Question

Milk sterilization kinetics is based on inactivation of index microorganism, bacillusstearothermophilus. The d-values at 121.1c and 139.1c are 1.2 min and 0.019 min, respectively.For 12 log-cycle reduction of this microorganism at 130c, the processing time in second is

Answer 1

Answer: 12 log-cycle reduction of this microorganism at 130°C is 132.23 seconds.

Explanation:

Given:  To find  value of z

            T1= 121°C

            T2= 139.1°C

             D value at T1= 1.2 minutes

             D value at T2= 0.019 minutes

            To find 12 log-cycle reduction of this microorganism at 130°C

             T1= 121°C

             T2=130°C

To find: 12 log-cycle reduction of this microorganism at 130°C(D 130)

Solution: We know the formula,

log₁₀ DT₂-log₁₀ DT₁= $\frac{1}{z}$ (T₁-T₂)

⇒  log₁₀ 0.019 - log₁₀ 1.2=   $\frac{1}{z}$ ( 121.1-139.1)

⇒ -1.8004=   $\frac{1}{z}$ × -18

⇒z= 9.997

Now,

To find 12 log-cycle reduction of this microorganism at 130°C,

We know the formula,

log₁₀ DT₂-log₁₀ DT₁= $\frac{1}{z}$ (T₁-T₂)

Here, T1= 121°C

          T2=130°C

⇒log₁₀ DT₂- log₁₀ 1.2=$\frac{1}{9.997}$ ( 121.1-130)

⇒log₁₀ DT₂- 0.0792=0.1ₓ ⁻8.9

⇒log₁₀ DT₂= -0.8108

⇒DT₂= 2.20398 minutes

As 1 minute= 60 seconds

⇒DT₂= 2.20398×60

⇒DT₂=132.23 seconds

Hence, 12 log-cycle reduction of this microorganism at 130°C is 132.23 seconds.

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