Question

The decimal reduction time D at \(121^\circ\text{C}\) and the value Z for a thermophilic spore in whole milk were recorded as 30 seconds and \(10.5^\circ\text{C}\), respectively. What is the value of D at \(150^\circ\text{C}\)?

• A. 0.00865 min
• B. 0.000865 min
• C. 0.0865 min
• D. 0.00785 min

Hint:

Always convert \(D\)-values to consistent units (usually minutes) before performing calculations.
A rapid increase in temperature (from \(121^\circ\text{C}\) to \(150^\circ\text{C}\)) causes an exponential decrease in the time required to kill the spores, enabling modern UHT processing.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The heat resistance of a microorganism varies logarithmically with temperature.
The \(z\)-value describes the temperature change required to cause a ten-fold change in the \(D\)-value.

Step 2: Key Formula or Approach:
The relationship between the \(D\)-values at two different temperatures is given by:
\[ \log\left(\frac{D_1}{D_2}\right) = \frac{T_2 - T_1}{z} \tag{1} \]

Step 3: Detailed Explanation:
Given:
- \(T_1 = 121^\circ\text{C}\)
- \(D_1 = 30\text{ seconds} = 0.5\text{ minutes}\)
- \(z = 10.5^\circ\text{C}\)
- \(T_2 = 150^\circ\text{C}\)
Let us solve for \(D_2\) at \(150^\circ\text{C}\):
Substituting the values into equation (1):
\[ \log\left(\frac{0.5}{D_2}\right) = \frac{150 - 121}{10.5} \]
\[ \log\left(\frac{0.5}{D_2}\right) = \frac{29}{10.5} \approx 2.7619 \]
Taking the antilog (base 10) of both sides:
\[ \frac{0.5}{D_2} = 10^{2.7619} \approx 577.965 \]
\[ D_2 = \frac{0.5}{577.965} \approx 0.0008651\text{ minutes} \]

Step 4: Final Answer:
The value of D at \(150^\circ\text{C}\) is 0.000865 min, which corresponds to option (B).