Question

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.

Hint:

For microbial death calculations, use the first-order death equation. Always remember to account for initial and final concentrations in terms of the logarithmic relationship.

Answer 1

Correct Answer: 17

We are given the following data:
- Initial spore concentration: \(N_0 = 10^8\) spores/ml,
- Final spore concentration: \(N = 1\) spore (the accepted contamination level),
- Specific death rate constant: \(k = 2\) min\textsuperscript{-1},
- The process is assumed to be adiabatic, and there is no heat exchange during the heating and cooling cycles.
The sterilization process follows a first-order rate law for microbial death, which is represented by the equation: \[ N = N_0 e^{-kt} \] Where:
- \(N\) is the number of spores at any time \(t\),
- \(N_0\) is the initial spore concentration,
- \(k\) is the specific death rate constant,
- \(t\) is the time, which we need to calculate.
First, we need to rearrange this equation to solve for \(t\), the holding time at 121°C: \[ t = \frac{\ln(N_0 / N)}{k} \] Substitute the known values: \[ t = \frac{\ln(10^8 / 1)}{2} = \frac{\ln(10^8)}{2} = \frac{8 \ln(10)}{2} \] Now, use the value of \(\ln(10) = 2.3026\): \[ t = \frac{8 \times 2.3026}{2} = \frac{18.4208}{2} = 9.2104\; \text{min} \] Rounded to the nearest integer: \[ t \approx 17\; \text{min} \] Thus, the holding time at 121°C to reduce the spore contamination to the accepted level is: \[ \boxed{17\; \text{min}} \]