《材料科学与工程基础》双语教学实践探索材料科学与工程基础第二版答案

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材料科学与工程基础第二版考试必备宝典.doc

第1章绪论1．材料科学与工程的四个基本要素解：制备与加工、组成与结构、性能与应用、材料的设计与应用2．金属﹑无机非金属材料﹑高分子材料的基本特性解：①金属材料的基本特性： a. 金属键； b. 常温下固体，熔点较高； c. 金属不透明，具有光泽； d. 纯金属范性大、展性、延性大； e. 强度较高； f. 导热性、导电性好； g. 多数金属在空气中易氧化。

②无机非金属材料的基本性能：a. 离子键、共价键及其混合键；b. 硬而脆；c. 熔点高、耐高温，抗氧化； d. 导热性和导电性差； e. 耐化学腐蚀性好； f. 耐磨损； g. 成型方式：粉末制坯、烧结成型。

③高分子材料的基本特性： a. 共价键，部分范德华键； b. 分子量大，无明显熔点，有玻璃化转变温度（ Tg）和粘流温度（ Tf ）； c. 力学状态有三态：玻璃态、高弹态和粘流态； d. 质量轻，比重小； e. 绝缘性好； f. 优越的化学稳定性； g. 成型方法较多。

第 2 章物质结构基础1 ．在多电子的原子中，核外电子的排布应遵循哪些原则解：泡利不相容原理、能量最低原理、洪特规则2．电离能及其影响电离能的因素解：电离能：从孤立原子中，去除束缚最弱的电子所需外加的能量。

影响因素：①同一周期，核电荷增大，原子半径减小 , 电离能增大；②同一族，原子半径增大，电离能减小；③电子构型的影响，惰性气体；非金属；过渡金属；碱金属；3．混合键合实例解：石墨：同一层碳原子之间以共价键结合，层与层之间以范德华力结合；高分子：同一条链原子之间以共价键结合，链与链之间以范德华力结合。

4.将离子键，共价键，金属键按有无方向性进行分类，简单说明理由有方向性：共价键无方向性：离子键，金属键③ 金属键：正离子排列成有序晶格，每个原子尽可能同更多的原子相结合，形成低能量的密堆结构，正离子之间相对位置的改变不破坏电子与正离子间的结合力，无饱和性又无方向性。

②共价键：共用电子云最大重叠，有方向性③离子键：正负离子相间排列，构成三维晶体结构，无方向性和饱和性5. 简述离子键，共价键，金属键的区别6. 为什么共价键材料密度通常要小于离子键或金属键材料金属密度高的两个原因：第一，金属有较高的相对原子质量。

《材料科学基础》第二版 (张联盟著)课后习题答案武汉理工大学出版社

2-13 根据半径比关系，说明下列离子与O2—配位时的配位数各是多少？已知rO2－＝0.132nm，rSi4＋＝0.039nm，rK＋＝0.131nm，rAl3＋＝0.057nm，rMg2＋＝0.078nm。
2-14 为什么石英不同系列变体之间的转化温度比同系列变体之间的转化温度高得多？
2-15 有效离子半径可通过晶体结构测定算出。在下面NaCl型结构晶体中，测得MgS和MnS的晶胞参数均为a＝0.520nm（在这两种结构中，阴离子是相互接触的）。
为NaCl型结构时，体积变化的百分数是多少？已
知CN＝6时，rMn
2＋＝0.08nm，rS2
－＝0.184nm；CN
＝4时，rMn
2＋＝0.073nm，r
2 S
－＝0.167nm。
2-25 钛酸钡是一种重要的铁电陶瓷，其晶型是钙钛矿结构，试问：（1）属于什么点阵？（2）这个结构中离子的配位数为若干？（3）这个结构遵守鲍林规则吗？请作充分讨论。
2-5 依据结合力的本质不同，晶体中的键合作用分为哪几类？其特点是什么？
2-6 等径球最紧密堆积的空隙有哪两种？一个球的周围有多少个四面体空隙、多少个八面体空隙？
2-7 n个等径球作最紧密堆积时可形成多少个四面体空隙、多少个八面体空隙？不等径球是如何进行堆积的？
2-8 写出面心立方格子的单位平行六面体上所有结点的坐标。
2-14 为什么石英不同系列变体之间的转化温度比同系列变体之间的转化温度高得多？
答：石英同一系列之间的转变是位移性转变，不涉及晶体结构中键的破裂和重建，仅是键长、键角的调整、需要能量较低，且转变迅速可逆；而不同系列之间的转变属于重建性转变，都涉及到旧键的破裂和新键的重建，因而需要较的能量，且转变速度缓慢；所以石英不同系列之间的转化温度比同系列变体之间转化的温度要高的多。

《材料科学基础》第二版 (张联盟著)课后习题答案武汉理工大学出版社

2-29 根据Mg2[SiO4]在（100）面的投影图回答：（1）结构中有几种配位多面体，各配位多面体间的连接方式怎样？（2）O2－的电价是否饱和？（3）晶胞的分子数是多少？（4）Si4＋和Mg2＋所占的四面体空隙和八面体空隙的分数是多少？
2-30 石棉矿如透闪石Ca2Mg5[Si4O11]（OH）2具有纤维状结晶习性，而滑石Mg3[Si4O10]（OH）2却具有片状结晶习性，试解释之。
2-21 （1）画出O2－作面心立方堆积时，各四面体空隙和八面体空隙的所在位置（以一个晶胞为结构基元表示出来）；（2）计算四面体空隙数、八而休空隙数与O2－数之比
2-22 根据电价规则，在下面情况下，空隙内各需填入何种价数的阳离子，并对每一种结构举出—个例子。（1）所有四面体空隙位置均填满；（2）所有八面体空隙位置均填满；（3）填满—半四面体空隙位置；（4）填满—半八面体空隙位置。
2-3 在立方晶系晶胞中画出下列晶面指数和晶向指数：（001）与[ ]，（111）与[ 与[111]，（）与[236]，（257）与[ ]，（123）与[ ]，（102），[
2-4 定性描述晶体结构的参量有哪些？定量描述晶体结构的参量又有哪些？
答：最紧密堆积原理是建立在质点的电子云分布呈球形对称以及无方向性的基础上的，故只适用于典型的离子晶体和金属晶体，而不能用最密堆积原理来衡量原子晶体的稳定性。另外，金刚石的单键个数为4，即每个原子周围有4个单键（或原子），由四面体以共顶方式共价结合形成三维空间结构，所以，虽然金刚石结构的空间利用率很低（只有34.01％），但是它也很稳定。
答：定性：对称轴、对称中心、晶系、点阵。定量：晶胞参数。 2-5 依据结合力的本质不同，晶体中的键合作用分为哪几类？其特点是什么？

《材料科学与工程基础》双语教学实践探索

《材料科学与工程基础》双语教学实践探索摘要：结合我校的情况分析了材料科学与工程双语教学在本学科中教学的重要性，简要介绍了其基本模式及存在的问题，介绍了根据具体的考与学过程，在材料科学与工程课程教材的选择，课堂的设计以及考试等方面的一些改革与探索，提出了在材料专业用双语进行材料科学与工程课程教学实行“以学生为中心的”双语教学的教学模式。

关键词：双语教学材料科学与工程教学模式在世界进入21世纪的今天，随着中国加入WTO，我国改革开放日趋深化，中国正走向世界，世界也在向中国走来，国家和社会发展使得对双语人才的需求程度迅速提高。

如何在材料学科本科教学中培养学生专业英语的应用能力，使之具备综合素质，是当前高分子材料与复合材料专业教育与教学改革中需要深入探讨的问题之一。

我校自1997年开始招收复合材料专业，每年基本上招生2个本科班，在原有的高分子材料基础上进行有机整合，分为高分子材料与工程专业与复合材料与工程专业。

从共性知识体系中提炼出基本问题，建立起材料科学与工程的基础教学体系，从宽基础角度加强专业基础课教学，使一部分专业课趋于向专业基础课的调整，在这样的教改思路下，原来各专业方向的专业外语课程面临着向大材料专业的过渡，从而产生了一门新的课程——《材料科学与工程基础》，并设为双语教学示范课程，作为高分子材料与工程专业与复合材料与工程专业六大平台课程之一，材料科学与工程基础的内容因双语教学而偏向于基本概念和基本应用，为避免重复，材料科学与工程基础将不涉及各材料方向的具体理论。

一、材料科学与工程基础双语教学的必要性与目的专业外语(实际上多为专业英语)是以往的专业设置下各专业开设的一门专业必修课，其目的在于提高学生对外文文献的阅读与理解能力，而主要以专业英语作为整班授课内容。

目前，由于高校近年来开设了双语教学课程，所以专业外语面临着被取消的可能或过渡为双语教学课程。

我校高分子材料科学与工程专业即以双语教学的《材料科学与工程基础》课程来代替专业外语，但两者之间又存在着本质的区别。

材料科学与工程专业英语第二版课文翻译(1,2,3,10)

United 1 材料科学与工程材料在我们的文化中比我们认识到的还要根深蒂固。

如交通、房子、衣物，通讯、娱乐和食物的生产，实际上，我们日常生活中的每一部分都或多或少地受到材料的影响。

历史上社会的发展、先进与那些能满足社会需要的材料的生产及操作能力密切相关。

实际上，早期的文明就以材料的发展程度来命名，如石器时代，铜器时代。

早期人们能得到的只有一些很有限的天然材料，如石头、木材、粘土等。

渐渐地，他们通过技术来生产优于自然材料的新材料，这些新材料包括陶器和金属。

进一步地，人们发现材料的性质可以通过加热或加入其他物质来改变。

在这点上，材料的应用完全是一个选择的过程。

也就是说，在一系列非常有限的材料中，根据材料的优点选择一种最适合某种应用的材料。

直到最近，科学家才终于了解材料的结构要素与其特性之间的关系。

这个大约是过去的60 年中获得的认识使得材料的性质研究成为时髦。

因此，成千上万的材料通过其特殊的性质得以发展来满足我们现代及复杂的社会需要。

很多使我们生活舒适的技术的发展与适宜材料的获得密切相关。

一种材料的先进程度通常是一种技术进步的先兆。

比如，没有便宜的钢制品或其他替代品就没有汽车。

在现代，复杂的电子器件取决于所谓的半导体零件.材料科学与工程有时把材料科学与工程细分成材料科学和材料工程学科是有用的。

严格地说，材料科学涉及材料到研究材料的结构和性质的关系。

相反，材料工程是根据材料的结构和性质的关系来设计或操纵材料的结构以求制造出一系列可预定的性质。

从功能方面来说，材料科学家的作用是发展或合成新的材料，而材料工程师是利用已有的材料创造新的产品或体系，和/或发展材料加工新技术。

多数材料专业的本科毕业生被同时训练成材料科学家和材料工程师。

“structure”一词是个模糊的术语值得解释。

简单地说，材料的结构通常与其内在成分的排列有关。

原子内的结构包括介于单个原子间的电子和原子核的相互作用。

在原子水平上，结构包括原子或分子与其他相关的原子或分子的组织。

《材料科学与工程基础》英文影印版习题及思考题及答案

《材料科学与⼯程基础》英⽂影印版习题及思考题及答案《材料科学与⼯程基础》英⽂习题及思考题及答案第⼆章习题和思考题Questions and Problems2.6 Allowed values for the quantum numbers ofelectrons are as follows:The relationships between n and the shell designationsare noted in Table 2.1.Relative tothe subshells,l ＝0 corresponds to an s subshelll ＝1 corresponds to a p subshelll ＝2 corresponds to a d subshelll ＝3 corresponds to an f subshellFor the K shell, the four quantum numbersfor each of the two electrons in the 1s state, inthe order of nlmlms , are 100(1/2 ) and 100(-1/2 ).Write the four quantum numbers for allof the electrons inthe L and M shells, and notewhich correspond to the s, p, and d subshells.2.7 Give the electron configurations for the followingions: Fe2+, Fe3+, Cu+, Ba2+,Br-, andS2-.2.17 (a) Briefly cite the main differences betweenionic, covalent, and metallicbonding.(b) State the Pauli exclusion principle.2.18 Offer an explanation as to why covalently bonded materials are generally lessdense than ionically or metallically bonded ones.2.19 Compute the percents ionic character of the interatomic bonds for the followingcompounds: TiO2 , ZnTe, CsCl, InSb, and MgCl2 .2.21 Using Table 2.2, determine the number of covalent bonds that are possible foratoms of the following elements: germanium, phosphorus, selenium, and chlorine.2.24 On the basis of the hydrogen bond, explain the anomalous behavior of waterwhen it freezes. That is, why is there volume expansion upon solidification?3.1 What is the difference between atomic structure and crystal structure?3.2 What is the difference between a crystal structure and a crystal system?3.4Show for the body-centered cubic crystal structure that the unit cell edge lengtha and the atomic radius R are related through a =4R/√3.3.6 Show that the atomic packing factor for BCC is 0.68. .3.27* Show that the minimum cation-to-anion radius ratio for a coordinationnumber of 6 is 0.414. Hint: Use the NaCl crystal structure (Figure 3.5), and assume that anions and cations are just touching along cube edges and across face diagonals.3.48 Draw an orthorhombic unit cell, and within that cell a [121] direction and a(210) plane.3.50 Here are unit cells for two hypothetical metals:(a)What are the indices for the directions indicated by the two vectors in sketch(a)?(b) What are the indices for the two planes drawn in sketch (b)?3.51* Within a cubic unit cell, sketch the following directions:.3.53 Determine the indices for the directions shown in the following cubic unit cell:3.57 Determine the Miller indices for the planesshown in the following unit cell:3.58Determine the Miller indices for the planes shown in the following unit cell:3.61* Sketch within a cubic unit cell the following planes:3.62 Sketch the atomic packing of (a) the (100)plane for the FCC crystal structure, and (b) the (111) plane for the BCC crystal structure (similar to Figures 3.24b and 3.25b).3.77 Explain why the properties of polycrystalline materials are most oftenisotropic.5.1 Calculate the fraction of atom sites that are vacant for lead at its meltingtemperature of 327_C. Assume an energy for vacancy formation of 0.55eV/atom.5.7 If cupric oxide (CuO) is exposed to reducing atmospheres at elevatedtemperatures, some of the Cu2_ ions will become Cu_.(a) Under these conditions, name one crystalline defect that you would expect toform in order to maintain charge neutrality.(b) How many Cu_ ions are required for the creation of each defect?5.8 Below, atomic radius, crystal structure, electronegativity, and the most commonvalence are tabulated, for several elements; for those that are nonmetals, only atomic radii are indicated.Which of these elements would you expect to form the following with copper:(a) A substitutional solid solution having complete solubility?(b) A substitutional solid solution of incomplete solubility?(c) An interstitial solid solution?5.9 For both FCC and BCC crystal structures, there are two different types ofinterstitial sites. In each case, one site is larger than the other, which site isnormally occupied by impurity atoms. For FCC, this larger one is located at the center of each edge of the unit cell; it is termed an octahedral interstitial site. On the other hand, with BCC the larger site type is found at 0, __, __ positions—that is, lying on _100_ faces, and situated midway between two unit cell edges on this face and one-quarter of the distance between the other two unit cell edges; it is termed a tetrahedral interstitial site. For both FCC and BCC crystalstructures, compute the radius r of an impurity atom that will just fit into one of these sites in terms of the atomic radius R of the host atom.5.10 (a) Suppose that Li2O is added as an impurity to CaO. If the Li_ substitutes forCa2_, what kind of vacancies would you expect to form? How many of thesevacancies are created for every Li_ added?(b) Suppose that CaCl2 is added as an impurity to CaO. If the Cl_ substitutes forO2_, what kind of vacancies would you expect to form? How many of thevacancies are created for every Cl_ added?5.28 Copper and platinum both have the FCC crystal structure and Cu forms asubstitutional solid solution for concentrations up to approximately 6 wt% Cu at room temperature. Compute the unit cell edge length for a 95 wt% Pt-5 wt% Cu alloy.5.29 Cite the relative Burgers vector–dislocation line orientations for edge, screw, andmixed dislocations.6.1 Briefly explain the difference between selfdiffusion and interdiffusion.6.3 (a) Compare interstitial and vacancy atomic mechanisms for diffusion.(b) Cite two reasons why interstitial diffusion is normally more rapid thanvacancy diffusion.6.4 Briefly explain the concept of steady state as it applies to diffusion.6.5 (a) Briefly explain the concept of a driving force.(b) What is the driving force for steadystate diffusion?6.6Compute the number of kilograms of hydrogen that pass per hour through a5-mm thick sheet of palladium having an area of 0.20 m2at 500℃. Assume a diffusion coefficient of 1.0×10- 8 m2/s, that the concentrations at the high- and low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.6.7 A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at 1200℃and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 6×10-11m2/s, and the diffusion flux is found to be 1.2×10- 7kg/m2-s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 kg/m3. How far into the sheet from this high-pressure side will the concentration be 2.0 kg/m3?Assume a linear concentration profile.6.24. Carbon is allowed to diffuse through a steel plate 15 mm thick. Theconcentrations of carbon at the two faces are 0.65 and 0.30 kg C/m3 Fe, whichare maintained constant. If the preexponential and activation energy are 6.2 _10_7 m2/s and 80,000 J/mol, respectively, compute the temperature at which the diffusion flux is 1.43 _ 10_9 kg/m2-s.6.25 The steady-state diffusion flux through a metal plate is 5.4_10_10 kg/m2-s at atemperature of 727_C (1000 K) and when the concentration gradient is _350kg/m4. Calculate the diffusion flux at 1027_C (1300 K) for the sameconcentration gradient and assuming an activation energy for diffusion of125,000 J/mol.10.2 What thermodynamic condition must be met for a state of equilibrium to exist? 10.4 What is the difference between the states of phase equilibrium and metastability?10.5 Cite the phases that are present and the phase compositions for the followingalloys:(a) 90 wt% Zn–10 wt% Cu at 400℃(b) 75 wt% Sn–25wt%Pb at 175℃(c) 55 wt% Ag–45 wt% Cu at 900℃(d) 30 wt% Pb–70 wt% Mg at 425℃(e) 2.12 kg Zn and 1.88 kg Cu at 500℃(f ) 37 lbm Pb and 6.5 lbm Mg at 400℃(g) 8.2 mol Ni and 4.3 mol Cu at 1250℃.(h) 4.5 mol Sn and 0.45 mol Pb at 200℃10.6 For an alloy of composition 74 wt% Zn–26 wt% Cu, cite the phases presentand their compositions at the following temperatures: 850℃, 750℃, 680℃, 600℃, and 500℃.10.7 Determine the relative amounts (in terms of mass fractions) of the phases forthe alloys and temperatures given inProblem 10.5.10.9 Determine the relative amounts (interms of volume fractions) of the phases forthe alloys and temperatures given inProblem 10.5a, b, and c. Below are given theapproximate densities of the various metalsat the alloy temperatures:10.18 Is it possible to have a copper–silveralloy that, at equilibrium, consists of a _ phase of composition 92 wt% Ag–8。

《材料学导论》双语教学初探

《材料学导论》双语教学初探作者：马瑞梁宇万明攀来源：《教育教学论坛》2013年第03期摘要：文章依据作者开展双语教学的实践，阐述了《材料科学导论》双语教学课程的教学目标，从教材选择、教学活动组织方式等方面探索了开展双语教学课程的方法。

关键词：材料科学；双语教学；教学目标；教学活动中图分类号：G642.4？摇文献标志码：A ？摇文章编号：1674-9324（2013）03-0085-022001年，教育部在《关于加强高等学校本科教学工作提高教学质量的若干意见》中明确要求：为适应经济全球化和科技革命的挑战，本科教育要创造条件使用英语等外语进行公共课和专业课教学。

高校应积极推动使用英语进行教学，对部分自然科学专业和社会科学专业，力争3年内，英语教学课程（双语课程）达到所开课程的5%～10%[1]。

双语教学（Bilingual Teaching）是指用两种语言作为教学媒介语言，即使用母语和另外一种不同的语言同时教授一门课程，通过学习科目知识达到掌握语言的目的，在我国主要指的是同时使用汉语和英语[2]。

基于此，我院在材料科学与工程专业针对《材料科学导论》课程进行了双语教学改革与实践，探索利用多元化的教学资源，培养学生利用英文理解和解决相关专业问题的能力，提高学生综合素质的双语教学新模式。

一、教学目标定位《材料科学导论》（双语课程）是材料科学与工程本科专业的专业限选课。

课程设在第五学期，是学生入学后接触的第一批专业课程之一，也是采用双语教学的第一门课。

因此该课程定位为导论性的学科专业课和完成基础英语学习后向在专业学习中运用英语的过渡桥梁。

作为导论性的学科专业课，要使学生初步了解材料科学与工程研究的主要领域，熟悉基本的材料科学与工程领域的概念、术语和原则，了解材料科学与工程研究领域的发展动向，能够运用所学的知识解决简单的材料科学与工程领域研究和生产中的实际问题。

强调适应学生本身的经验和经历，控制课程内容的深度和范围。

《材料科学与工程基础》习题和思考题及答桉.doc

《材料科学与工程基础》习题和思考题及答案第二章2.1.按照能级写出N、0、Si、Fe、Cu、Br原子的电子排布（用方框图表示）。

2・2.的镁原子有13个中子，11.17%的镁原子有14个中子，试计算镁原子的原子量。

2.3.试计算N壳层内的最大电子数。

若K、L、M、N壳层中所有能级都被电子填满时，该原子的原子序数是多少？2-4.计算O壳层内的最大电子数。

并定出K、L、N、O壳层中所有能级都被电子填满时该原子的原子序数。

2-5.将离子键、共价键和金属键按有方向性和无方向性分类，简单说明理由。

2-6.按照杂化轨道理论，说明下列的键合形式：（1）CO?的分子键合（2）甲烷CH’的分子键合（3）乙烯C2H4的分子键合（4）水HQ的分子键合（5）苯环的分子键合（6）城基中C、。

间的原子键合2.7.影响离子化合物和共价化合物配位数的因素有那些？2-8.试解释表2.3-1中，原子键型与物性的关系？2-9.0°C时，水和冰的密度分别是1.0005 g/cm3和0.95g/cm3,如何解释这一现象？2.10.当CN=6时,K+离子的半径为0.133nm（a）当CN=4时，半径是多少? （b）CN=8时，半径是多少？2-11 .（a）利用附录的资料算出一•个金原子的质量？（b）每mm3的金有多少个原子？（c）根据金的密度，某颗含有10由个原子的金粒，体积是多少？（d）假设金原子是球形（E=0.1441nm）, 并忽略金原子之间的空隙，则10刀个原子占多少体积？（e）这些金原子体积占总体积的多少百分比？12.—个CaO的立方体晶胞含有4个Ca*离子和4个O?-离子，每边的边长是0.478nm, 则CaO的密度是多少？-硬球模式广泛的适用于金属原子和离子，但是为何不适用于分子？14.计算（a）面心立方金属的原子致密度；（b）面心立方化合物NaCl的离子致密度（离子半径r Na+=0.097,心=0.181）；（C）由计算结果，可以引出什么结论？2-15.铁的单位品胞为立方体，品格常数a=0.287nm,请由铁的密度算出每个单位品胞所含的原子个数。

材料科学与工程导论双语教学实践与思考

材料科学与工程导论双语教学实践与思考摘要：阐述了材料科学与工程导论双语教学的教学研究现状，探讨了制约其双语教学开展的不利状况，分析了双语教学存在的问题并给出了解决方案。

教学实践结果表明：合理组织教学资源，选用合适的教材，充分发挥多媒体技术的积极作用，在我校有针对性地开展双语教学是可行的。

关键词：材料科学与工程导论；双语教学；多媒体随着我国经济的高速发展，中国加入WTO和经济全球化时代的到来，加之科学技术的飞速发展，各学科知识体系不断交叉融合。

为了缩短与他国的信息交流距离，尽快与世界接轨，我国迫切需要既有很强的专业知识又精通外语的复合型高素质人才［1］。

双语教学已经成适应时代需要的教学改革的必然趋势，成为当前高校教学改革的一个热点［2］。

2001年9月教育部高等教育司颁发了《关于加强高等学校本科教学工作提高教学质量的若干意见》，明确提出要在高校积极推动使用英语等外语进行教学，其中“本科教育要创造条件使用英语等外语进行公共课和专业课教学”，还特别强调“高新技术领域的生物技术、信息技术、新材料技术等专业，更要先行一步，力争三年内，外语教学课程要达到所开课程的5%至10%”。

这份文件的颁发，不仅使高校开展双语教学的目的、意义、必要性和重要性得以升华，而且对推动全国高校开展双语教学提高教学质量将发挥重要的指导价值［3］。

国内的一流院校已率先引进国外先进的原版教材，在研究生和本科生中推行双语教学。

而地方院校也应该积极科学地应对高等教育的国际化趋势，根据自身的条件和特点，选择合适的学科和模式，积极探索和开展双语教学。

一制约材料科学与工程导论双语教学开展的现状长期以来，我国高等院校理工专业课程教学一般仅使用母语授课，只在有限的公共英语课和专业英语课上采用英语教学。

造成这种局面的主要原因一方面是高校对双语教学重视不够，另一方面是专业课教师的英语水平有限。

尽管理工专业课教师多数具备很强的专业知识背景，并且在长期跟踪领域前沿的过程中形成了较优秀的英语读写能力，但由于缺乏英语交流环境，其英语口语能力普遍较弱，甚至有的发音还不够标准。

材料科学基础第二版

材料科学基础第二版材料科学是一门研究材料的性质、结构、制备和应用的学科，它涉及到物质的基本特性和相互作用，对于现代工业和科技的发展起着至关重要的作用。

本书《材料科学基础第二版》旨在系统介绍材料科学的基本理论和知识，帮助读者全面了解材料科学的基本概念和原理，为相关专业的学生和科研人员提供一本全面而深入的参考书籍。

第一章从材料科学的基本概念和发展历程入手，介绍了材料科学的研究对象、基本特征以及其在工程技术中的应用。

通过对材料科学的起源和发展进行梳理，读者可以更好地理解材料科学的学科内涵和研究意义。

第二章主要介绍了材料的结构与性能。

材料的性能直接受其结构的影响，因此了解材料的结构对于预测和改善材料的性能至关重要。

本章详细介绍了晶体结构、非晶态结构以及材料的力学性能、热学性能等方面的知识，为读者提供了全面的材料结构与性能的基础知识。

第三章涉及了材料的制备与加工技术。

材料的制备和加工是材料科学的重要内容之一，它直接影响着材料的性能和应用。

本章主要介绍了材料的制备方法、加工工艺以及相关的材料表征技术，为读者提供了全面了解材料制备与加工技术的知识基础。

第四章讨论了材料的性能测试与评价。

材料的性能测试是材料科学研究的重要手段，通过对材料性能的测试和评价，可以全面了解材料的特性和应用潜力。

本章详细介绍了材料性能测试的方法、技术以及测试结果的分析与评价，为读者提供了全面了解材料性能测试与评价的知识基础。

第五章介绍了材料的应用与发展。

材料的应用是材料科学研究的最终目的，本章主要介绍了材料在工程技术、电子材料、光学材料、生物材料等方面的应用，并展望了材料科学的未来发展方向。

通过对材料科学基础的系统介绍，本书旨在帮助读者全面了解材料科学的基本理论和知识，为相关专业的学生和科研人员提供一本全面而深入的参考书籍。

希望本书能够成为读者学习和研究材料科学的重要工具，为材料科学的发展做出贡献。

材料科学与工程导论课程双语教学实践初探

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材料科学基础第二版答案

材料科学基础第二版答案材料科学基础是材料科学与工程专业的入门课程，它为学生提供了材料科学的基本概念、原理和知识体系。

本文档将为您提供材料科学基础第二版的答案，希望能够对您的学习和教学有所帮助。

第一章，材料科学基础概论。

1. 什么是材料科学？材料科学是研究材料的结构、性能、制备和应用的学科，它涉及金属、陶瓷、高分子材料等各种材料的研究和开发。

2. 材料的分类有哪些？材料可以分为金属材料、无机非金属材料和有机高分子材料三大类，每一类又可以进一步细分。

3. 材料的性能指标有哪些？材料的性能指标包括力学性能、物理性能、化学性能、热学性能等多个方面。

第二章，晶体结构。

1. 什么是晶体？晶体是由原子或分子按一定的规则排列而成的固体，具有规则的几何形状和周期性的结构。

2. 晶体结构的分类有哪些？晶体结构可以分为离子晶体、共价晶体、金属晶体和分子晶体四种类型，每一种类型都有其特定的结构特点和性质。

3. 晶体缺陷对材料性能有何影响？晶体缺陷会对材料的机械性能、热学性能、电学性能等产生影响，了解晶体缺陷对材料设计和制备具有重要意义。

第三章，材料的物理性能。

1. 材料的密度如何影响其性能？材料的密度直接影响其质量和体积，对材料的力学性能、热学性能等有重要影响。

2. 材料的热膨胀系数是什么？材料的热膨胀系数是材料在温度变化时长度变化的比例，对材料的热胀冷缩性能有重要影响。

3. 材料的导热性能和电导率有何关系？材料的导热性能和电导率都与材料内部的电子、原子结构密切相关，了解二者之间的关系对材料的应用具有指导意义。

第四章，材料的力学性能。

1. 材料的弹性模量是什么？材料的弹性模量是材料在受力时表现出的弹性变形能力，是衡量材料刚度的重要参数。

2. 材料的屈服强度和抗拉强度有何区别？材料的屈服强度是材料在受力时开始产生塑性变形的应力值，而抗拉强度是材料在拉伸断裂时所承受的最大应力值。

3. 材料的硬度测试方法有哪些？材料的硬度测试方法包括布氏硬度、洛氏硬度、维氏硬度等多种方法，每种方法都有其适用的范围和特点。

材料科学基础课程双语教学的实践与研讨

材料科学基础课程双语教学的实践与研讨作者：昌霞张小彬来源：《卷宗》2013年第01期摘要：介绍了本系开课的“材料科学基础”双语教学的意义以及面临的有利和不利条件，讨论了双语教学中教学模式的选择、教学内容的增删、教学方法的选择和考核方式等问题，并对未来的教学方法提出建议和展望。

关键字：双语教学；材料科学基础中图分类号：G642 文献标识码：A双语教育在很多国家受到重视，特别是美国、加拿大、新加坡等一些移民国家。

随着世界经济的一体化和我国加入WTO，我国对精通外语的专业人才的需求在迅速增加。

我国高等院校开展双语教学无疑是提高我国专业人才英语水平的一条重要途径。

为此，近年来在重庆理工大学也开展了材料科学基础课程双语教学的探索与实践。

但双语教学中第二语言的介入，对于一直以汉语母语为主的教学模式是很大的冲击，为探索双语教学的一般规律和教学特点，解决双语交叉的矛盾，并提高双语教学质量，为此对《材料科学基础》进行双语教学中取得的经验和存在的不足介绍给大家，以供同行参考与共勉。

一、开展双语教学的重要意义双语教学不是外语教学，而是学科教学。

双语教学的做法是用外语作为教学媒介进行其他学科的教学，融学科教学与外语教学为一体，为外语教学创造了良好的实践环境，为学生提供更多的实践机会。

外语教学是力求解决学生的语言基本问题，如发音、常用的单词、语法、句式等，使学生具备听、说、读和写的能力。

理想的学科英语的教学目标是让学生很好地掌握英语并使用英语，但现实中往往被各种考试所左右，如中学会考、高考、大学英语四六级考试、研究生入学考试、GRE等；而双语教学是将英语作为一种语言工具应用于课堂教学，目标是通过使用外语进行教学使学生理解和掌握各学科的知识和理论，通过应用外语而更好地掌握和使用外语。

语教学是在非语言类学科中用外语进行教学，通过在非语言类学科知识的学习来习得外语。

一方面，双语教学对外语学习有积极的促进作用；另一方面，外语学习为双语教学创造条件。

工程材料科学与设计原书第2版课后习题答案4—8章

工程材料科学与设计原书第2版课后习题答案4—8章Solutions to Chapter 41. FIND: What material has a property that is hugely affected by a small impurity level?SOLUTION: Electrical conductivity spans a wide range. Incorporation of a few parts per million impurities can change electrical conductivity orders of magnitude. Small cracks in brittle materials decrease their tensile strength by orders of magnitude. Small additions of impurity can change the color of gems. COMMENTS: These are but a few examples. 2. COMPUTE: The temperature at which the vacancy concentration is one half thatof 25o C.GIVEN: C 2 = C C 25v C 35vo oEQUATION:⎪⎪⎭⎫⎝⎛RT Q - = C fv v expwhere C v = vacancy concentrationQ fv = activation energy for vacancy information R = gas constant 8.314 J/mole-KT = absolute temperatureIn the present problem C)25(C = C C);35(C = C o v 2v o v 1vand T 1 = 35 + 273 = 308KT 2 = 25 + 273 = 298K⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛RT Q + RT Q - = C CRT Q - RT Q -= C CRT Q = CRT Q = C 2fv 1fv 2v 1v2fv 1fv 2v1v 2fv 2v1fv 1vexp exp exp exp expalso C v(35o C) = 2C v(25o C)Thus, Solving for Q fv we get Q fv = 52893.5 J/mole.Using this value of Q fv , the C v (25o C) can be calculatedThe problem requires us to calculate the temperature at which the vacancy concentration is ½ C v (25o C).½ C v (25o C) = 2.675 x 10-10Thusfor solving T, we get: T = 288.63K or 15.63o C.3. COMPUTE:C)80( C 3 = (T) C ov vGIVEN: C) 80( C 41 = C) 25( C o v ov EQUATION:⎪⎭⎫⎝⎛298.R Q - C) 25( C Sv o v expDividing (1) by (2) we get:⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎥⎦⎤⎢⎣⎡91784Q 308 + Q 298- R 1 = 2R(298)Q + R(308)Q - = C)25( C C)25( C 2fv fv fv fv ov o v exp exp10 x 5.35 = C)25(C 298 x 8.3152893.5- = C)25(C 10-o v ov ⎪⎭⎫ ⎝⎛exp ⎪⎭⎫⎝⎛T x 8.3152893.5- = 10 x 2.67510-ex p⎪⎭⎫ ⎝⎛353.R Q - = C) 80( C ov ex pSolving for Q, we get:Q = 22033.56 J/mole= exp(-7.511)= 5.46 x 10-4The problem requires computing a temperature at which C v = 3C v (80o C).3C v (80o C) = 3 x 5.46 x 10-4= 1.63 x 10-3⎪⎭⎫⎝⎛T x 8.3122033.56- = 10 x 1.633-ex psolving for T, we get:T = 413.05K or 140.05o C4.5. FIND: Are Al and Zn completely soluble in solid solution?If Al-Zn system obeys all the Hume-Rothery rules. Then it is expected to show complete solubility.(i) The atomic radii of Al and Zn are 0.143nm and 0.133 nm respectively. Thedifference in their radii is 7.5% which is less than 15%.(ii) The electronegativities of Al and Zn are 1.61 an 1.65 respectively which arealso very similar.(iii) The most common valence of Al is +3 and +2 for Zn.(iv) Al has an FCC structure where Zn has a HCP structure.It appears that Al-Zn system obeys 3 out of 4 Hume-Rothery rules. In this case they are not expected to be completely soluble.⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛3531 - 2981 R Q - = 41= C) 80( C C) 25( C ov ov exp ⎪⎭⎫ ⎝⎛353 x 8.3122033.56- = C) 80(C ov ex p6. SHOW: The extent of solid solution formation in the following systems usingHume-Rothery Rules.(a) Al in NiSize: r(Ni) = 0.125nm; r(Al) = 0.143nm difference = 14.4%Electronegativity: Al = 1.61; Ni = 1.91Most Common Valence: Al3+; Ni2+Crystal Structure: Al: FCC; Ni:FCCThe crystal structure of Al and Ni are the same and the most common valencies are also comparable. However, the size difference is close to 15% and the difference is electronegativities is rather significant.Based on this, it appears that Ni and Al would not form a solid solution over theentire compositional range.(b) Ti in NiSize: r(Ti) = 0.147 nm, r(Ni) = 0.125nm difference = 17.6%Electronegativity: Ti: 1.54; Ni: 1.91Valence: Ti4+; Ni2+Crystal Structure: Ti:HCP; Ni FCCTi in Ni would not exhibit extensive solid solubility(c) Zn in FeSize r(Zn) = 0.133nm; r(Fe) - 0.124nm difference = 7.25%Electronegativity: Zn = 1.65; Fe = 1.83Most Common Valence: Zn2+; Fe2+Crystal Structure: An: HCP; Fe: BCCSince electronegativities and crystal structures are very different, Zn - Fe will notexhibit extensive solid solubility.(d) Si in AlSize r(Si) = 0.117 nm; r(Al) = 0.143nm; difference = 22.2%Electronegativity: (Si) = 1.90; Al = 1.61Valence: Si4+; A;3+Crystal Structure: Si: Diamond Cubic; Al: FCCSince the size difference is greater than 15%, and the crystal structures are different, Si-Al would not exhibit extensive solid solubility.(e) Li in AlSize r(li): 0.152, r(Al): 0.143; difference - 6.29%Electronegativity: Li: 0.98; Al: 1.61Most Common Valence: Li1+; Al3+Crystal Structure: Li:BCC; Al: FCCSince electronegativity and crystal structures are very different, Li-Al will not exhibit extensive solid solubility.(f) Cu in AuSize r(Cu) = 0.125nm; r(au) = 0.144nm; difference = 12.5% Electronegativity: Cu = 1.90; Au = 1.93Most Common Valence: Cu+; Au+Crystal Structure: Cu:FCC; Au:FCCCu-Au will exhibit extensive solid solubility.(g) Mn in FeSize r(Mn) = 0.112, r(Fe) = 0.124 difference = 10.71%Electronegativity: Mn 1.55; Fe 1.83Most Common Valence: Mn2+; Fe2+Crystal Structure: Mn:BCC; Fe BCCThe difference in electronegativity is high but Mn-Fe does obey the other 3Hume-Rothery rules. Therefore, it will form solid solutions but not over the entire compositional range.(h) Cr in FeSize r(Cr) = 0.125nm, Fe = 0.144nm difference = 12.5%Electronegativity: Cr = 1.66; Fe = 1.83Most Common Valence: Cr3+; Fe2+Crystal Structure: Cr:BCC; Fe:BCCCr in Fe will exhibit extensive solid solubility but not over the entire compositional range since it obeys only 3 of 4 Hume-Rothery rules.(i) Ni in FeSize r(Ni) = 0.125nm, r(Fe) = 0.124nm difference = 0.8%Electronegativity: Ni: 1.91; Fe 1.83Most Common Valence: Ni3+; Fe3+Crystal Structure: Ni:FCC; Fe: BCCNi and Fe obeys 3 of the 4 Hume-Rothery rules therefore, extensive solid solutionwill be exhibited but not over the entire compositional range.7. (a) When one attempts to add a small amount of Ni to Cu, Ni is the solute and Cuis the solvent.(b) Based on the relative sizes of Ni and Cu, radius of Ni = 0.128nm, radius of Cu =0.125nm, these two are expected to form substitutional solid solutions.(c) Ni and Cu will be completely soluble in each other because they obey all fourHume-Rothery rules.8. FIND: Predict how Cu dissolves in Al.DATA: Cu Alatomic radius (A) 1.28 1.43electronegativity 1.90 1.61valence 1+,2+ 3+crystal structure FCC FCCSOLUTION: All of Hume-Rothery's rules must be followed for a substitutionalsolution. In this case, the valences do not match. Cu will not go into substitutional positions in Al to a large extent.COMMENTS: This principle is often used to precipitation harden Al using Cu.9. What type of solid solution is expected to form when C is added to Fe?The radius of carbon atom is 0.077nm and that of an Fe atom is 0.124nm. The size difference between these two is ~61% which is much grater than ~15%. Thus,these two are not expected to form substitutional solid solution.If we compare the size ratio of C to Fe atoms with the size of tetrahedral andoctahedral interstitial sites in BCC iron, we find that C does not easily fit into either type of interstitial position. C, however, forms an interstitial solid solution with Fe but the solubility is limited.10. FIND: Calculate the activation for vacancy formation in Fe.GIVEN: The vacancy concentration at 727 C = 1000K is 0.00022.SOLUTION: We use equation 4.2-2 to solve this problem:C v = exp (-Q fv/RT)Solving for Q fv:Q fv = -RT ln C v = -(8.31 J/mole-K)(1000K) ln 0.00022 = 7.0 x 104 J/mole11. SHOW: A Schottky and Frenkel defect in MgF2 structuresA 2-D representation of the MgF2 structure containing a Schottky defect and aFrenkel defect is shown below.12. Explain why the following statement is incorrect: In ionic solids the number ofcation vacancies is equal to the number of anion vacancies.In ionic crystals, even in the presence of vacancies, the charge neutrality must bemaintained. Therefore, single vacancies do not occur in ionic crystals sinceremoval of a single ion would lead to charge imbalance. Instead the vacanciesoccur in a manner such that the anion: cation vacancy ratio render the solidelectrically neutral. This, however, does not mean that the anion vacancies areequal to cation vacancies. For example, a Schottky defect in MgCl2 or MgF2involves two Cl- or F- cation vacancies for every Mg2+ anion vacancy to maintainelectrical neutrality.The number of cation vacancies equals the number of anion vacancies only for thelimiting case where the chemical formula of the compound is MX.13. Calculate the number of defects created when 2 moles of NiO are added to 98 molesof SiO2. Also, determine the type of defect created.GIVEN: Neglect interstial vacanciesWe have 2 moles of NiO and 98 moles of SiO2. Since NiO is a 1:1 compound there are 2 moles of Ni2+ ions and 2 moles of O2- ions present. SiO2 on the other hand is a 1:2 compound; therefore, there are 98 moles of Si4+ and 196 moles of O2-. Thetotal number of each type of ion isN Ni = 2 molesN Si = 98 molesN O2 = 196 molesThe total number of moles of ions in the system isN T = N Ni + N Si + N O = 2 + 98 + 196 = 196 molesEach substitution of an Ni2+ for Si4+ results in a loss of 2 positive charges. If nointerstitials are created, this loss of positive charge is balanced by the creation ofanion vacancies. Charge neutrality requires one oxygen vacancy created for every Ni 2+ ion. Therefore, the number of oxygen vacancies isN Ov = N Ni = 2 molesThere are 2 moles of oxygen ion vacancies created with the addition of 2 moles of NiO to 98 moles of SiO 2. 14. Calculate the number of defects created when 1 mole of MgO is added to 99 moles ofAl 2O 3.MgO is a 1:1 compound, therefore there is 1 mole of Mg 2+ ions and 1 mole of O 2- ions in the system.From Al 2O 3, there are 198 moles of Al 3+ ions and 297 moles of O 2- ions in the system.Each substitution of an Mg 2+ ion for Al 3+ ion results in a loss of one positive charge. This loss of positive charge is balanced by oxygen vacancy. Charge neutrality requires one oxygen vacancy to be created for every two Mg 2+ ion 3. Therefore thenumber of oxygen ion vacancies created is0.5 moles of oxygen ion vacancies are created by the addition of 1 mole of MgO to 99 moles of Al 2O 3. 15.COMPUTE: Relative concentration of cation vacancies, anion vacancies and cation interstitials.GIVEN: Q Cv = 20kJ/moleQ Av = 40kJ/mole Q CI = 30kJ/moleASSUMPTION: assume room temperatureT = 298KConcentration of cation vacancies, C Cv is given by0.5moles = 21= 2N= N Mg O vSimilarly for anion vacancies and for cation interstitials16. (a) Describe a Schottky defect in U 2(b) Would you expect to find more cation or anion Frenkel defects in this compound? Why?UO 2 has a fluorite structure with U 4+ ions occupying FCC lattice sites and O 2- occupying tetrahedral interstitial sites.(a) A Schottky defect in UO 2 will involve one U 4+ cation vacancy and 2 O 2- anion vacancies.(b) In general cation Frenkel defects are more common than anion Frenkel defects because cations are usually smaller. In this case, the radii of U 4+ is 0.106nm and that of O 2- is 0.132nm. The U 4+ cation is smaller than the O 2- anion. However, the size difference is not very high. Still, cation Frenkel defects are expected to be more.17. Ionic compound Li 2O(a) Describe a Schottky defect (b) Describe a Frenkel defectLi 2O has an antifluorite structure. O 2- ions occupy FCC lattice sites and Li + occupies tetrahedral interstitial sites.10 x 3.108 = (-8.0763) = C 298 x 8.3120,000- = RT Q - = C 4-Cv cv Cv exp exp exp ⎪⎭⎫⎝⎛⎪⎭⎫⎝⎛10 x 9.6 = (-16.152) = 298 x 8.3140,000- = C 8-AV ex p ex p ⎪⎭⎫⎝⎛ 10 x 5.48 = (-12.114) = 298 x 8.3130,000- = C 6-CIex p ex p ⎪⎭⎫⎝⎛(a) A Schottky defect in Li2O involves 2 Li2+ cation vacancies and one O2- anion vacancy(b) The ionic radii of Li+ and O2- are 0.078nm and 0.132nm respectively. Thismaterial is most likely to exhibit cation Frenkel defect since the size of the cation is much smaller than the anion.18. DETERMINE:(a) Interstitial Na+ ions(b) Interstitial O2- ions(c) Vacant Na+ sites(d) Vacant O2- sites in Na2OGIVEN: r(Na+) = 0.098nmr(O2-) = 0.132nmNa2O structure is similar to antifluorite structure. Na+ ions occupy tetrahedralinterstitial sites and O2- ions occupy FCC lattice sites.Since the ratio of Na:0 is 2:1 for this materials, a Schottky defect results in 2 cation vacancies for every one anion vacancy.no. of vacant Na+ sites = 2 x no. of vacant O2- sitesA cation Frenkel defect is more likely to occur in this material(a) Interstitial Na+ ions = 1(b) Interstitial O2- ions = 0(c) Vacant Na+ sites = 2(d) Vacant O2- sites = 119. SOLVENT: AuSOLUTE: N, Ag or CsDETERMINE: (a) which element is most likely to form an interstitial solidsolution.(b) which element is most likely to form a substitutional solid solution.r(Au) = 0.144nmr(N) = 0.071nmr(Ag) = 0.144nmr(Cs) = 0.265nm(a) Based on atomic radii N is most likely to form are interstitial solid solutionwith Au as solvent.(b) Ag is most likely to form a substitutional solid solution because the size difference between Au & N and Au & Cs is more than 15%.In addition, Au and Ag have similar valence, and crystal structure. Theelectronegativities are not quite similar, but since Ag-Au system obeys 3 out of 4 of the Hume-Rothery rules, Ag is the most likely element with which Au forms a substitutional solid solution.Section 4.4 Diffusion20. Under what condition can Fick’s first law be used to solve diffusion problems.The Fick’s first law can be used to solve diffusion problems provided the concentration gradient does not change with time.21. GIVEN: 1 wt% B is added to Fe.FIND: (a) if B would be present as an interstitial impurity or substitutionalimpurity, (b) fraction of sites occupied by B atoms, (c) if Fe containing B were to be gas carburized, would the process be faster or slower than for Fe which has no B? Explain.r(B) = 0.097nm r(Fe) = 0.124nm(a) Based on the atomic radii B would be present as an interstitial impurity(b) amount of B present = 1 wt%As a basis of calculation assume 100gms of material.Determine the no. of moles of Fe and B present.Total no. of moles of Fe and B = 1.773 + 0.092 = 1.865 moles.Fraction of sites occupied by B atoms = 1.8650.092= mole fraction of B = 0.049Thus, B roughly occupies 5% of the sites.mole 0.092 = 10.8111= B of molesmoles 1.773 = 55.8599=Fe of wt mol.Feof gms= Fe of moles(c) If Fe containing B were to be gas carburized the process would be slower than for Fe which has no B simply because the presence of B atoms already in interstitial sites leave fewer sites for interstitial C to diffuse through. 22.Determine which type of diffusion would be easier (a) C in HCP Ti (b) N in BCC Ti (c) Ti in BCC Tir(C) << r(Ti) so we can predict that diffusion occurs via an interstitial mechanism r(N)<<r(Ti). In this case the diffusion also occurs via interstitial mechanism.Ti in BCC Ti is a case of self-diffusion and self-diffusion occurs via a vacancy mechanism. In general the activation energy for self diffusion is higher than interstitial mechanism because vacancy mechanism involves two steps. One is to create a vacancy and second is to promote a vacancy/atom exchange. Thus Ti in BCC Ti will be the slowest.The activation energy for diffusion via interstitial mechanism is just the energy necessary to move an atom into a neighboring interstitial site. An open crystal structure, as opposed to a dense structure, should have a lower activation energy. Between BCC Ti and HCP Ti, BCC Ti has a more open structure (lower APF) than HCP Ti.Thus, N in BCC Ti diffusion would be the easiest by virtue of its lowest activation energy.23. GIVEN:C 1 = 0.19 at % at surfaceC 2 = 0.18 at % at 1.2mm below the surfaceD = 4 x 10-14 m 2/sec a o = 4.049 A oCOMPUTE: Flux of copper atoms from surface to interior.We must first calculate the concentration gradient in terms of [copperatoms/cm 3/cm]. It can be calculated as follows:The concentration gradient is thencm / atoms 10 x 4.60 = 10 x 6.02 x 63.54] / 2.70) x [(0.0018= C cm / atoms 10 x 4.86 = 10 x 6.02 x 63.54] / 2.70) x [(.0019= CN x Cu] wt at / FCCAl) of density x [(a/oCu = C 319232319231AV24.FIND: Predict whether diffusion is faster in vitreous or crystalline silica.GIVEN: Diffusion is the movement of atoms through the material one step at a time. The ease of movement is in part determined by the amount of space that surrounds each atom. In more open or less dense structures, atoms have an increased chance of being able to squeeze past a neighbor into a new position. SOLUTION: Diffusion can be thought of as an Arrhenius process. The activation energy is that required to move an atom from one position to another, as shown in Fig. 2.3-2. In a crystal the activation energy will be greater than in a glass, since the density is higher and there is less free, or unoccupied, volume. Thus, we expect diffusion to be slower in crystal than in glasses at the same temperature.COMMENTS: When a noncrystalline material is raised to a temperature above the glass transition temperature, diffusion increases enormously. In metals this brings about rapid crystallization. In some ceramic and polymer systems, crystallization may be slow or absent.25.FIND: Do textile dyes more readily penetrate crystalline or noncrystalline regions? GIVEN: Most textile fibers are semicrystalline, containing both crystalline andnoncrystalline regions. The density of the noncrystalline regions is less than that of the crystalline regions. Often dyeing is conducted at a temperature at which the noncrystalline regions are above their glass transition temperature.SOLUTION: Dye penetration through the glass will be greater than that through the crystal; however, the rate of dyeing is not sufficiently high to be commercially feasible. The temperature must be raised so that the noncrystalline polymer is in the rubber state. Diffusion becomes rapid (radially inward) into the small fibers.secsec cm atoms10 x 8.5 = Jcm atoms 10 x 2.125-m cm 10 x m 10 x 4- =dx dc D - = Jcm / atoms 10 x 2.125- = 0.1210 x 2.55- =0.1210 x 4.86) - (4.60 = dx dc 29419224214-4191819⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛COMMENTS: One of the key lessons that dye houses learn is that a sufficient amount of noncrystalline poorly oriented polymer must be present in the fiber. The temperature of the dye bath needs to be above the glass transition temperature. Sometimes water and carriers are used to swell the noncrystalline regions to get yet a greater diffusion rate. The dyes may attach to the polymer using ionic bonds or covalent bonds. Unattached dye may wash out later. 26.CALCULATE: The factor by which the diffusion coefficient of Al in Al 2O 3 change when temperature is increased from 1800o C to 2000o C GIVEN: T 1 = 1800o C = 2073KT 2 = 2000o C = 2273KEQUATION: ⎪⎭⎫⎝⎛RT Q - D = D o ex pdividing (1) by (2), we getfrom table 4.4-1 of the text Q = 477kJ/mole and R = 8.31 J/mole-K⎪⎭⎫⎝⎛RT Q - D = D 1o 1exp ⎪⎭⎫⎝⎛RT Q - D = D 2o 2exp ⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛ T 1 - T 1 R Q - = RT Q + RT Q - = D D 212121exp expThus, the diffusion coefficient of Al in Al 2O 3 changes by a factor of 11.43 when thetemperature is increased from 1800o C to 2000o C.27. FIND: Temperature at which a specimen of Fe must becarburized for two hours to achieve the same diffusion result asat 900o C for 15 hrs.GIVEN: T 1 = 900o C = 1173K; Q = 84000 J/molet 1 = 15 hrs; D o = 2.00 x 10-6 m 2/sec. t 2 = 2 hrs; R = 8.31 J/mole-KThe value of flux J is in units of cm 2 per sec.Flux per cm 2 J f = Jx time 3600 x 15 x dxdcD - = J 11f(1)We need the same result in 2 hours.D 11.43 = D 0.0871= D 0.087 = D D0.087= [-2.436]=22731 - 20731 8.31477000- = D D 1122121exp exp ⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛dxdcD - = J./ m 10 x 3.62 =1173 x 8.3184000- 10 x 2.00 = RT Q - D = D 1210-6-o1sec exp exp ⎪⎭⎫ ⎝⎛⎪⎭⎫⎝⎛ 3600 x 2 x dx dc D - = J 22f,J = J 2f 1f dividing (1) by (2).28. GIVEN:D = 4 x 10-4 m 2/s @ 20o CC 1 = 2.2 x 10-3 k mol/m 3wall thickness = 3mm, diameter = 50cm height = 10cmCOMPUTE: Initial rate of mass loss through cylinder.Initially the concentration of He outside the cylinder, C 2, is zero. First, we need to convert the concentration of He from kmol/m 3 into (atoms/cm 3)/cm.C 1 = 2.2 x 10-3 kmol/m 3 = 2.2mol/m 3 = 2.2 x 10-6 mol/cm 3 = 2.2 x 10-6 mole/cm 3In terms of (atoms/cm 3)/cm0.00135 = T 10108.3-T 8.3184000- 10 x 2.00 = 10 x 2.7RT Q - D = D / m 10 x 72. = D x 7.5 = D 7.5 = D D7.5 x D D = 1226-9-2o229-2121221⎪⎭⎫⎝⎛⎪⎭⎫⎝⎛⎪⎭⎫⎝⎛exp exp exp sec C.1258 =1531K = T 6.60310108.3 = T6.603- = )1n(0.00135 = T 10108.3-o 222The concentration gradient isThe flux of atoms per second per cm 2 is obtained by using Fick’s first law ofdiffusionThe rate of mass loss is 1.766 x 1019atoms/cm 2 sec. The total surface area of the cylinder is 2πr(r+h) where r = radius and h = height.Total surface area = 2π x 25 (25 + 10) = 5497.79 cm 2The rate of mass loss per secondNote:(i ) The steady state mass loss is calculated because the initial rate of mass loss (i.e., rate of mass loss at time t = 0) is 0. (ii ) It is assumed that the curvature of the cylinder is large enough to calculate J using the expression for plate geometry. 29. Diffusion across a polymer membrane depends not only on size of the diffusing species but also the polarity of the diffusing species. A polar membrane may pass nonpolar species but serve as a barrier to polar species.Saran wrap contains highly polar atoms making it a polar membrane which serves as a barrier to water which is a polar compound. thus, there is no diffusion of water through the package unlike polyethylene, which is a nonpolar membrane and allows diffusion of water molecules which form ice. 30.COMPUTE: Temperature required to yield a carbon content of 0.5% at a depthcmatoms 10 x 1.324 = mole atoms 10 x 6.02 x cm mole102.2x = C 3182336-1 cm atoms/ 10 x 4.415- = 0.310 x 1.324 - 0 = dx dcxC - C = dx dc 4181812 .cm atoms/ 10 x 1.766 )10 x (-4.415 )10 x 10 x (4- = dxdcD - = J 2191844-sec secsec secmoles0.16 = .atoms 10 x 9.709 =cm 5497.79 x - cm atoms10 x 1.766 =222219of 0.4mn below the surface of the rod in 48 hours. GIVEN: Carbon concentration the interior = 0.2w/oCarbon concentration in the furnace = 1.0w/oBase material: HCP TiEQUATION: In this problem c(x, t) = 0.5wt%c o = 0.2 wt%c s = 1.0 wt%From figure 4.4-11, when⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛Dt 2x erf - 1 = ]c - c []c - t) (x, [cRT Q- D = D o s o o ex p0.625=0.375 - 1 = Dt .04 erf Dt .04 erf - 1 = 0.375Dt 2.04cm erf - 1 = 0.2 - 1.00.2 - 0.5⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡From Metals Handbook, Desk Edition, Pg. 28.66 for C diffusion in Ti, D o = 3.02 x 10-3 cm 2/sec, Q = 20,000 cal/mole = 83682 J/mole. 31.The diffusion process through vacancy-interchange mechanism depends on creation of vacancies and vacancy/atom interchange.At comparable homologous temperatures, for Ge and Cu the diffusion coefficientfor that material which has a higher vacancy concentration would be higher.A covalent bond as opposed to a metallic bond is stronger and directional. It isalso difficult to create vacancies in a covalently bonded material due to its strongbonding. Therefore, the activation energy for vacancy creation in a covalentlybonded material such as Ge is larger than Cu which has a weak metallic bond.The directional nature of a covalent bond places geometrical restrictions on thevacancy atom interchange which again results in an increase in the activationenergy.sec /cm 10 x 2.33 = DDt = 0.630.040.630 = Dt0.040.630 = Z0.625 = (Z) erf 28-2⎪⎭⎫ ⎝⎛ C 582 = T855K = T11.18-10065.2 = T11.18- = T10065.2-10 x 7.72 = T 10065.2- T x 8.31483682- 10 x 3.02 = 10 x 2.33 = D e 83682J/mol = mole 20,000cal/ = Q /cm 10 x 3.02 = D o 6-3-8-2-3o ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛exp exp secTherefore, at comparable temperatures the diffusion coefficient for Ge will belarger.32. FIND: Describe the energy and entropy in Fig. 4.4-5a, b, and c.SOLUTION: The order in part a is high. The materials is perfect. There is only one way to arrange the atoms in such a system. The entropy is low. In part b there is less order, more disorder, and the entropy has increased. Part c is nearly random. It has low order and high entropy. Energy contains a contribution from entropy: E = H -TS, where E is energy, T is absolute temperature, and S is entropy. Assuming all other contributions to energy change negligibly (T and H), the energy of part c is the low, part a is high and part b is intermediate.COMMENTS: What is shown in going from a to c is the entropy of mixing.33. GIVEN: After 10 hrs at 550o C an oxide layer of thickness 8 μm is formed.COMPUTE: Thickness after 100 hrs. Using the definition of effective penetration distance and equation 4.4-11 of text,with γ = 2 we have Dt 2 x eff ≈.In this case34. GIVEN: D w = 1.0 x 10-12 m 2/s (water)D dc = 1.0 x 10 (dye carrier)D d = 1.0 x 10-14 m 2/s (dye)COMPUTE:(a) Times required for the water, dye and carrier to penetrate to the center of the fiber. m25.3 = x 100hr 10hr = x m8t t = t D t D = x x100hr = t 10hr = t? = x m 8 = x 2eff 2eff2122111eff 1eff 212eff 1eff μμμ(b) Same as (a) but fiber diameter doubles(c) If thermal diffusivity of PET is secm 10 x 828-how long will it take for the heat to penetrate to the center of a 50μm diameter fiber.(a) using equation 4.4-11 of text with γ = 2.for water,for dye carriersimilarly for dye t = 6.25secs.(b) If the diameter fiber is doubled x eff = 50 x 10-6 mfor water, Dt 2 = x eff minutes2.60 or secs 156 = tt x 10 x 1.0 = 10 x 625.00t x 10 x 1.0 = 2)10 x (25.0m 10 x 25 = x/s m 10 x 1.0 = D 12-12-12-6-26-eff 2-12w ⎪⎪⎭⎫⎝⎛ minutes26.04 = secs 15.63 = tm 10 x 25 = x/s m 10 x 1.0 = D 6-eff 2-13dcsimilarly for dye carriert = 6250 secsand for dyet = 6.25 x 104 secs(c) Substituting D with D th , we can use the same equation to calculate the timerequired for heat to penetrate the center of fiber diameter = 50μm.Note: The units of thermal diffusivity is m 2/sec and notK - m Watt as printed in text35. FIND: How long will it take to case carburize a steel chain to a depth of 1/16 inch?GIVEN: It requires 4 hours to carburize a plate of similar composition to a depthof 1/16 inch.ASSUMPTIONS: All carburization conditions are the same in both treatments.SOLUTION: Equation 4.4-11 is used to solve the problem:secs.625 = 10 x 1.0)10 x 50 ( = tt x D = 2)10 x (5010 x 50 = x/s m 19 x 1.0 = D 12-26-26-6-eff 2-12wsecs0.00195 = t10x 81 x 210 x 25 = _tt x 10 x 8 = 210 x 25t x D 2 = x 8-6-28-6-2th eff ⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛。

“材料工程基础”课程全英语教学的实践与思考

【探索与实践】
“ 材料工程基础 ” 课程全英语教学的实践与思考
赵
（上海理工大学
斌
２０００９３）
材料科学与工程学院，上海
摘要：介绍了在材料科学与工程专业低年级本科生中开设 “ 材料工程基础 ” 全英语教学课程的实践与经验。通过教师的精心准备，激发学生的学习兴趣，采取多媒体教学并结合讨论、互动等灵活教学手段，是保障全英语授课教学质量的重
仅要选用国外优秀的英文原版教材作为主教材，制作使用英文课件，用英语讲授课程内容，而且需要布置并批阅英文作业，考试采用英文命题并要求学生用英文答题。因此，全英语教学与双语授课有显著的不同，教学难度比双语课程大，要求也更高。另外，全英语课程的设立，目的在于传授专业知识，而非学习英语本身，因此，其与专业英语课程也
择全英语课程的关键ｌｌ】。而有的研究者在教学实践中发现，即使通过英语六级考试的学生，也不一定能适应全英语教学。对此，复旦大学蔡基刚教授认为，具有大学英语四级水平以上的学生，具备上全英语课程的能力，关键在于坚持和信心ｌ。在全英语课堂教学中，如何克服学生对语言本身的畏难情绪，提高其对课程内容的兴趣，是影响教学效果的重

《材料科学基础》课程双语教学探索与实践

热点和重点。为适应教学形势的发展，我们提出了在材料成型及控制工程专业开展《材料科学基础》课程的双语教学。根据应用型本科人才的培养目标，结合本课程的特点开展包括课程改革、理论与
实验教学方法改革、教学手段改革等在内的一系列研究与实践工作实现教学内容更加丰富、生动、理论联系实际和易于接受的教学目的，使学生更好地理解和掌握材料科学的基础知识。２双语教学的必要－陛双语教学，对于加深概念的理解，掌握最新的知识，提高综合素质等方面都具有显著的作用。通过教学方法的改革与实践，以使可
多、覆盖面广、理论和概念比较集中。课程中引入的语言ｗｒｓｎｏｄｄａｌｎｕｇ）概念（ｏｃｐｓ、理（ｒｃｌ）性能（ｒｐｒｅａｄａｇａｅ、ｃｎｅｔ）原ｐｉｉｅ、ｎｐｓｐｏｅｔｓｎｉｐｒｒｎｅ等，ｅｆｍａｃ）需要学生透彻理解，ｏ牢固掌握。对于我校材料成型及控制工程专业所培养的学生，毕业后绝大
多数充实到工厂、企业生产的第一线，少数将进入科研院所或继续深造。学生具有较高的科技英语及外文资料查阅的能力对学习和工作都将有很大的帮助，采用英语教学方式传授给学生的不是英语，
而是专业的核心知识和核心理论，这是因为材料科学基础课程的概念、理论、术语都来自西方，有时翻译会有偏差，有些较新的专业名
Ｆｎａｎａｓｕｄｍｅｔ，Ｍｉｒ — ｔｃｕａｖｌｐｎ，ＰｏｅｔｓｌｃｏｓｒｔｒｌＤｅｅｏｍｅｔｒｐｒｉ，Ｍａｅｉｌｕｅｔｒｓａ
教育部２０年明确提出要在高校积极推动使用英语等外语进０１行教学，中本科教育要首先创造条件，其力争外语教学课程达到所开课程的５１％％～０。双语教学是我国高等教育与国际接轨的重要手段，是我国高等教育改革发展的必然趋势，也是当前教育改革的

武汉理工大学材料科学基础(第2版)课后习题和答案

武汉理工大学材料科学基础（第2版）课后习题和答案第一章绪论1、仔细观察一下白炽灯泡，会发现有多少种不同的材料？每种材料需要何种热学、电学性质？2、为什么金属具有良好的导电性和导热性？3、为什么陶瓷、聚合物通常是绝缘体？4、铝原子的质量是多少？若铝的密度为2.7g/cm3，计算1mm3中有多少原子？5、为了防止碰撞造成纽折，汽车的挡板可有装甲制造，但实际应用中为何不如此设计？说出至少三种理由。

6、描述不同材料常用的加工方法。

7、叙述金属材料的类型及其分类依据。

8、试将下列材料按金属、陶瓷、聚合物或复合材料进行分类：黄铜钢筋混凝土橡胶氯化钠铅-锡焊料沥青环氧树脂镁合金碳化硅混凝土石墨玻璃钢9、Al2O3陶瓷既牢固又坚硬且耐磨，为什么不用Al2O3制造铁锤？第二章晶体结构1、解释下列概念晶系、晶胞、晶胞参数、空间点阵、米勒指数（晶面指数）、离子晶体的晶格能、原子半径与离子半径、配位数、离子极化、同质多晶与类质同晶、正尖晶石与反正尖晶石、反萤石结构、铁电效应、压电效应.2、（1）一晶面在某、y、z轴上的截距分别为2a、3b、6c，求出该晶面的米勒指数；（2）一晶面在某、y、z轴上的截距分别为a/3、b/2、c，求出该晶面的米勒指数。

3、在立方晶系的晶胞中画出下列米勒指数的晶面和晶向：（001）与[210]，（111）与[112]，（110）与[111]，（322）与[236]，（257）与[111]，（123）与[121]，（102），（112），（213），[110]，[111]，[120]，[321]4、写出面心立方格子的单位平行六面体上所有结点的坐标。

5、已知Mg2+半径为0.072nm，O2-半径为0.140nm，计算MgO晶体结构的堆积系数与密度。

6、计算体心立方、面心立方、密排六方晶胞中的原子数、配位数、堆积系数。

7、从理论计算公式计算NaC1与MgO的晶格能。

MgO的熔点为2800℃，NaC1为80l℃,请说明这种差别的原因。

材料科学与工程专业英语第二版翻译答案老师整理的知识分享

2.英译汉材料科学石器时代肉眼青铜器时代光学性质集成电路机械(力学)强度热导率1.材料科学指的是研究存于材料的结构和性能的相互关系。

相反，材料工程指的是，在基于材料结构和性能的相互关系的基础上，开发和设计预先设定好具备若干性能的材料。

2. 实际上，固体材料的所有重要性质可以概括分为六类：机械、电学、热学、磁学、光学和腐蚀降解性。

3. 除了结构和性质，材料科学和工程还有其他两个重要的组成部分：即加工和性能。

4. 工程师与科学家越熟悉材料的结构-性质之间的各种相互关系以及材料的加工技术，根据这些原则，他或她对材料的明智选择将越来越熟练和精确。

5. 只有在极少数情况下材料在具有最优或理想的综合性质。

因此，有必要对材料的性质进行平衡。

3. 汉译英Interdispline dielectric constantSolid materials heat capacityMechanical properties electro-magnetic radiationMaterials processing elasticity modulus1.直到最近，科学家才终于了解材料的结构要素与其特性之间的关系。

It was not until relatively recent times that scientists came to understand the relationship between the structural elements of materials and their properties .2.材料工程学主要解决材料的制造问题和材料的应用问题。

Material engineering mainly solve the problems of materials processing and materials application.3.材料的加工过程不但决定了材料的结构，同时决定了材料的特征和性能。

《材料科学与工程基础》习题和思考题及答案

《材料科学与工程基础》习题和思考题及答案《材料科学与工程基础》习题和思考题及答案第二章2-1.按照能级写出N、O、Si、Fe、Cu、Br原子的电子排布（用方框图表示）。

2-2.的镁原子有13个中子，11.17%的镁原子有14个中子，试计算镁原子的原子量。

2-3.试计算N壳层内的最大电子数。

若K、L、M、N壳层中所有能级都被电子填满时，该原子的原子序数是多少？2-4.计算O壳层内的最大电子数。

并定出K、L、M、N、O壳层中所有能级都被电子填满时该原子的原子序数。

2-5.将离子键、共价键和金属键按有方向性和无方向性分类，简单说明理由。

2-6.按照杂化轨道理论，说明下列的键合形式：（1）CO2的分子键合（2）甲烷CH4的分子键合（3）乙烯C2H4的分子键合（4）水H2O的分子键合（5）苯环的分子键合（6）羰基中C、O间的原子键合2-7.影响离子化合物和共价化合物配位数的因素有那些？2-8.试解释表2-3-1中，原子键型与物性的关系？2-9.0℃时，水和冰的密度分别是1.0005 g/cm3和0.95g/cm3，如何解释这一现象？2-10.当CN=6时，K+离子的半径为0.133nm(a)当CN=4时，半径是多少？(b)CN=8时，半径是多少？2-11.(a)利用附录的资料算出一个金原子的质量？(b)每mm3的金有多少个原子？(c)根据金的密度，某颗含有1021个原子的金粒，体积是多少？(d)假设金原子是球形(rAu=0.1441nm),并忽略金原子之间的空隙，则1021个原子占多少体积？(e)这些金原子体积占总体积的多少百分比？2-12.一个CaO的立方体晶胞含有4个Ca2+离子和4个O2-离子，每边的边长是0.478nm，则CaO的密度是多少？2-13.硬球模式广泛的适用于金属原子和离子，但是为何不适用于分子？2-14.计算（a）面心立方金属的原子致密度；（b）面心立方化合物NaCl的离子致密度（离子半径rNa+=0.097，rCl-=0.181）；（C）由计算结果，可以引出什么结论？2-15.铁的单位晶胞为立方体，晶格常数a=0.287nm，请由铁的密度算出每个单位晶胞所含的原子个数。

《材料科学与工程基础》双语教学实践探索 材料科学与工程基础第二版答案

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