Allied & ICAR Exam Bank • ALLIED_0972 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Thermal Processing and Sterilization

Subtopic: Pasteurization and Sterilization

Question Statement:

For *Clostridium botulinum* with $z = 10^\circ\text{C}$ and reference temperature $T_{ref} = 121.1^\circ\text{C}$, the lethal rate at retort temperature $T = 124.11^\circ\text{C}$ is $L = 10^{(124.11 - 121.1)/10} = 10^{0.30103}$. Since $10^{0.30103} = 2.0$, holding at this temperature for $t = 3.0\text{ minutes}$ delivers an equivalent lethality $F_0 = L \times t = 2.0 \times 3.0$ in minutes:

Answer Key:

6.0

Step-by-Step Solution & Explanation

$F_0 = 2.0 \times 3.0 = 6.0\text{ min}$.
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