Question

Carbon is allowed to diffuse through a steel plate 9.7-mm thick. The concentrations of carbon at the two faces are 0.664 and 0.339 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

844°C

Step-by-step explanation:

The problem can be easily solve by using Fick's law and the Diffusivity or diffusion coefficient.

We know that Fick's law is given by,

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

Where
`(\Delta c)/(\Delta x)` is the concentration of gradient

D is the diffusivity coefficient

and J is the flux of atoms.

In the other hand we have, that

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

Where
`D_0` is the proportionality constant,

R is the gas constant, T the temperature and
`E_d` is the activation energy.

Replacing the value of diffusivity coefficient in Fick's law we have,

`J = -D_0 ^{(E_d)/(RT)}(\Delta c)/(\Delta x)`

Rearrange the equation to get the value of temperature,

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

We have all the values in our equation.

`\Delta c = 0.664-0.339 = 0.325 C. cm^(-1)`

`\Delta x = 9.7*10^(-3)m`

`E_d = 82000J`

`D_0 = 6.5*10^(-7)m^2/s`

`J = 3.2*10^(-9)m^2/s`

`R= 8.31Jmol^(-1)K`

Substituting,

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

`T=-(-82000)/((8.31)ln((3.2*10^(-9)(9.7*10^(-3)))/(6.5*10^(-7) (0.325))))`

`T=1118.07K=844\°C`