Question

A food product is heated at 115 C for 5 minutes. This treatment results in the reduction of aStreptococcus species from an initial count of 1 x 10⁷ CFU/g to 10 CFU/g. Assume the initial count of the Streptococcus species was lowered to 1 x 10⁴ CFU/g. How long would the product with the reduced initial count have to be heated at 105 C to achieve the same amount of surviving bacteria (10 CFU/g)? For your calculations assume a z-value of 5 C and instantaneous heating and cooling.

Answer 1

The question is

A food product is heated at 115 C for 5 minutes. This treatment results in the reduction of a Streptococcus species from an initial count of 1 x 10⁷ CFU/g to 10 CFU/g. Assume the initial count of the Streptococcus species was lowered to 1 x 10⁴ CFU/g. How long would the product with the reduced initial count have to be heated at 105 C to achieve the same amount of surviving bacteria (10 CFU/g)? For your calculations assume a z-value of 5 C and instantaneous heating and cooling.

Answer: Lets start with two definitions

D-Value is the time required at a given condition to achieve 1-log ( or 90% ) reduction in CFU count.

Z-Value is the temperature change required to change the D-value by a factor of 10 ( 1-log)

In the above example the

D-value at 115 C = 5/6 Min = 0.83 min ( since it takes 5 minutes for 6-log reduction from 1X10⁷ to 10 CFU/g)

Z value is 5 C, so 5 degree reduction in temperature increase the D value by a factor of 10

At 105 C, there is 10 degree reduction in temperature which will lead increase in the D value by a factor of 100 ( 2-logs)

Therefore - D value at 105 C = 0.83 X 100 = 83 min

for three log reduction i.e. reduction of an initial count of 1 x 10⁴ CFU/g to 10 CFU/g the time required will be 83 * 3 = 249 minutes

So the product would have to be heated for 249 minutes at 105 C to achieve 10 CFU/g final count