Question

Food cans are sterilized in a retort to inactivate Clostridium botulinum. The process time \(F_0\) is 150 s and z value is 10°C. The temperatures at the slowest heating region are measured as 71.1°C, 98.9°C, and 110°C. Find the actual process time required for sterilization at 121.1°C.

Hint:

Use the sterilization formula to calculate the process time at different temperatures.

Answer 1

Correct Answer: 2.01

Sterilization formula: \[ F = F_0 \times 10^{\frac{T_1 - T_2}{z}} \] Where \(T_2 = 110°C\) and \(T_1 = 121.1°C\): \[ F = 150 \times 10^{\frac{121.1 - 110}{10}} = 150 \times 10^{1.11} \] \[ F = 150 \times 12.9 = 1935\ \text{seconds} \] Thus: \[ \boxed{1935} \]