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

Answer:

The temperature is 2584.5 K

Explanation:

Given:

Activation energy $Q = 80000$ $\frac{J}{mol}$

Preexponential $D= 6.2 \times 10^{-7}$ $\frac{m^{2} }{s}$

Diffusion flux $J = 6.4 \times 10^{-10}$ $\frac{kg}{m^{2} s}$

Thickness of plate $\Delta x = 11 \times 10^{-3}$ m

Concentration of carbon at two faces $\Delta C = (0.88 - 0.41 ) = 0.47$ $\frac{kg}{m^{3} }$

From the formula of temperature in terms of diffusion flux,

  $T = (\frac{Q}{R} ) \frac{1}{\ln (\frac{D\Delta C}{J\Delta x} )}$

Where $R =$ 8.314 $\frac{J}{mol.K}$ ( gas constant )

Put the values and find the temperature,

  $T = (\frac{80000}{8.314} ) \frac{1}{\ln (\frac{6.2 \times 10^{-7} \times 0.47 }{6.4 \times 10^{-10}\times 11 \times 10^{-3} } )}$

  $T = 2584.5$ K

Therefore, the temperature is 2584.5 K