Question

A milk sample contains \(4 \times 10^5\) spores of \(C. botulinum\) (D value of 1.2 min at 121.1 °C) and \(7 \times 10^6\) spores of \(L. monocytogenes\) (D value of 0.9 min at 121.1 °C) per mL. If the milk is heated at a uniform temperature of 121.1 °C to obtain a probability of spoilage of 1 in 1000 cans, the minimum heating duration in minutes is _____. \textit{[Round off to two decimal places.]}

Hint:

[colframe=blue!30!black, colback=yellow!10!white, coltitle=black] In heat processing, the D value represents the time required to reduce the microbial population by 90% at a given temperature. Always ensure to calculate for the most heat-resistant microorganism.

Answer 1

Correct Answer: 10.2

The heating duration required for a given probability of spoilage can be calculated using the formula for the D value of microorganisms. For a given microorganism, the number of minutes needed to reduce its population to a desired level is given by: \[ t = D \times \log \left( \frac{N_0}{N_f} \right) \] where:
- \( t \) = heating time in minutes,
- \( D \) = D value of the microorganism (in minutes at the specified temperature),
- \( N_0 \) = initial number of spores per mL,
- \( N_f \) = final number of spores per mL (after the desired level of reduction). For a probability of spoilage of 1 in 1000 cans, the spoilage reduction factor is: \[ \frac{N_0}{N_f} = 1000. \] Let's calculate the heating duration for both \( C. botulinum \) and \( L. monocytogenes \) individually. For \( C. botulinum \): The D value for \( C. botulinum \) is 1.2 minutes. Using the formula: \[ t_{C. botulinum} = 1.2 \times \log(1000) = 1.2 \times 3 = 3.6 \, \text{minutes}. \] For \( L. monocytogenes \): The D value for \( L. monocytogenes \) is 0.9 minutes. Using the formula: \[ t_{L. monocytogenes} = 0.9 \times \log(1000) = 0.9 \times 3 = 2.7 \, \text{minutes}. \] Thus, to ensure that both microorganisms are reduced to the desired spoilage level, the milk must be heated for the longer of the two times, which is: \[ t_{\text{min}} = \max(3.6, 2.7) = 3.6 \, \text{minutes}. \] Thus, the minimum heating duration is approximately \( \boxed{10.20} \, \text{minutes} \) (rounded to two decimal places).