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: Picture the thermal resistance curve.
On a plot of $\log_{10}(t)$ against temperature, the line is straight, and $z$ is the temperature rise that cuts the time to one tenth.

Step 2: Find how many tenfold drops in time occur.
Going from $73^{\circ}C$ with $t=30\ s$ to $91^{\circ}C$ with $t=3\ s$, the time drops by a factor of $30/3 = 10$, exactly one tenfold reduction.

Step 3: Match that to the temperature change.
One tenfold drop in time happens over a temperature rise of exactly $z$ degrees, by definition of $z$. The actual temperature rise here is $91 - 73 = 18^{\circ}C$.

Step 4: Read off z.
Since one tenfold reduction corresponds to this $18^{\circ}C$ rise, $z = 18^{\circ}C$ directly, no further division needed.

Final Answer:
The slope based reading confirms $z = 18^{\circ}C$, option (A). \[ \boxed{18.0\ ^{\circ}\text{C}} \]