Q.63
A pilot sterilization was carried out in a vessel containing 100 m³ medium with
an initial spore concentration of 10⁸ spores/ml. The accepted level of
contamination after sterilization is 1 spore in the entire vessel. The specific death
rate constant for the spore is 2 min⁻¹ at 121 °C. Assuming no death takes place
during the heating and cooling cycles, the holding time at 121 °C (rounded off to
nearest integer) is ____________ min.

Sterilization in bioprocessing ensures media is free from contaminants like spores using first-order death kinetics. This GATE question requires finding the holding time at 121°C to reduce spores to 1 per vessel.

Problem Breakdown

A 100 m³ vessel holds medium with 10⁸ spores/mL initially. Target: 1 spore total after sterilization. Death rate constant k = 2 min⁻¹. Ignore heating/cooling deaths. Holding time t (nearest integer) in minutes.

First, compute total initial spores N₀:

Volume = 100 m³ = 100 × 10⁶ L = 10⁸ L = 10¹¹ mL.
N₀ = 10⁸ spores/mL × 10¹¹ mL = 10¹⁹ spores.

Final spores N = 1.

First-Order Kinetics Formula

Spores follow N = N₀ e^-kt, so t = (1/k) ln(N₀ / N).

Here, ln(N₀ / N) = ln(10¹⁹ / 1) = 19 ln(10) ≈ 19 × 2.3026 = 43.749.

t = 43.749 / 2 ≈ 21.875 min ≈ 22 min.

Del Factor Approach (Alternative View)

Sterilization design often uses Del (∫ k dt) = ln(N₀ / N) = 43.749. With constant k during holding, Del_holding = k t = 43.749, confirming t = 22 min.

Common Mistakes Explained

• Wrong units (e.g., treating 100 m³ as 100 L): Yields t ≈ 12 min (underestimates).
• Probability confusion (e.g., 10^-3): Changes ln to ~43, still ~22 min but mismatches “1 spore”.
• D-value mix-up: D₁₂₁ = 1/k = 0.5 min; log reductions = 19, t = 19 × 0.5 × 2.303 ≈ 22 min (equivalent).

No options given; numerical answer is 22.

Why 22 Minutes?

This achieves 99.999999999999999999% kill (19-log reduction), standard for pilot-scale sterility assurance in biotech fermentation.