Merv takes his potato salad with him when he goes camping but doesn’t pack it in a cooler. When he sits down to eat, it has been at 75 ⁰F for 6 hours. At that warmer temperature the Salmonella in the potato salad reproduces much faster, with a growth rate of 8. Bacterial Count for Merv’s potato salad after 6 hours?

Question

Merv takes his potato salad with him when he goes camping but doesn’t pack it in a cooler. When he sits down to eat, it has been at 75 ⁰F for 6 hours. At that warmer temperature the Salmonella in the potato salad reproduces much faster, with a growth rate of 8. Bacterial Count for Merv’s potato salad after 6 hours?

Answer 1

To calculate the bacterial count after 6 hours at a growth rate of 8, we can use the formula:

Bacterial Count = Initial Bacterial Count * (Growth Rate)^(Time / Doubling Time)

In this case, the initial bacterial count is assumed to be 1 (this is just a unitless value used to demonstrate growth, it could be 1 bacterial cell or 1 million bacterial cells). The doubling time for Salmonella is typically around 20 minutes, so we can convert 6 hours to minutes (6 hours * 60 minutes/hour = 360 minutes).

So, plugging in the values:

Bacterial Count = 1 * (8)^(360 / 20)
Bacterial Count = 1 * (8)^18
Bacterial Count = 1 * 68719476736
Bacterial Count = 68719476736

Therefore, after 6 hours at 75 ⁰F, Merv's potato salad would have a bacterial count of 68,719,476,736.