In commercial pre-cooling of harvested oranges, cooling follows Newton's law of cooling: $\frac{T - T_a}{T_0 - T_a} = e^{-k t}$. The half-cooling time ($Z$) required to cool oranges from $30^\circ\text{C}$ to $17^\circ\text{C}$ with chilled air at $4^\circ\text{C}$ is $60\text{ minutes}$. The seven-eighths cooling time ($S$, the time required to cool the fruit from $30^\circ\text{C}$ to $7.25^\circ\text{C}$) is ________ minutes (answer in integer).
Answer Key:
180
Step-by-Step Solution & Explanation
Multi-Approach Solution:
**Method 1: Direct Exponential Half-Life Scaling**
In Newtonian cooling:
At $t = Z$ (half-cooling time), unaccomplished temperature ratio $\theta = \frac{1}{2} = 0.5$.
At $t = 2 Z$, $\theta = (0.5)^2 = 0.25$ (three-quarters cooling).
At $t = 3 Z$, $\theta = (0.5)^3 = 0.125 = \frac{1}{8}$.
Since seven-eighths cooling means $1 - \frac{7}{8} = \frac{1}{8}$ of the initial temperature difference remains:
$S = 3 \times Z = 3 \times 60\text{ min} = 180\text{ minutes}$
**Method 2: Analytical Rate Constant**
$k = \frac{\ln(2)}{Z} = \frac{0.69315}{60} = 0.011552\text{ min}^{-1}$
$S = -\frac{\ln(1/8)}{k} = \frac{\ln(8)}{k} = \frac{3 \ln(2)}{\ln(2)/60} = 3 \times 60 = 180\text{ minutes}$