An FCC iron-carbon alloy initially containing 0.20 wt% C is carburized at an elevated temperature and in an atmosphere wherein the surface carbon concentration is maintained at 1.0 wt%. If after 48 h the concentration of carbon is 0.35 wt% at a position 3.9 mm below the surface, determine the temperature at which the treatment was carried out. You will need to use data in the two tables below to solve this problem.
Answers
Answer:
Hello the table required is missing attached below is the missing table
Answer : 1299.05 k
Explanation:
Given data :
concentration of carbon ( Cx ) at a position 3.9mm below surface = 0.35
concentration of carbon ( Co ) = 0.20
surface carbon ( Cs ) = 1
Determine the temperature at which the treatment was carried out
first we will determine the value of Z in the table attached
given that the value of erf ( Z ) for Z = 0.8125 from the table
= [tex]\frac{Z- 0.9}{0.95-0.9} = \frac{0.8125-0.797}{0.8209-0.797}[/tex]
make Z subject of the equation
Z = 0.932
next calculate the diffusion coefficient using the relation below
Z = [tex]\frac{x}{2\sqrt{Dt} }[/tex] ----- ( 1 ) where ; z = 0.932 , x = 3.9 mm , t = 48h
0.932 =
x = 3.9 mm = 0.0039 m
t = 48 h = ( 48 *60* 60 ) = 172800 secs
Insert values into equation 2
0.932 = [tex]\frac{0.0039}{2\sqrt{D * 172800} }[/tex]
[tex]\sqrt{D*172800}[/tex] = (0.0039 / 0.932 ) / 2
172800 * D =( 0.0021 )^2
therefore D = 2.55 * 10^-11 m^2/s
Finally calculate the temperature at which the treatment was carried out
D = [tex]Do exp(\frac{-Qd}{RT} )[/tex] ----- ( 3 )
D = 2.55 * 10^-11 m^2/s ,
Do = 2.3*10^-5 m^2/s ( gotten from Diffusion data table )
Qd = 148000 J/mol ( gotten from Diffusion data table )
R = 8.31 J/mol k ( gotten from Diffusion data table )
back to equation 3
D / Do = exp ( -17810 / T )
1.11 * 10^-6 = exp ( -17810 / T )
therefore T = -17810 / ln( 1.11 * 10^-6 )
= - 17810 / -13.71 = 1299.05 k
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Answers
Answer:
See explanation
Explanation:
The percentage by mass of oxygen is defined as;
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Answers
Answer:
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Answers
Answer:
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Answers
Answer:
See Explanation
Explanation:
Given:
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From the question, the metal is the limiting reactant and must be used to obtain the required results which are;
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b) There are 40g of M in MF2.
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Answers
Answer:
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When a chemical reaction is run in aqueous solution inside a calorimeter, the temperature change of the water (and Ccal) can be used to calculate the enthalpy change for the reaction. Here, we will study the reaction of hydrochloric acid with sodium hydroxide in the calorimeter from problem 3. Equal volumes (50.0 mL) of 1.00 M sodium hydroxide and 1.00 M hydrochloric acid are mixed.
HCl+NaOH→NaCl+H2O
1. What is the total change in enthalpy (in Joules) for the reaction?
2. Where initial temperature is 21.2 °C and final temperature is 28.0 °C. Ccal is 1234.28 j
Answers
Answer:
The total change in enthalpy for the reaction is - 81533.6 J/mol
Explanation:
Given the data in the question;
Reaction;
HCl + NaOH → NaCl + H₂O
Where initial temperature is 21.2 °C and final temperature is 28.0 °C. Ccal is 1234.28 J
Moles of NaOH = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol
Moles of HCl = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol
so, 0.0500 moles of H₂O produced
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now ,
Heat energy of Solution q= (mass × specific heat capacity × temp Δ) + Cal
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so we substitute
q_soln = (100g × 4.18 × ( 28.0 °C - 21.2 °C) ) + 1234.28
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Enthalpy change for the neutralization is ΔH[tex]_{neutralization}[/tex]
ΔH[tex]_{neutralization}[/tex] = -q_soln / mole of water produced
so we substitute
ΔH[tex]_{neutralization}[/tex] = -( 4076.68 J ) / 0.0500 mol
ΔH[tex]_{neutralization}[/tex] = - 81533.6 J/mol
Therefore, the total change in enthalpy for the reaction is - 81533.6 J/mol
