Cooling Coffee and the Half-Life Trap
A forensic analyst models the temperature of a cup of coffee left in a 20 °C room using Newton's law of cooling: T(t) = 20 + 75·e^(−kt), where T is in °C and t in minutes. A thermometer reading at t = 0 gives 95 °C, and at t = 10 min gives 66 °C. A junior analyst proposes using the half-life formula t½ = ln(2)/k to find when the coffee reaches "half its starting temperature," and computes a value using 95/2 = 47.5 °C. The senior analyst warns that half-life applies to the excess above ambient, not to the raw temperature. A second reading at t = 20 min is recorded as 50 °C. The analyst must decide which reading is consistent with the model and how to report the time to reach 47.5 °C.
1. Determine k from the t = 10 min reading and check whether the t = 20 min reading is consistent.
Read the worked answer
Using T(10) = 66:
66 = 20 + 75·e^(−10k), so 46 = 75·e^(−10k), e^(−10k) = 46/75 ≈ 0.61333.
−10k = ln(0.61333) ≈ −0.48885, so k ≈ 0.048885 min⁻¹.
Predict T(20) = 20 + 75·e^(−20·0.048885) = 20 + 75·e^(−0.9777) ≈ 20 + 75(0.3762) ≈ 20 + 28.2 = 48.2 °C.
The recorded 50 °C is about 1.8 °C above the prediction, within plausible thermometer error but not exact. The judgement is to state the discrepancy and decide whether it is measurement noise or model failure; a single 1.8 °C deviation does not justify abandoning the model, but it should be reported.
2. Explain why the junior analyst's half-life calculation is wrong and compute the correct time to reach 47.5 °C.
Read the worked answer
Half-life in exponential decay applies to the decaying quantity, which here is the excess temperature above ambient: E(t) = T(t) − 20 = 75·e^(−kt). The raw temperature T is not exponential because of the +20 offset, so halving 95 to 47.5 is not a half-life event.
The half-life of the excess is t½ = ln2/k ≈ 0.6931/0.048885 ≈ 14.18 min. After one half-life the excess is 37.5, so T ≈ 57.5 °C, not 47.5 °C.
To reach T = 47.5: 47.5 = 20 + 75·e^(−kt), so 27.5 = 75·e^(−kt), e^(−kt) = 0.36667, −kt = ln(0.36667) ≈ −1.00330, t ≈ 20.52 min.
The correct answer is about 20.5 min, not the 14.2 min the junior would have reported. The judgement call is recognizing that an additive constant breaks the half-life shortcut and that the decay must be measured on the excess.