Sterilization Time Calculation Using First-Order Kinetics
The thermal deactivation of microbial cells in a batch sterilizer follows a first-order process. The time taken for sterilization can be calculated using the first-order integrated rate law.
First-Order Deactivation Formula
For a first-order process, the relationship between initial concentration ($C_0$), final concentration ($C$), the death rate constant ($k$), and time ($t$) is given by:
$ \ln\left(\frac{C_0}{C}\right) = k \cdot t $
To find the time ($t$), we rearrange the formula:
$ t = \frac{1}{k} \ln\left(\frac{C_0}{C}\right) $
Applying Given Values
Given:
- Initial concentration, $C_0 = 10^{10} \text{ cells m}^{-3}$
- Final concentration, $C = 10 \text{ cells m}^{-3}$
- Death rate constant at 121$^{\circ}$C, $k = 0.69 \text{ min}^{-1}$
Substitute these values into the formula:
$ t = \frac{1}{0.69 \text{ min}^{-1}} \ln\left(\frac{10^{10} \text{ cells m}^{-3}}{10 \text{ cells m}^{-3}}\right) $
Calculating Sterilization Time
- Simplify the concentration ratio:
$ \frac{C_0}{C} = \frac{10^{10}}{10} = 10^9 $
- Calculate the natural logarithm of the ratio:
$ \ln(10^9) = 9 \ln(10) \approx 9 \times 2.3026 = 20.7234 $
- Calculate the time ($t$):
$ t = \frac{1}{0.69} \times 20.7234 \text{ min} $
$ t \approx 30.0339 \text{ min} $
- Round the result to the nearest integer:
$ t = 30 \text{ min} $
The time taken to reduce the microbial load from $10^{10} \text{ cells m}^{-3}$ to $10 \text{ cells m}^{-3}$ at 121$^{\circ}$C is approximately 30 minutes.
Important Questions from Sterilization of Air and Media
The decimal reduction time of a microbe during sterilization at $120 \text{ °C}$ with a first order thermal death rate constant of $1 \text{ min}^{-1}$ will be _______________ min (rounded off to 1 decimal place).
- A pilot sterilization was carried out in a vessel containing $100 \text{ m}^3$ medium with an initial spore concentration of $10^8 \text{ spores/ml}$. The accepted level of contamination after sterilization is 1 spore in the entire vessel. The specific death rate constant for the spore is $2 \text{ min}^{-1}$ at $121 \text{ }^\circ C$. Assuming no death takes place during the heating and cooling cycles, the holding time at $121 \text{ }^\circ C$ (rounded off to nearest integer) is ________________ min.
Decimal reduction time of a bacterial strain is $20$ min. Specific death rate constant in $min^{-1}$ (rounded off to two decimal places) is____.
Moist heat sterilization of spores at $121 \text{ } ^\circ C$ follows first order kinetics as per the expression:
$ \frac{dN}{dt} = -k_d N $
where, N is the number of viable spores, t is the time, $k_d$ is the rate constant and $ \frac{dN}{dt} $ is the rate of change of viable spores.
If $k_d$ value is $1.0 \text{ min}^{-1}$, the time (in minutes) required to reduce the number of viable spores from an initial value of $10^{10}$ to a final value of 1 is (up to two decimal places)______.
- Decimal reduction time of bacterial spores is 23 min at $121 \ °C$ and the death kinetics follow first order. One liter medium containing $10^9$ spores per mL was sterilized for 10 min at $121 \ °C$ in a batch sterilizer. The number of spores in the medium after sterilization (assuming destruction of spores in heating and cooling period is negligible) will be ____________________ $ \times 10^7$.
