Question

Pasteurisation of milk can be carried out either at \(91\,^{\circ}\text{C}\) for 3 s or at \(73\,^{\circ}\text{C}\) for 30 s to inactivate vegetative cells of microorganism(s). The sterilisation value is 10 in both the cases. Thermal death time constant (z value) for reference temperatures of \(73\,^{\circ}\text{C}\) and \(91\,^{\circ}\text{C}\), in \(^{\circ}\text{C}\), is nearest to

• A. 18.0
• B. 1.8
• C. 9.0
• D. 0.9

Hint:

Use the thermal death time formula linking the log of the time ratio to the temperature difference and the z value.

Answer 1

The Correct Option is A

Step 1: Recall the thermal death time relation.
The time to reach the same lethal effect at two different temperatures follows \( \log_{10}\left(\dfrac{t_2}{t_1}\right) = \dfrac{T_1 - T_2}{z} \), where \(T_1\) is the higher temperature with the shorter time \(t_1\).

Step 2: Identify the two time temperature pairs.
Higher temperature: \( T_1 = 91\,^{\circ}\text{C} \), \( t_1 = 3\ s \). Lower temperature: \( T_2 = 73\,^{\circ}\text{C} \), \( t_2 = 30\ s \).
Since both give the same sterilisation value (10), they lie on the same thermal death time curve, so the z value relation applies directly.

Step 3: Compute the log ratio of times.
\( \log_{10}(30/3) = \log_{10}(10) = 1 \).

Step 4: Solve for z.
\( 1 = \dfrac{91-73}{z} = \dfrac{18}{z} \), so \( z = 18\,^{\circ}\text{C} \).

Final Answer:
The z value works out to exactly 18 °C, matching option (A). \[ \boxed{z \approx 18.0\ ^{\circ}\text{C}} \]