Subtopic: Kinetics of microbial death – pasteurization and sterilization of milk and other liquid foods
Question Statement:
The reaction rate for destruction of Clostridium botulinum increases 11 times for temperature rise of 10 °C from 121.1 °C. The decimal reduction time of this organism is 5.7 s at 121.1 °C. The minimum sterilization time in seconds at 135 °C, for eight log cycle reduction of the organism, is
Options:
(A) 0.20
(B) 0.31
(C) 1.63
(D) 2.49
Official Answer Key:
C
Step-by-Step Solution & Explanation
• A rate increase of 11× for a 10 °C rise means the D-value decreases by a factor of 11 for the same rise, which defines the z-value via D2/D1 = 10^(–ΔT/z): 1/11 = 10^(–10/z) ⇒ z = 10/log10(11) = 10/1.0414 ≈ 9.60 °C.
• ΔT from 121.1 to 135 °C = 13.9 °C.
• D(135) = D(121.1) × 10^(–ΔT/z) = 5.7 × 10^(–13.9/9.60) = 5.7 × 10^(–1.4475) = 5.7 × 0.0357 ≈ 0.2033 s.
• Minimum sterilization time for an 8-log reduction = 8 × D(135) = 8 × 0.2033 ≈ 1.63 s.
• Final Answer: (C) 1.63
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