Respuesta :
Answer:
975 K
Explanation:
Here, given :
The alloy is a FCC nickel-carbon.
[tex]$D_o= 2.3 \times 10^{-5} \ m^2/s$[/tex]
[tex]$Q_d=111,000 \ J/mol$[/tex]
Boltzmann Constant, [tex]$k=8.617 \times 10^{-5} \ eV/(atom-K)$[/tex]
Therefore,
[tex]$\frac{C_x-C_o}{C_s-C_o}= \frac{035-0.20}{1.0-0.20}$[/tex]
= 0.1875
[tex]$=1-\text{erf}\left(\frac{x}{2\sqrt{Dt}}\right)$[/tex]
So, [tex]$\text{erf}\left(\frac{x}{2\sqrt{Dt}}\right) = 0.8125$[/tex]
Therefore,
w erf w
0.92 0.80677
y 0.8125
0.96 0.82542
Now,
[tex]$\frac{y-0.92}{0.96-0.92} = \frac{0.8125-0.80677}{0.82542-0.80677}$[/tex]
y = 0.93228
[tex]$\text{erf}\left(\frac{x}{2\sqrt{Dt}}\right) = 0.93228$[/tex]
[tex]$D=\frac{x^2}{4t(0.93228)^2}$[/tex]
[tex]$D=\frac{(4\times 10^{-3})^2}{4\times 49.5 \times 3600 \times (0.93228)^2}$[/tex]
[tex]$= 2.58 \times 10^{-11}$[/tex]
[tex]$T=\frac{Q_d}{R(\ln D_o - \ln D)}$[/tex]
[tex]$T=\frac{111000}{8.31(\ln (2.3 \times 10^{-5}) - \ln (2.58 \times 10^{-11}))}$[/tex]
T = 974.84 K
T = 975 K
