Subtopic: Calculation of power, torque, fuel consumption, heat load and power losses
Question Statement:
A suspension contains 3.6×10⁵ spores of C. botulinum having a D-value of 1.5 minute at 121.1 °C and 8.5×10⁶ spores of B. subtilis having a D-value of 0.9 minute at the same temperature. The suspension is heated at a constant temperature of 121.1 °C. The heating time needed in minutes for the suspension to obtain a survival probability of 10⁻³ for the most heat resistant organism in it is _____.
Official Answer Key:
12 to 14
Step-by-Step Solution & Explanation
Thermal death time (to reach a target survivor level Nᶠ) for each organism: t = D×log₁₀(N₀/Nᶠ), with target Nᶠ corresponding to survival probability 10⁻³.
For C. botulinum (D=1.5 min, N₀=3.6×10⁵): t = 1.5×log₁₀(3.6×10⁵/10⁻³) = 1.5×log₁₀(3.6×10⁸) = 1.5×8.556 = 12.83 min.
For B. subtilis (D=0.9 min, N₀=8.5×10⁶): t = 0.9×log₁₀(8.5×10⁶/10⁻³) = 0.9×log₁₀(8.5×10⁹) = 0.9×9.929 = 8.94 min.
The organism requiring the longer heating time (higher D-value, i.e. more heat-resistant — C. botulinum) governs the required process time: t = 12.83 min. This lies within the accepted range 12–14. (Choosing B. subtilis instead, which has the larger initial population but lower D-value, gives the shorter, under-protective time of 8.94 min — a common error, since ‘more heat resistant’ is governed by D-value, not initial spore count.)
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