Subtopic: Kinetics of microbial death – pasteurization and sterilization of milk and other liquid foods
Question Statement:
A spore suspension of initial population $N_0 = 5.0 \times 10^6\text{ spores/mL}$ is thermally processed at $121.1^\circ\text{C}$ ($D_{121} = 1.20\text{ minutes}$). The first-order reaction rate constant for thermal death ($k = \frac{\ln(10)}{D}$) in $\text{min}^{-1}$ is ________ (round off to two decimal places).
Answer Key:
1.92
Step-by-Step Solution & Explanation
Multi-Approach Solution:
**Method 1: Relationship Between k and D**
First-order survivor curve:
$\ln\left(\frac{N}{N_0}\right) = -k t \implies \log_{10}\left(\frac{N}{N_0}\right) = -\frac{k}{\ln(10)} t = -\frac{t}{D}$
$k = \frac{\ln(10)}{D} = \frac{2.302585}{1.20\text{ min}} = 1.9188\text{ min}^{-1} \approx 1.92\text{ min}^{-1}$
**Method 2: Ratio Verification**
After 1.20 minutes, survival is $10\%$: $\ln(0.10) = -k(1.20) \implies -2.3026 = -1.20 k \implies k = 1.919\text{ min}^{-1}$.