Carbon is allowed to diffuse through a steel plate 14-mm thick. The concentrations of carbon at the two faces are 0.680 and 0.129 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: 1135.58 K

Explanation:

Given

Q (d) = 82000 J/mol

D = 6.5*10^-7 m/s²

ΔC = 0.680 - 0.129 kgC / m³ Fe = 0.551

Δx = 14*10^-3 m

J = 3.2*10^-9 kg/m²s

J = - D * (ΔC/Δx) * e^[Q (d) / RT]

If we make T subject of formula, we have

T = Q (d) / R * In [-DΔC / JΔx]

T = 82000 / 8.31 * In [-6.5*10^-7 * 0.551 / 3.2*10^-9 * 14*10^-3]

T = 82000 / 8.31 * In [-3.582*10^-7 / 4.2*10^-11]

T = 82000 / 8.31 * 9.05

T = 82000 / 72.21

T = 1135.58 K

So, the temperature at which the diffusion flux is stated is 1135.58 K

Temperature = 1092 K

Explanation:

The equation for diffusion flux is given as;

J = D (ΔC/Δx) •e^ (Q_d/RT)

Where;

J is diffusion flux

D is pre-exponential energy

ΔC is change in concentration

Q_d is Activation energy

R is gas constant

T is temperature

The question wants us to find the temperature. Thus, let's make Temperature T the subject.

Thus,

T = Q_d/[RIn (D (ΔC/JΔx)) ]

Now, we are given that;

Q_d = 82 KJ/mol = 82,000 J/mol

R = 8.31 J/mol. k

D = 6.5 x 10^ (-7) m²/s

ΔC = 0.68 - 0.129 = 0.579kg C per m³

J = 3.2 x 10^ (-9) kg / (m²s)

Δx = 14mm = 14 x 10^ (-3) m

Thus, plugging in all of the above values into the equation, we have;

T = 82,000/[8.31In (6.5 x 10^ (-7) (0.579 / (3.2 x 10^ (-9) •14 x 10^ (-3))) ]

T = 82,000 / (8.31In (8400.67))

T = 82,000/75.09 = 1092K