Question

The process time required at 115°C for 1 log cycle reduction of a particular microorganism having D\(_{121}\) = 3 min and z = 15°C is :

• A. 10.04 min
• B. 11.94 min
• C. 17.15 min
• D. 7.53 min

Hint:

When using the \( z \)-value formula, ensure the temperature difference is divided by \( z \), and carefully compute \( 10^{\text{ratio}} \) using a calculator or basic log rules.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The thermal death time of a microorganism changes with temperature, which is described by its \( z \)-value.
The \( z \)-value is the temperature change required to alter the \( D \)-value by a factor of 10.
The process time required for a 1 log cycle reduction is exactly equal to one \( D \)-value at that operating temperature.
Key Formula or Approach:
The relationship between \( D \)-values at two different temperatures is given by:
\[ \log\left(\frac{D_1}{D_2}\right) = \frac{T_2 - T_1}{z} \]

Step 2: Detailed Explanation:
Let:
\( T_1 = 115^\circ\text{C} \)
\( T_2 = 121^\circ\text{C} \)
\( D_2 = D_{121} = 3\text{ min} \)
\( z = 15^\circ\text{C} \)
We need to find the process time for a 1 log reduction at \( 115^\circ\text{C} \), which is equal to \( D_1 \) (\( D_{115} \)):
\[ \log\left(\frac{D_{115}}{3}\right) = \frac{121 - 115}{15} \]
\[ \log\left(\frac{D_{115}}{3}\right) = \frac{6}{15} = 0.4 \]
Taking the antilog of both sides:
\[ \frac{D_{115}}{3} = 10^{0.4} \approx 2.5119 \]
\[ D_{115} = 3 \times 2.5119 = 7.5357\text{ min} \]
Therefore, the required process time is approximately \( 7.53\text{ min} \).

Step 3: Final Answer:
The process time required at \( 115^\circ\text{C} \) is \( 7.53\text{ min} \).