Question Bank • QB_DFE_032 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Unit Operations in Dairy and Food Engineering

Subtopic: Sterilization

Question Statement:

During thermal processing of a low-acid canned food, the cold-point temperature is monitored at 5-minute intervals during a holding period: at $t = 0\text{ min}$, $T = 115.1^\circ\text{C}$; at $t = 5\text{ min}$, $T = 121.1^\circ\text{C}$; and at $t = 10\text{ min}$, $T = 118.1^\circ\text{C}$. Taking $z = 10.0^\circ\text{C}$ and reference temperature $T_{ref} = 121.1^\circ\text{C}$, the lethal rates ($L = 10^{(T - 121.1)/10}$) are $0.251$ at $115.1^\circ\text{C}$, $1.000$ at $121.1^\circ\text{C}$, and $0.501$ at $118.1^\circ\text{C}$. Using the trapezoidal numerical integration rule ($F_0 = \frac{\Delta t}{2} [L_0 + 2 L_1 + L_2]$), the accumulated $F_0$ value in minutes over this 10-minute period is ________ (round off to 2 decimal places).

Answer Key:

6.88

Step-by-Step Solution & Explanation

Using the trapezoidal numerical integration rule with time interval $\Delta t = 5\text{ min}$: $F_0 = \frac{\Delta t}{2} [L_0 + 2 L_1 + L_2]$ $F_0 = \frac{5}{2} [0.251 + 2(1.000) + 0.501]$ $F_0 = 2.5 \times [0.251 + 2.000 + 0.501] = 2.5 \times 2.752 = 6.88\text{ minutes}$
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