Question

Factors that Influence Diffusion III Carbon is allowed to diffuse through a steel plate 13-mm thick. The concentrations of carbon at the two faces are 0.702 and 0.110 kg of carbon per m³ of iron, which are maintained constant. If the preexponential and activation energy are 6.5 x 10^-⁷ m²/s and 82 kJ/mol, respectively, calculate the temperature (in K) at which the diffusion flux is 3.2 x 10-⁹ kg/(m²-s).

Answer 1

`T=899.849K=626.7\°c`

Step-by-step explanation:

We need to use Fick's Law to resolve this problem.

We know for this Law that:

`J=-D(\Delta c)/(\Delta x)`

And we know as well that the diffusivity coefficient can be expressed as follows,

`D=D_0 e^{-(Ed)/(RT)}`

Where J is the flux of atoms, D is the diffusivity, R is the gas constant, Ed is the activation energy and \frac{\Delta c}{\Delta x} is the concentration of gradient.

To calculate the temperature we need to remplace the equation of diffusivity coefficient in the Fick's law equation.

`J=-D(\Delta c)/(\Delta x) = -D_0 e^(-E_d/RT)(\Delta c)/(\Delta x)`

Rearrange the equation to get the value of temperature

`ln((J\Delta x)/(D_0 \Delta c))=(-(E_d)/(RT))`

`T=-(E_d)/(Rln((J\Delta x)/(D_0 \Delta c)))`

We have all the values,

`\Delta x= 10*10^(-3)m\\\Delta c= 0.85-0.40 =0.45kg.c.cm^(-1)`

`Ed=-80000Jmol^(-1)K^(-1)\\R=8.31Jmol^(-1)K^(-1)\\J=6.3*10^(-10)kg.m^(-2)\\D_0 = 6.2*10^(-7)m^2s^(-1)`

So substituting,

`T=(-80000)/(8.31*ln(((6.3*10^(-10))(10*10^(-3)))/(6.2*19^(-7)*0.45)))`

`T=899.849K=626.7\°c`