Answer: The temperature of the system will be 1622 K
Explanation:
The equation relating the pre-exponential factor and activation energy follows:
[tex]\log D=\log D_o-\frac{E_a}{2.303RT}[/tex]
where,
D = diffusion coefficient = [tex]2.7\times 10^{-14}m^2/s[/tex]
[tex]D_o[/tex] = pre-exponential constant = [tex]1.7\times 10^{-5}m^2/s[/tex]
[tex]E_a[/tex] = activation energy of iron in cobalt = 273,300 J/mol
R = Gas constant = 8.314 J/mol.K
T = temperature = ?
Putting values in above equation, we get:
[tex]\log (2.7\times 10^{-14})=\log (1.7\times 10^{-5})-\frac{273,300}{2.303\times 8.314\times T}\\\\T=1622K[/tex]
Hence, the temperature of the system will be 1622 K
