Subtopic: Kinetics of microbial death – pasteurization and sterilization of milk and other liquid foods
Question Statement:
A milk sample contains 4×10⁵ spores of C. botulinum (D value of 1.2 min at 121.1 °C) and 7×10⁶ spores of L. monocytogenes (D value of 0.9 min at 121.1 °C) per mL. If the milk is heated at a uniform temperature of 121.1 °C to obtain a probability of spoilage of 1 in 1000 cans, the minimum heating duration in minutes is ___________. [round off to two decimal places]
Official Answer Key:
10.20 to 10.40
Step-by-Step Solution & Explanation
• Required process time t = D × log₁₀(N₀/Nf), where Nf = target survivor probability = 10⁻³.
• For C. botulinum: t1 = 1.2 × log₁₀(4×10⁵/10⁻³) = 1.2 × log₁₀(4×10⁸) = 1.2 × 8.602 = 10.32 min
• For L. monocytogenes: t2 = 0.9 × log₁₀(7×10⁶/10⁻³) = 0.9 × log₁₀(7×10⁹) = 0.9 × 9.845 = 8.86 min
• The process must be safe for BOTH organisms, so the governing (larger) time controls: t1=10.32 min > t2=8.86 min.
Computed value: 10.32 min (within official range 10.20–10.40; C. botulinum is the governing organism)
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