Question

A batch sterilizer is being operated at 121 ^oC for sterilizing a medium containing microbial cells. Assume that the thermal deactivation of cells is a first order process with a death rate constant of 0.69 min\(^{-1}\) at 121 ^oC. If the initial concentration of microbes in the medium is \(10^{10}\) cells m\(^{-3}\), the time taken to reduce the microbial load to a final concentration of 10 cells m\(^{-3}\) is min. (rounded off to the nearest integer)

Hint:

Use the first order death kinetics equation \(\ln(N_0/N) = k_d t\) with \(N_0/N = 10^9\) and \(k_d = 0.69\) min\(^{-1}\).

Answer 1

Correct Answer: 30

Step 1: Understanding the Concept.
Thermal death of microbial cells during sterilization is treated as a first order process: the number of viable cells falls exponentially with time, at a rate proportional to the number of cells present at that instant. This is the same mathematical form as radioactive decay.

Step 2: Key Formula.
For a first order death process with rate constant \(k_d\), the cell concentration \(N\) at time \(t\) is related to the initial concentration \(N_0\) by
\[ N = N_0\, e^{-k_d t} \quad \text{or equivalently} \quad \ln\left(\frac{N_0}{N}\right) = k_d t \]

Step 3: Substitute the given values.
Here \(N_0 = 10^{10}\) cells m\(^{-3}\), \(N = 10\) cells m\(^{-3}\), and \(k_d = 0.69\) min\(^{-1}\). The ratio to reduce is
\[ \frac{N_0}{N} = \frac{10^{10}}{10} = 10^{9} \]
So
\[ \ln(10^{9}) = 9 \ln(10) = 9 \times 2.3026 = 20.723 \]

Step 4: Solve for the time \(t\).
\[ t = \frac{\ln(N_0/N)}{k_d} = \frac{20.723}{0.69} = 30.03 \ \text{min} \]

Final Answer:
Rounded to the nearest integer, the sterilization time needed is
\[ \boxed{t = 30 \ \text{min}} \]