Coffee cup cooling: time constant, drinking time, the cold hour
MHF4U Grade 12 Advanced Functions · Exponential Models
A student pours coffee at 82.0 °C, stands the mug on the desk in a 21.0 °C room, and starts a stopwatch. At 12.0 min the probe thermometer reads 60.0 °C. Treating the mug as one lump obeying Newton's law of cooling — the exponential decay of its excess temperature over the room — they want the mug's time constant, the moment the coffee passes their preferred 45.0 °C, and how cold it will be at the full hour.
- T₀ = 82 °C — Coffee at the pour
- T_room = 21 °C — Room temperature
- T(12 min) = 60 °C — Reading at 12.0 min
- T_drink = 45 °C — Preferred drinking temperature
- (a)the mug's thermal time constant
- (b)the time when the coffee reaches 45.0 °C
- (c)the temperature after a full hour
What decays is the EXCESS over the room, so the operative ratio is (82 − 21)/(60 − 21) = 61/39, never 82/60 — subtracting the ambient first is the whole discipline of Newton's law. One timed reading then pins τ = t/ln(61/39) ≈ 26.8 min: the time to fall 63% of whatever excess remains.
Carried onward at full precision, not this rounded figure.
Same law, solved for the clock: t = τ·ln(61/24) ≈ 25.0 min from the pour. Note it is NOT 12 min plus another proportional stretch — a linear interpolation from the first reading would say 45 °C arrives near 20 min, and the curve's slowing flank makes that guess five minutes optimistic.
Carried onward at full precision, not this rounded figure.
The hour mark is 2.24 time constants in, so only e^(−2.24) ≈ 11% of the original 61 K excess survives: about 27.5 °C, tepid. The asymptote is the room itself — the mug never crosses below 21.0 °C, and any model that predicts it has lost a minus sign in the exponent.
Carried onward at full precision, not this rounded figure.
Therefore the mug's time constant is 26.8 min, the coffee is at its drinkable 45.0 °C about 25.0 min after the pour, and by the hour it has sagged to 27.5 °C — six and a half degrees of excess left from the original sixty-one.
Why this order
One measurement, then two predictions: that is the entire architecture, and it is why τ must be earned first. Newton's law says the mug forgets its history at a constant fractional rate, so a single timed reading fixes the only free parameter, and every later question — when is it 45°? what is it at the hour? — is the same exponential read in a different direction. The error the chain is built to expose is treating temperatures instead of temperature EXCESSES: the decaying quantity is T − 21, so the ratio in every logarithm is 61-to-something, and a student who forms 82/60 has fit a curve that decays toward absolute zero rather than toward the desk it sits on. The second habit worth drilling is part (b)'s refusal to interpolate: exponentials cover their first degrees quickly and their last degrees achingly slowly, which is why the mug spends 12 minutes reaching 60 °C but needs another 13 to shed just 15 more.
This is also where the advanced-functions classroom and the heat-transfer lab shake hands. The same T = T_room + 61·e^(−t/τ) is, to the mathematician, a shifted exponential with a horizontal asymptote — and τ is physical: ρVc/hA, thermal mass over surface conductance, which is why a wide shallow cup (more A, less V) goes cold faster than a tall narrow one holding the same coffee. Everything cools on this curve — forensic body-temperature estimates, quenched steel, a house after the furnace fails — and each is the same three-step chain: one calibrated reading, then the clock, then the forecast. The half-life shortcut is worth carrying: τ·ln 2 ≈ 18.6 min here, so the excess halves roughly every nineteen minutes — 61, 30, 15 K — a mental model that lands within a degree of the exact curve.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.