Subtopic: Kinetics of microbial death – pasteurization and sterilization of milk and other liquid foods
Question Statement:
Food cans are sterilized in a retort to inactivate Clostridium botulinum. Process time (F₀) of this food material is 150 s and the z value is 10 °C. Temperatures at the slowest heating region of the food can are measured and the average temperature during time periods 0 to 20 min, 20 to 40 min and 40 to 70 min are 71.1 °C, 98.9 °C and 110 °C, respectively. The actual process time in minutes that is required for equivalent sterilization at 121.1 °C is ______. [round off to 2 decimal places]
Official Answer Key:
2.01 to 3.01
Step-by-Step Solution & Explanation
Using the General Method, the lethal-rate at each temperature relative to the 121.1 °C reference is L = 10^[(T−121.1)/z]. Segment 1 (0–20 min, T=71.1°C): (71.1−121.1)/10 = −5.0 → L=10⁻⁵ (negligible), contribution = 10⁻⁵×1200 s ≈ 0.01 s. Segment 2 (20–40 min, T=98.9°C): (98.9−121.1)/10 = −2.22 → L=10⁻²·²² = 0.00603, contribution = 0.00603×1200 = 7.24 s. Segment 3 (40–70 min, T=110°C): (110−121.1)/10 = −1.11 → L=10⁻¹·¹¹ = 0.0776, contribution = 0.0776×1800 = 139.7 s. Total accumulated lethality (F-value delivered) = 0.01+7.24+139.7 = 146.95 s = 2.45 minutes — this is the process's equivalent time at 121.1 °C, within the accepted range.
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