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)______.
Moist Heat Sterilization Time Calculation
This question requires calculating the time needed for spore reduction under moist heat sterilization, based on first-order kinetic principles.
First-Order Kinetics Equation
The rate of viable spore inactivation ($N$) over time ($t$) is described by the first-order kinetic expression:
$ \frac{dN}{dt} = -k_d N $
Given values are:
- Rate constant, $ k_d = 1.0 \text{ min}^{-1} $
- Initial viable spores, $ N_0 = 10^{10} $
- Final viable spores, $ N_f = 1 $
Integrated Rate Law for Time
To find the time ($t$) for this reduction, we integrate the rate equation:
$ \int_{N_0}^{N_f} \frac{dN}{N} = \int_{0}^{t} -k_d dt $
This leads to the integrated rate law:
$ \ln\left(\frac{N_f}{N_0}\right) = -k_d t $
Rearranging to solve for $t$:
$ t = \frac{1}{k_d} \ln\left(\frac{N_0}{N_f}\right) $
Calculating Sterilization Time
Substitute the provided values into the equation:
$ t = \frac{1}{1.0 \text{ min}^{-1}} \ln\left(\frac{10^{10}}{1}\right) $
Simplify and compute:
$ t = 1 \text{ min} \times \ln(10^{10}) $
Using $ \ln(10^{10}) = 10 \times \ln(10) $ and $ \ln(10) \approx 2.302585 $:
$ t \approx 10 \times 2.302585 \text{ min} $
$ t \approx 23.02585 \text{ min} $
Rounded to two decimal places, the time required is approximately 23.03 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____.
- 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$.
- An industrial fermentor containing $10,000 \text{ L}$ of medium needs to be sterilized. The initial spore concentration in the medium is $10^6 \text{ spores mL}^{-1}$. The desired probability of contamination after sterilization is $10^{-3}$. The death rate of spores at $121 \text{ °C}$ is $4 \text{ min}^{-1}$. Assume that there is no cell death during heating and cooling phases. The holding time of the sterilization process is __________ min (rounded off to the nearest integer).
