Question

A food product contains 1 x 10⁶ CFU/g of Escherichia coli. Heating at 80 C for 10 minutes will reduce this population to 1 x 10⁴ CFU/g. The z-value of this E. coli strain is 3 C. How long would this food product have to be heated at 89 C to achieve the same microbial destruction?

Answer 1

The population of E. Coli initially (a) = 1 x 10⁶ CFU/g

The population of E. Coli after heating (b) = 1 x 10⁴ CFU/g.

time = 10 minutes

temperature = 80 degrees Celcius.

D₈₀ = time/ (log a - log b) = 10/ (log 10⁶ - log 10⁴) = 10/ (6 -4) = 10/2 = 5.

The D-value at 80 degrees Celcius is 5 minutes.

Z value = (T₂ - T₁)/ (log D₂ - log D₁)

3 = (89 - 80)/ (log D₂ - 5)

log D₂ - 5 = 9/3 = 3

log D₂ = 3 + 5 = 8

hence, to achieve the same microbial destruction, the food would have to be heated at 89 degrees celcius for 8 minutes.