Allied & ICAR Exam Bank • ALLIED_0709 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Thermal Processing and Kinetics

Subtopic: Thermal Processing and Kinetics

Question Statement:

For *Clostridium botulinum* spores, $D_{121.1} = 0.21\text{ min}$ and $D_{111.1} = 2.1\text{ min}$. The temperature difference is $121.1 - 111.1 = 10.0^\circ\text{C}$, and $\log_{10}(2.1 / 0.21) = \log_{10}(10) = 1.0$. The $z$-value in $^\circ\text{C}$ is:

Answer Key:

10.0

Step-by-Step Solution & Explanation

$z = 10.0 / 1.0 = 10.0^\circ\text{C}$.
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