Question

The \( F_{121} \) value of a known microorganism with \( Z \) value of \( 11^\circ C \) is 2.4 min for 99.9999% inactivation. For a 12D inactivation of the said microorganism at \( 143^\circ C \), the \( F \) value (in min) is .......... {(rounded off to 3 decimal places)}

Hint:

The \( F \)-value decreases exponentially with increased temperature due to the negative exponent in the log term.

Answer 1

Correct Answer: 0.046

The thermal death time (\( F_T \)) at any temperature \( T \) is given by the formula: \[ F_T = F_{ref} \times 10^{\left( \frac{T_{ref} - T}{Z} \right)} \] Given:
\quad \( F_{121} = 2.4 \) min, \( Z = 11^\circ C \), \( T_{ref} = 121^\circ C \), \( T = 143^\circ C \), \( D = 12 \) Now compute: \[ F_{143} = 2.4 \times 10^{\left( \frac{121 - 143}{11} \right)} = 2.4 \times 10^{-2} = 0.024 \] For 12D inactivation: \[ F_{143} = 12 \times 0.024 = 0.048 \, {min} \]