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Boss lesson · The Retort Run
Exam-hall rules, last cook of the shift
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Exam-hall rules, last cook of the shift

This is the boss. No calculator. Tonight zz is pinned at 10 C∘10\ \mathrm{C^\circ}, the D-value at the reference is a whole 1 or 2 minutes, and every temperature the vessel might hold is exactly one zz away from 121.1 ∘C121.1\ ^\circ\mathrm{C}. So every multiplier in the run is a single power of ten, and every line is a shift of the decimal point.

One cook, four steps, and every answer feeds the next: the two counts give the logs required, the logs times the D give the F₀ owed, the F₀ through the z-shift gives the real hold at the vessel's temperature, and that hold meets the booking. Write each line down as you go — a slip in line one rides all the way to the verdict, magnified by a factor of ten on the way.

Watch two traps the chapter has been building toward. The target is below one spore per can, so its exponent is negative, and subtracting a negative widens the gap rather than narrowing it. And the exponent between the reference and the vessel moves the answer by a full power of ten in one step — the direction of that step is the single most common error on the F-value equation, and it is worth a deliberate pause.

Note what the last line actually asks. Not another number: a decision. Does the booked hold deliver the specified process, yes or no. The arithmetic exists only to earn you the right to answer it.

A bonus mark follows, one line long. Some nights it prices one more log in the same vessel. Other nights it crosses to the brewery next door, where the scale is the pasteurisation unit: reference 60 °C and a z of 7 C°, with the beer held a whole number of 7 C° steps from the reference. The multiplier is still a power of ten.

Everything in the chapter rides in this one cook. 5★ under par here is the chapter crown. Clock in.

DN/N0tLR

D=tLRD = \frac{t}{\mathrm{LR}}

  • DD= D-value (time)
  • tt= Holding time (time)
  • LR\mathrm{LR}= Log reduction achieved
D-Value (Decimal Reduction Time) solver →
D1D2zT1T2DT

z=T2−T1log⁡10D1−log⁡10D2z = \frac{T_2 - T_1}{\log_{10} D_1 - \log_{10} D_2}

  • zz= z-value (temperature difference)
  • T1T_1= Lower temperature (temperature)
  • T2T_2= Higher temperature (temperature)
  • D1D_1= D-value at T₁ (time)
  • D2D_2= D-value at T₂ (time)
z-Value (Thermal Resistance Constant) solver →
TrefTFtt

F=t×10 (T−Tref)/zF = t \times 10^{\,(T - T_{ref})/z}

  • FF= F-value (equivalent time) (time)
  • tt= Actual time held (time)
  • TT= Process temperature (temperature)
  • TrefT_{\mathrm{ref}}= Reference temperature (temperature)
  • zz= z-value (temperature difference)
F-Value (Equivalent Time at Reference Temperature) solver →
PLRN

P=1−10−LRP = 1 - 10^{-\mathrm{LR}}

  • PP= Percent kill
  • LR\mathrm{LR}= Log reduction
Log Reduction to Percent Kill solver →
no calculator in the boss room — the numbers are chosen to fit in your head