Subtopic: Kinetics of microbial death – pasteurization and sterilization of milk and other liquid foods
Question Statement:
Peas (0.5 kg/can) are processed at 121°C. Clostridium spp. spores: initial level 5×10¹⁰/can, D-value 2.5 min. Desired spoilage probability = 5/100 cans. Bacillus spp. D-value 6.0 min, same lethality applied to both. Find the initial number of Bacillus spores per g of peas. (Answer in integer)
Official Answer Key:
10
Step-by-Step Solution & Explanation
Process time set by Clostridium requirement: Nt = N0×10^(−t/D); desired Nt = 0.05 (spoilage probability) ⇒ t = D×log10(N0/Nt) = 2.5×log10(5×10¹⁰/0.05) = 2.5×12 = 30 min. Applying the same process time (t=30 min) to Bacillus with the same target spoilage probability (Nt=0.05): N0,Bacillus = Nt×10^(t/D) = 0.05×10^(30/6) = 0.05×10⁵ = 5000 spores/can. Per gram of peas (500 g/can): 5000/500 = 10 spores/g.
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.