Question

The \(F_{121}\) value of a known microorganism with \(Z\) value of \(11~^\circ\mathrm{C}\) is \(2.4\) min for \(99.9999%\) inactivation. For a \(12D\) inactivation of the said microorganism at \(143~^\circ\mathrm{C}\), the \(F\) value (in min) is \underline{\hspace{2cm}. (rounded off to 3 decimal places)}

Hint:

Use \(D_T = D_{T_{\mathrm{ref}}}\,10^{(T_{\mathrm{ref}}-T)/Z}\) and remember: \(F = nD\). A jump of \(Z\) degrees reduces \(D\) by a factor of \(10\) (for bases \(>1\)).

Answer 1

Correct Answer: 0.046

Step 1: Relate \(F\) and \(D\).
For a given temperature, \(F = (\text{number of log reductions})\times D\). Here, \(99.9999% = 6D\), so \[ F_{121}(6D)=2.4~\text{min} \Rightarrow D_{121}=\frac{2.4}{6}=0.4~\text{min}. \]

Step 2: Temperature dependence via \(Z\)-value.
\(D_T = D_{T_{\mathrm{ref}}}\,10^{\frac{T_{\mathrm{ref}}-T}{Z}}\). With \(T_{\mathrm{ref}}=121^\circ\mathrm{C}\), \(T=143^\circ\mathrm{C}\), \(Z=11^\circ\mathrm{C}\): \[ D_{143}=0.4\times 10^{\frac{121-143}{11}} =0.4\times 10^{-2} =0.004~\text{min}. \]

Step 3: Required \(12D\) process at \(143^\circ\mathrm{C}\).
\[ F_{143}(12D)=12\times D_{143}=12\times 0.004=0.048~\text{min}. \] Final Answer:
\[ \boxed{0.048~\text{min}} \]