Question Bank • QB_DFE_UOP_020 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Unit Operations in Dairy and Food Engineering

Subtopic: Sterilization

Question Statement:

A thermal sterilization process at $121.1^\circ\text{C}$ is designed for canned meat to achieve a target lethality of $F_0 = 6.0\text{ minutes}$ against \textit{Clostridium botulinum}. If the initial spore load is $N_0 = 10^4\text{ spores/can}$ and the decimal reduction time is $D_0 = 0.21\text{ minutes}$, the decimal logarithm of the survival probability per can ($\log_{10} N$) is ________ (round off to two decimal places).

Answer Key:

-24.57

Step-by-Step Solution & Explanation

Multi-Approach Solution: **Method 1: First-Order Log Survival Law** $F_0 = D_0 [\log_{10}(N_0) - \log_{10}(N)]$ $6.0 = 0.21 [\log_{10}(10^4) - \log_{10}(N)]$ $6.0 = 0.21 [4 - \log_{10}(N)]$ $4 - \log_{10}(N) = \frac{6.0}{0.21} = 28.5714$ $\log_{10}(N) = 4 - 28.5714 = -24.5714 \approx -24.57$ **Method 2: Number of Decimal Reductions ($n$)** $n = \frac{F_0}{D_0} = \frac{6.0}{0.21} = 28.57$ $N = N_0 \times 10^{-n} = 10^4 \times 10^{-28.57} = 10^{-24.57}$ $\log_{10}(N) = -24.57$
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