Question

The "z" value of a micro-organism is 6°C and the temperature of processing is 105°C then the lethal rate of microorganism is

• A. 21.5
• B. 0.215
• C. $2.15 \times 10^{-3}$
• D. $2.15 \times 10^2$

Hint:

To verify lethal rate calculations quickly:
At \(T = T_{\text{ref}}\), the lethal rate \(L = 10^0 = 1\).
As the temperature falls below \(121^\circ\text{C}\), the lethal rate decreases exponentially.
For \(T = 105^\circ\text{C}\), the value must be very small (\(10^{-3}\) range), which helps identify the correct option.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In the thermal sterilization of food and dairy products, the lethal rate (\(L\)) represents the sterilizing intensity of a specific process temperature relative to a standard reference temperature.
The reference temperature (\(T_{\text{ref}}\)) used globally for sterilization calculations is 121°C (corresponding to 250°F).
Key Formula or Approach:
The lethal rate (\(L\)) of a microorganism at a processing temperature \(T\) is calculated using the formula:
\[ L = 10^{\frac{T - T_{\text{ref}}}{z}} \]
where:
- \(T\) is the operating processing temperature in °C.
- \(T_{\text{ref}}\) is the standard reference temperature, taken as \(121^\circ\text{C}\).
- \(z\) is the temperature change required to cause a tenfold change in the decimal reduction time (\(D\)-value) of the microorganism.

Step 2: Detailed Explanation:
Let us identify the given parameters from the problem:
- Processing temperature, \(T = 105^\circ\text{C}\).
- \(z\)-value of the microorganism \(= 6^\circ\text{C}\).
- Reference temperature, \(T_{\text{ref}} = 121^\circ\text{C}\).
Substitute these parameters into the lethal rate formula:
\[ L = 10^{\frac{105 - 121}{6}} \]
Calculate the exponent:
\[ \frac{105 - 121}{6} = \frac{-16}{6} = -2.667 \]
Now, calculate the value of \(L\):
\[ L = 10^{-2.667} \]
Using logarithmic calculations:
\[ L \approx 0.002154 = 2.154 \times 10^{-3} \]
This matches Option (C) perfectly.

Step 3: Final Answer:
The lethal rate of the microorganism at 105°C is \(2.15 \times 10^{-3}\).