Question

Carbon is allowed to diffuse through a steel plate 11 mm thick. The concentrations of carbon at the two faces are 0.88 and 0.41 kg C/m3 in Fe, which are maintained constant. If the preexponential and activation energy are 6.2 × 10-7 m2/s and 80,000 J/mol, respectively, compute the temperature (in K) at which the diffusion flux is 6.4 × 10-10 kg/m2-s. Enter your answer in accordance to the question statement K.

Answer 1

The temperature is 2584.5 K

Step-by-step explanation:

Given:

Activation energy
`Q = 80000`
`(J)/(mol)`

Preexponential
`D= 6.2 * 10^(-7)`
`(m^(2) )/(s)`

Diffusion flux
`J = 6.4 * 10^(-10)`
`(kg)/(m^(2) s)`

Thickness of plate
`\Delta x = 11 * 10^(-3)` m

Concentration of carbon at two faces
`\Delta C = (0.88 - 0.41 ) = 0.47`
`(kg)/(m^(3) )`

From the formula of temperature in terms of diffusion flux,

`T = ((Q)/(R) ) (1)/(\ln ((D\Delta C)/(J\Delta x) ))`

Where
`R =` 8.314
`(J)/(mol.K)` ( gas constant )

Put the values and find the temperature,

`T = ((80000)/(8.314) ) (1)/(\ln ((6.2 * 10^(-7) * 0.47 )/(6.4 * 10^(-10)* 11 * 10^(-3) ) ))`

`T = 2584.5` K

Therefore, the temperature is 2584.5 K