71. If the decimal reduction time for spores of a certain bacterium at 121∘C is 12 seconds, the
time required (in minutes) to reduce 1010 spores to one spore by heating at 121∘C is ____

Decimal reduction time, or D-value, measures the time required at a specific temperature to reduce a microbial population by 90% (one log cycle).
For spores with D₁₂₁ = 12 seconds, reducing 10¹⁰ spores to approximately one survivor requires 10 log reductions.

Therefore, the total sterilization time is:

Time = 10 × 12 seconds = 120 seconds = 2 minutes

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D-Value Fundamentals

The D-value at 121°C (D₁₂₁) is 12 seconds, meaning that every 12 seconds of heating reduces the surviving spore population by one logarithmic cycle (90% reduction).

This process follows first-order kinetics, expressed as:

log(N / N₀) = −t / D

Where:

• N₀ = initial population = 10¹⁰
• N = final population ≈ 1 spore

The number of log reductions required is:

log₁₀(10¹⁰ / 1) = 10

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Step-by-Step Calculation

1. D-value = 12 seconds = 0.2 minutes
2. Required log reductions = 10
3. Total time = 10 × 12 seconds = 120 seconds
4. Converted to minutes = 2 minutes

In food microbiology, this aligns with the well-known 12D concept used for Clostridium botulinum sterilization
(D₁₂₁ ≈ 0.2 min for 12 log reductions), scaled here to 10 logs.

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Why Exactly 2 Minutes?

The reduction sequence is:

• Initial: 10¹⁰ spores
• After 1 D (12 s): 10⁹
• After 2 D (24 s): 10⁸
• …
• After 10 D (120 s): 10⁰ = 1 spore

The probability of survival becomes 10⁻¹⁰, ensuring sterilization safety.

The governing equation confirms this:

t = D × log₁₀(N₀ / N)

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Common Misconceptions

• 1.2 minutes – results from incorrect unit conversion
• 20 minutes – overestimates the required number of log reductions
• No z-value adjustment is needed because the temperature is fixed at 121°C

Always verify units carefully—12 seconds × 10 = exactly 2 minutes.