Scheduling a Disinfectant Release from a Log-Linear Decay Model
A technician at a water-treatment plant monitors the concentration C, in mg per litre, of a decaying disinfectant t minutes after dosing. The plant's sensor logged: t = 0, C = 48.0; t = 10, C = 34.1; t = 20, C = 24.2; t = 30, C = 17.2. Operating rules allow the water to be released once C falls below 8.0 mg/L. The technician proposes the model C = C₀e^(−kt) and wants to use it both to justify the value of k and to schedule the release. A supervisor objects that no model is needed: the concentration halves roughly every twenty minutes, so 8.0 mg/L will be reached in about an hour, give or take. The technician has only a scientific calculator and must write a log entry that an auditor will check line by line, including evidence that the exponential model actually fits the four readings rather than being assumed.
1. Show that the model implies a straight-line relationship between ln C and t, then use the readings at t = 0 and t = 10 to obtain C₀ and k (4 significant figures). Test the fitted model against the readings at t = 20 and t = 30.
Read the worked answer
Taking logs: ln C = ln C₀ − kt, so a plot of ln C against t has gradient −k and intercept ln C₀. Equivalently, for any two readings, k = [ln C(t₁) − ln C(t₂)]/(t₂ − t₁).
Constants: C₀ = 48.0 mg/L (the t = 0 reading).
k = (ln 48.0 − ln 34.1)/10 = ln(1.40762)/10 = 0.341901/10 = 0.03419 min⁻¹ (4 s.f.).
Check at t = 20: predicted C = 48.0e^(−0.03419 × 20) = 48.0e^(−0.68380) = 48.0 × 0.50470 = 24.23 mg/L, against a measured 24.2 mg/L - a discrepancy of about 0.1%.
Check at t = 30: predicted C = 48.0e^(−1.02570) = 48.0 × 0.35855 = 17.21 mg/L, against a measured 17.2 mg/L - again about 0.1%.
The residuals are far smaller than the rounding of the recorded data, so the exponential model is supported by all four readings, not merely fitted to the first two. Note what has been established: consistency with four points, not proof of an underlying first-order mechanism.
2. Use the model to find the release time, and state the answer in a form suitable for the log entry.
Read the worked answer
Solve 48.0e^(−0.03419t) = 8.0.
Divide by 48.0: e^(−0.03419t) = 1/6.
Take logs: −0.03419t = −ln 6 → t = 1.791759/0.034190 = 52.42 minutes.
So release can begin 52 minutes after dosing (52.4 min to 3 s.f.). Because the rule is a threshold that must be passed, the safe scheduling statement is 'no earlier than 52.4 minutes'; rounding down to 52 minutes would release water while C is still marginally above 8.0 mg/L. Rounding up to 53 minutes is the defensible operational choice.
3. Evaluate the supervisor's 'about an hour' argument. Where exactly does it go wrong, and by how much?
Read the worked answer
The supervisor's reasoning is a half-life argument, so first extract the half-life from the model: t½ = ln 2/k = 0.693147/0.034190 = 20.27 min.
The halving chain is 48.0 → 24.0 (20.3 min) → 12.0 (40.5 min) → 6.0 (60.8 min). The supervisor is right that 8.0 mg/L lies between the second and third of these, and therefore right that the answer is 'under an hour'. The error is quantitative: three half-lives overshoot to 6.0 mg/L, so an hour is not the time to reach 8.0, it is the time to reach roughly 6.0.
The exact bridge from 12.0 to 8.0 is a factor of 1.5, not 2, so it takes ln 1.5/k = 0.405465/0.034190 = 11.86 min. Adding to 40.54 min gives 52.4 min, matching the direct solution.
Judgement: an approximate argument is acceptable as a sanity check on the exact answer, but it must not be the basis for scheduling. Here the loose version is wrong by about 8 minutes of holding time - a real cost in tank capacity, and in the opposite direction from the safety margin.