Answer:
x = 2.78x10⁻³ m
Explanation:
According to Fick's law of diffusion:
[tex] J = -D\frac{\Delta C}{\Delta x} = -D\frac{C_{s} - C_{x}}{s - x} [/tex]
Where:
J: is the diffusion flux = 1.0x10⁻⁷ kg/(s*m²)
D: is the diffusion coefficient = 1.85x10⁻¹⁰ m²/s
[tex]C_{s}[/tex]: is the nitrogen concentration in the surface of steel = 2 kg/m³
[tex]C_{x}[/tex]: is the nitrogen concentration in the point x = 0.5 kg/m³
s: is the position at the steel's surface = 0
x: is the position into the sheet where the concentration is Cₓ
So, x is:
[tex] 1.0 \cdot 10^{-7} kg/s*m^{2} = -1.85 \cdot 10^{-10} m^{2}/s\frac{2 kg/m^{3} - 0.5 kg/m^{3}}{0-x} [/tex]
[tex] x = 1.85 \cdot 10^{-10} m^{2}/s\frac{2 kg/m^{3} - 0.5 kg/m^{3}}{1.0 \cdot 10^{-7} kg/s*m^{2}} = 2.78 \cdot 10^{-3} m [/tex]
Therefore, the nitrogen concentration will be 0.5 kg/m³ at 2.78x10⁻³ m in the steel sheet.
I hope it helps you!
