Q.54 Moist heat sterilization of spores at 121 °C follows first order kinetics as per the expression:

dN/dt = -k_dN

where, N is the number of viable spores, t is the time, k_d is the rate constant and dN/dt is the rate of change of viable spores.
If k_d value is 1.0 min^-1, the time (in minutes) required to reduce the number of viable spores from an initial value of 10¹⁰ to a final value of 1 is (up to two decimal places) _______

Moist Heat Sterilization: Calculating Time to Reduce Spores from 10¹⁰ to 1 at 121°C

Moist heat sterilization at 121°C follows first-order kinetics, requiring 23.03 minutes to reduce viable spores from 10¹⁰ to 1 with a death rate constant k_d = 1.0 min^-1. This GATE-style calculation demonstrates microbial death rate principles in bioprocessing.

First-Order Kinetics Basics

The microbial death follows:

dN/dt = –k_dN

Integrated form:

N = N₀ e^-k_dt

Rearranged to solve time:

t = (1/k_d) ln(N₀ / N)

For:

• N₀ = 10¹⁰
• N = 1
• k_d = 1.0 min^-1

Since ln(10¹⁰) = 10 ln 10 ≈ 23.02585, time becomes:

t = 23.03 minutes (to two decimals)

Step-by-Step Calculation

1. Start: t = ln(N₀/N) / k_d
2. N₀/N = 10¹⁰
3. ln(10¹⁰) = 23.02585
4. Divide by k_d=1 → t = 23.03 min

This matches the expected GATE answer (≈23.03–23.04 min), confirming a 10-log reduction.

Common Miscalculations Explained

• Zero-order error: t = N₀/k_d → gives 10¹⁰ min, ignores exponential decay
• Wrong log base: ln(10⁹) yields 20.72 min, missing one log cycle
• Unit mistakes: Using k_d in seconds inflates time
• Rounding too early: 23 min loses decimal precision

Bioprocess Applications

Steam sterilization at 121°C ensures contamination probability (<10^-3) by designing for 12–15 log reductions. Holding time refers to the isothermal plateau and excludes heat-up and cool-down periods.