒子为相互独立的，无相互作用。
（2）符合下列方程的溶体称为规则溶体：（形成（混合）热不为零，混合熵等于理想的混合熵）⎪⎪⎭⎪⎪⎬⎫=='='=2A B 2B A 2A B 2BA ln ln ln ln x x x RT x RT αγαγαγαγ其中，α’为常数，而α为(1/T)的函数，即α =α’/RT相同点：混合熵相等。
不同点：（1）理想熔体模型混合热为零，规则混合热不为零；（2）理想假设组元间粒子为相互独立的，无相互作用，规则考虑粒子间的相互作用。
四、固溶体的亚规则溶体模型中，自由能表示为m ii i i i i m G x x RT G x G E 0 ln ++=∑∑其中过剩自由能表示为 ∑=-=0B A AB B A E )(ννx x L x x G m实际测得某相中0L AB 和1L AB ，请分别给出组元A 和B 的化学位表达式。
解：该模型有A ，B 两相。
00(ln ln )E m A A B B A A B B m G x G x G RT x x x x G =++++过剩自由能表示为∑=-=0B A AB B A E )(ννx x L x x G m E 01m A B AB A B AB A B G =x x L +x x L x -x （）代入Gm 中 00(ln ln )m A A B B A A B B G x G x G RT x x x x =+++01A B AB A B AB A B x x L x x L x -x ++（）化学位 m m B A G G x x μ∂==+∂A A Gm B B m A B G G G x x μ∂==+∂ 解得：020ln (3)A A A B AB A B AB G RT x x L x x L μ⎡⎤=+++-⎣⎦020ln (3)B B B A AB A B AB G RT x x L x x L μ⎡⎤=+++-⎣⎦五、向Fe 中加入α形成元素会使γ区缩小，但无论加入什么元素也不能使两相区缩小到0.6at%以内，请说明原因。
解：当 1,γB αB <<x x 时γαF e 0γαA 0αB γB 11→→∆=∆≈-G RT G RT x x加入一种合金元素后，0B x γ≈，此时01ααγB Fe x G RT→-=∆ 在1400K （xγB 最大值点）时，0αγFe G →∆有最小值71.7J 此时B x γ≈0.6 at% 则：%1001400314.87.71⨯⨯-=γB x =0.6 at% 六、今有Fe-18Cr-9Ni 和Ni80-Cr20两种合金，设其中含碳量为0.1wt%，求T=1273︒C 时碳在这两种合金中活度。
解：对于Fe-20Cr-10Ni 合金，由x i 与y i 的关系可得00462.01CC C =-=x x y 21330.0Cr =y 09447.0N i =y 69223.0Fe =y 从表9-1查得 J γCr = -100964J/mol ，J γNi = 46000J/mol 而molJ 21701178.1946115])21([C C gr C 0Fe 0FeC 0=-=-+--T I y G G G γγγm ol J 35788555.1121079C -=--=T I γ58.1)]2(1ex p[C C gr C 0C C F e 0F eC 0C =∑+--+-=M M v v y J I y G I G G RTf γγγγ因此在Fe-20Cr-10Ni 合金%727.000727.0C C C ===x f a γ 对于 Ni80-Cr20合金，有%465.0N i γC=-a七、假如白口铁中含有3.96%C 及2.2%Si ，计算在900︒C 时发生石墨化的驱动力，以铸铁分别处于γ +渗碳体两相状态与γ +石墨两相状态时碳的活度差来表示此驱动力。
由于Si不进入Fe 3C 中，所以有K Si Cem/γ = 0。
在Fe-C 二元合金中，已知900︒C 时γ +渗碳体两相状态碳的活度为二a γC = 1.04；当γ与石墨平衡时a γC = 1。
解：要计算Fe-Si-C 三元合金中石墨化驱动力，首先要求出三元合金中x γC ，u γC ，x γSi 和u γSi 四个参数。
188.009.28/0.285.55/04.94011.12/96.31Si Fe C C C alloy C =+=+=-=x x x x x u 0406.009.28/0.285.55/04.9409.28/0.21Si Fe Si C Si alloy Si =+=+=-=x x x x x u 假定γ中的碳含量与二元系中相同，根据Fe-C 相图，900℃与渗碳体相平衡时奥氏体碳含量为1.23%。
因此有 0579.085.55/77.98011.12/23.1γC ==u渗碳体的分子式为Fe 3C ，因此x C Cem =0.25或u C Cem =0.333，利用杠杆定律计算γ相的摩尔分数528.00579.0333.0188.0333.0=--=γf 472.0Cem=f 因为K Si Cem/γ=0，由硅的质量平衡可得 alloy Si Cem Si 0u f f u =⋅+γγ0769.0528.0/0406.0S i ==γu279.01)()(ln C Cem C Cem Si B C TC =--=γγγγu u K a aa γC = 1.375二元合金中石墨化驱动力为 ()()04.0104.1Gr C Fe 3=-=-γγγγC Ca a 三元合金中石墨化驱动力为()()375.01375.1Gr C Fe 3=-=-γγγγC C a a八、通过相图如何计算溶体的热力学量如熔化热、组元活度。
解：熔化热以Bi-Cd 相图为例计算如含0.1摩尔分数的Cd 时，合金的熔点要降低，已知Bi 的熔点为T A * = 43.5K ，于是Bi 的熔化热0H Bi 可由以下方法计算得到：l s G G Bi Bi =l l s s a RT G a RT G Bi Bi 0Bi Bi 0ln ln +=+s l l s G G a a RT Bi 0Bi 0Bi Bi ln -=Bi 0Bi 0Bi 0Bi 0Bi 0S T H G G G s l ∆-∆=∆=-在纯Bi 的熔点温度T Bi *时，熔化自由能Δ0G Bi = 0，于是由式（10-4）可得纯Bi 的熔化熵为*∆=∆Bi Bi 0Bi 0T H S)1(Bi Bi 0Bi 0Bi 0Bi 0*-∆=-=∆T T H G G G s l由于Bi-Cd 为稀溶体，可近似取1Bi Bi ==s s x al l l x x a Cd Bi Bi 1-== l l x x Cd Cd )1ln(-≈-于是得l x T R T H Cd 2Bi Bi 0)(1*∆=∆ 将具体数据，T Bi *=543.5K ，R=8.314J/K*mol ，x Cd l =0.1 mol 代入得Δ0H Bi = 10.77 kJ/mol组元活度：设已知相图如图所示。
在温度为T 1时，a 点组成的α相与b 点组成的l 相平衡共存，所以l A αA μμ=αA αA 0l A l A 0ln ln a RT a RT +=+μμl A αA αA 0l A 0ln a a RT =-μμ RT G a a *∆=A 0l A αA lnαA 0l A 0A 0 μμ-=∆*G 为A 组分的摩尔熔化吉布斯自由能当固溶体α中A 浓度x A α 接近1时，可近似假定A 组元遵从拉乌尔定律，即用x A α代替a A α，则RT G x a *∆-=A 0αA l A ln ln ⎰⎰********∆-∆+∆-∆=∆T T p T T p T T C T T C T H T H G A A d d A ,A ,A A 0A 0A 0A ,A ,A ,≈-=∆*s p l p p C C C ***-∆+=A A A 0αA A ][ln ln RTT T T H x a l***-∆=A A A 0A ][ln RTT T T H a l （当固溶体α为极稀溶体，x A α→1）九、请说明相图要满足那些基本原理和规则。