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.

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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: $$ \boxed{1935} $$

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.

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Correct Answer: 2.01

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