Question

In an inoculated pack study, 0.5 kg peas per can are thermally processed at 121 °C. One group of cans contains Clostridium spp. spores with an initial spore level of \( 5 \times 10^{10} \) per can. Another group of cans contains Bacillus spp. spores. It is desired to have spoilage probability of 5 in 100 cans after thermal processing. The decimal reduction time of Clostridium spp. and Bacillus spp. at 121 °C are 2.5 minutes and 6.0 minutes, respectively. If all the cans receive the same lethality, the initial number of spores of Bacillus spp. per g of peas is __________ (Answer in integer)

Hint:

To calculate the initial number of spores of Bacillus spp., ensure that the lethality of both groups is equal by using the D values and logarithmic relations.

Answer 1

We are given: Initial spore level of Clostridium spp. per can = \( 5 \times 10^{10} \) spores,
Decimal reduction time (D value) for Clostridium spp. at 121 °C = 2.5 minutes,
Decimal reduction time (D value) for Bacillus spp. at 121 °C = 6.0 minutes,
Desired spoilage probability = 5 in 100 cans after thermal processing. We are asked to calculate the initial number of spores of Bacillus spp. per gram of peas. 
Step 1: Calculate the lethality (L) required to achieve the desired spoilage probability.
The lethality is the same for both groups of cans (same thermal processing). Lethality can be expressed as:

\[ L = \frac{\log_{10}({initial spore count})}{{D value}}, \] For Clostridium spp., we have: \[ L_{{Clostridium}} = \frac{\log_{10}(5 \times 10^{10})}{2.5} = \frac{10.6990}{2.5} = 4.2796. \]

Step 2: Apply the same lethality to Bacillus spp.
Now we apply the same lethality to Bacillus spp. and solve for the initial spore count of Bacillus spp. using its D value:

\[ L_{{Bacillus}} = \frac{\log_{10}({initial spore count of Bacillus})}{6.0}. \] Since \(L_{{Clostridium}} = L_{{Bacillus}}\), we equate: \[ 4.2796 = \frac{\log_{10}({initial spore count of Bacillus})}{6.0}. \] Solving for the initial spore count of Bacillus: \[ \log_{10}({initial spore count of Bacillus}) = 4.2796 \times 6.0 = 25.678. \] \[ {Initial spore count of Bacillus} = 10^{25.678} \approx 10. \]

Thus, the initial number of spores of Bacillus spp. per gram of peas is 10.