Answer:
option e
Explanation:
The avrami equation is given as follows
[tex]y = 1 -e^{-kt^{n} }[/tex]
the above equation can be re-written as
[tex]1-y = e^{-kt^{n} } \\\\\frac{1}{1-y} = e^{kt^{n}}\\\\ln(\frac{1}{1-y} )=kt^{n} \\\\hence\\\\t^{n} = \frac{ln(\frac{1}{1-y} )}{k}[/tex]
where y = the percentage
k = 4.46×10⁻⁴
t = time
n = constant
We determine n as follows
Using any of the cases, for instance case 1
when y = 40% or 0.4 , t = 13.1s
[tex]t^{n} = \frac{ln(\frac{1}{1-y} )}{k}[/tex]
[tex]13.1^{n} = \frac{ln(\frac{1}{1-0.4} )}{4.46\times 10^{-4} }[/tex]
solving the above equation, we have
[tex]13.1^{n} = 1145.35\\\\ln(13.1^{n}) = ln(1145.35)\\\\n\times ln(13.1) = ln(1145.35)\\\\n = \frac{ln(1145.35}{ ln(13.1)}= 2.74[/tex]
The value of n is the same if we use the second case
that is, y = 60% or 0.6 and t = 16.2s
Now when y = 95% or 0.95 and using n = 2.74
[tex]t^{2.74} = \frac{ln(\frac{1}{1-0.95} )}{4.46\times 10^{-4} }[/tex]
[tex]t^{2.74}= 6716.88\\ \\ln(t^{2.74})=ln(6716.88)\\\\2.74 \times ln(t)=ln(6716.88)\\\\ln(t)=\frac{ln(6716.88)}{2.74}\\ \\ln(t)=3.2162\\\\t = e^{3.2162}\\ \\t = 24.93s[/tex]
Hence
t = 25s
