Foundations of Materials Science and Engineering
6th Edition
ISBN: 9781259696558
Author: SMITH
Publisher: MCG
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Chapter 5.7, Problem 39SEP
The activation energy of nickel atoms in FCC iron is 280 kJ/mol and carbon atoms in FCC iron is 142 kJ/mol. (a) What does this tell you about the comparative diffusion of nickel and carbon in iron? (b) Can you explain why the activation energies are so drastically different? (c) Find a way to qualitatively explain how much energy is 142 kJ to a non-engineer or a non-scientist.
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why diffusion coefficients of carbon in BCC and FCC in iron at temperature 912 c are difference? explain ?
Question: 4
A sheet of copper is exposed to oxygen at 1000°C. After 100 h, 0.245 g of copper are lost per
cm' of surface area; after 250 h, 0.385 g/ cm² are lost; and after 500 h, 0.560 g/ em are lost.
i. Determine whether oxidation is parabolic, linear, or logarithmic
ii. Calculate the time required for a 0.76-cm sheet of copper to be completely oxidized. The
sheet of copper is oxidized from both sides.
Maximum solubility
in FCC austenite at
1147 °C is
2.04 wt%
2.14 wt%
14.2 wt% O
0.022 wt% O
Chapter 5 Solutions
Foundations of Materials Science and Engineering
Ch. 5.7 - Prob. 1KCPCh. 5.7 - Write an equation for the number of vacancies...Ch. 5.7 - Prob. 3KCPCh. 5.7 - Prob. 4KCPCh. 5.7 - Describe the substitutional and interstitial...Ch. 5.7 - Prob. 6KCPCh. 5.7 - What factors affect the diffusion rate in solid...Ch. 5.7 - Write the equation for Ficks second law of...Ch. 5.7 - Prob. 9KCPCh. 5.7 - Prob. 10KCP
Ch. 5.7 - (a) Calculate the equilibrium concentration of...Ch. 5.7 - Prob. 12AAPCh. 5.7 - Determine the diffusion flux of zinc atoms in a...Ch. 5.7 - The diffusion flux of copper solute atoms in...Ch. 5.7 - Prob. 15AAPCh. 5.7 - Prob. 16AAPCh. 5.7 - Prob. 17AAPCh. 5.7 - A gear made of 1020 steel (0.20 wt% C) is to be...Ch. 5.7 - Prob. 19AAPCh. 5.7 - The surface of a steel gear made of 1020 steel...Ch. 5.7 - Prob. 21AAPCh. 5.7 - If boron is diffused into a thick slice of silicon...Ch. 5.7 - Prob. 23AAPCh. 5.7 - Prob. 24AAPCh. 5.7 - Prob. 25AAPCh. 5.7 - Prob. 26AAPCh. 5.7 - Prob. 27AAPCh. 5.7 - Prob. 28AAPCh. 5.7 - Prob. 29AAPCh. 5.7 - Prob. 30AAPCh. 5.7 - The diffusivity of copper atoms in the aluminum...Ch. 5.7 - Prob. 32AAPCh. 5.7 - Prob. 33SEPCh. 5.7 - Prob. 34SEPCh. 5.7 - Prob. 37SEPCh. 5.7 - Prob. 38SEPCh. 5.7 - The activation energy of nickel atoms in FCC iron...Ch. 5.7 - Prob. 40SEPCh. 5.7 - The self-diffusion of iron atoms in BCC iron is...Ch. 5.7 - Would you expect the diffusion rate of copper...Ch. 5.7 - Would you expect the diffusion rate of copper...Ch. 5.7 - Prob. 44SEPCh. 5.7 - Prob. 45SEPCh. 5.7 - Prob. 46SEPCh. 5.7 - Prob. 47SEPCh. 5.7 - Prob. 48SEPCh. 5.7 - Prob. 49SEPCh. 5.7 - Prob. 50SEP
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- P.3 The diffusivity of manganese atoms in the FCC iron lattice is 1.50 x 10-14 m²/s at 1300°C and 1.50 x 10-15 m²/s at 400°C. Calculate the activation energy in kJ/mol for this case in this temperature range. Data: R= 8.314 J/(mol-K).arrow_forwardDetermine the activation energy for vacancy formation in aluminum, if theequilibrium number of vacancies at 500 °C is 7.57 × 1023 m-3. Given that theatomic weight and density for aluminum are 26.98 g/mol and2.62 g/cm3, respectively.arrow_forwardNickel experiences corrosion in an acid solution according to the reaction Ni + 2H† → Ni²+ + H, (a) The rates of both oxidation and reduction half-reactions are controlled by activation polarization. Compute the rate of oxidation of Ni (in mol/cm2-s) given the following activation polarization data: For Nickel For Hydrogen V(H/H,) = 0 V i, = 6 × 10-7 A/cm² B = -0.10 V(NINP") = -0.25 V io = 10¬8 A/cm² B = +0.12 (b) Compute the value of the corrosion potential.arrow_forward
- Iron (Fe) undergoes an allotropic transformation at 912°C: upon heating from a BCC (α phase) to an FCC (γ phase). Accompanying this transformation is a change in the atomic radius of Fe—from RBCC = 0.12584 nm to RFCC = 0.12894 nm—and, in addition, a change in density (and volume). Compute the percentage volume change associated with this reaction. Indicate a decreasing volume by a negative number.arrow_forwardConsider de-oxidation by the addition of ferromanganese (60 percent Mn) to molten steel at 1600°C.The initial oxygen content is 0.04 wt%. It has to be brought down to 0.02 wt%. Calculate the quantity of ferromanganese required per tonne of steel. The manganese content of steel before de- oxidation is 0.1 wt%. (MnO) + [Fe] = [Mn] + (FeO) Given: K Mn at 1873 K = 0.4 K Fe-Mn at 1873 K = 0.156 Assumptions: Fe and Mn form ideal solutionsarrow_forwardGive three examples of materials processes that rely on diffusion in solids and explain how diffusion plays a critical role for one of those processes.arrow_forward
- Q3/ A- What are the important reactions in( Fe- C) system? B- Calculate the activation energy for vacancy formation in aluminum ,given that the equilibrium number of vacancies at 500°C is 7.57 x10 m.The atomic weight and density for aluminum are respectively, 26.98g /mol and 2.62 g/em.,Avogadros number ( 6.022x10 atoms/mol ), and Boltzmanns constant (8.62x10 ev /atom.K).?arrow_forwardA NIO-20 mol % Mgo ceramic is allowed to solidify. (See the figure below.) Temperature (°C) 2800 2600 2400- 2200 2000 L (Ni, Mg)0 40 60 80 MgO Mole percent MgO NIO 20 (a) Determine the composition (in mol% MgO) of the first solid to form. 80 x (b) Determine the composition (in mol% MgO) of the last liquid to solidify under equilibrium conditions. 35 X%arrow_forwardAt 660°C the fractional concentration of vacancies in aluminum is 9 × 10-5. The energy for vacancy formation is therefore: a. 0.72 eV b. 1.44 eV c. 0.79 eV d. 0.75 eV e. 0.68 eVarrow_forward
- Metal A (Tm=860 oC) and metal B (Tm=620 oC) were fully dissolved in each other in the liquid state. The ultimate solubility of B in A is 12% at 460 oC which decreasing with decreasing temperature to 3% at room temperature while the maximum solubility of A in B is 7% at 460 oC still constant to room temperature. A and B form eutectic at 460 oC and 48 % of B. construct phase diagram for A-B alloying system, label all phases and temperatures, and answer the following: a) 28 wt % A alloy is slowly cooled from 720 oC at what temperature does the first solid phase is formed? What is the composition of this solid phase? For this alloy what is the mass fraction of the phases at, 600 oC, 300 oC ,and TR? c) What is the ?composition of the phases in (b)arrow_forward3. Why are dislocations not considered thermodynamic equilibrium defects like vacancies? Explain by describing the relative interplay between entropy and enthalpy affecting the Gibbs free energy of the dislocations-containing material system.arrow_forwardThe activation energy for the diffusion of atomic species A in metal B is 114 kJ/mol. Calculate the diffusion coefficient at 891°C, given that the value of D at 1170°C is 76.7 × 10^-12 m2/s.arrow_forward
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