Explanation:
The given data is as follows.
[tex]T_{o} = 25^{o}C[/tex], d = 70 mm = [tex]70 \times 10^{-3} m[/tex]
k = 0.62 [tex]W/m^{o}C[/tex], [tex]\alpha = 0.2 \times 10^{-6} m/sec[/tex]
[tex]T_{\infinity} = 1.8^{o}C[/tex], v = 4 m/sec
h = 25 [tex]W/m^{2}^{o}C[/tex], [tex]T_{c} = 5.5^{o}C[/tex]
First, we will calculate the value of [tex]B_{i}[/tex] as follows.
[tex]B_{i} = \frac{h \times d}{4k}[/tex]
= [tex]\frac{25 \times 70 \times 10^{-3}}{4 \times 0.62}[/tex]
= 0.7056
Since, 0.7056 is greater than 0.1.
As, [tex]\frac{T_{c} - T_{\infinity}}{T_{o} - T_{infinity}}[/tex] = 0.159
= 0.16
Also here,
[tex]\frac{\alpha T}{d^{\gamma}}[/tex] = 1.9
[tex]\frac{0.2 \times 10^{-6} \times T}{\frac{(70 \times 10^{-3}}{4})^{\gamma}}[/tex] = 1.9
T = 46.18 hrs
Thus, we can conclude that it will take 46.18 hrs for the center temperature of the potato to drop to [tex]5.5^{o}C[/tex].
