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Carbon is allowed to diffuse through a steel plate 13-mm thick. The concentrations of carbon at the two faces are 0.654 and 0.377 kg C/m3 Fe, which are maintained constant. If the preexponential and activation energy are 6.5 × 10-7 m2/s and 82 kJ/mol, respectively, calculate the temperature (in K) at which the diffusion flux is 3.2 × 10-9 kg/(m2-s).

Respuesta :

Answer:

T = 1177.95 K

Explanation:

Equation for diffusion flux is given as;

J = D(ΔC/Δx)e^(Qd/RT)

Where J = 3.2 × 10^(-9) kg/m².s

Qd = 82Kj/mol = 82,000 J/mol

D = 6.5 × 10^(-7) m²/s

ΔC = 0.654 - 0.377 = 0.277 kg.C/m³

Δx = 13mm = 0.013m

R = 8.314 J/mol.k

Let's make T the subject of the equation;

T = Qd/(R In D(ΔC/JΔx))

T = 82,000/(8.314 In 6.5 × 10^(-7)(0.277/(3.2 × 10^(-9) x 0.013))

T = 82000/(8.314In 4328.125) = 1177.95 K

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