What halves every few minutes as soup cools?

1. 1. The rate follows the gap

   Soup at 82 °C is left in a room at 18 °C. Newton's law of cooling says the gap between them, $D = T - 18$, shrinks at a rate proportional to itself: $D' = -kD$. The figure shows the gap, from 64 °C, and its tangent. Drag the point along the curve.

   Question: At the start the gap is 64 °C and it closes at about 7.4 °C a minute. How fast does it close when the gap is 32 °C?

   About 3.7 °C a minute, half as fast: the rate is always $k$ times the gap, here with $k \approx 0.116$ per minute. A big gap closes fast and a small one slowly, so soup cools quickly at first and then lingers near the room's temperature.

   The figure: The gap between the soup and the room over 30 minutes, from 64 °C down towards 0. At 6.0 minutes it is 32.0 °C, falling at 3.70 °C a minute along the tangent.

2. 2. Halving on a steady clock

   The marks step along the curve 6 minutes at a time. In the first 6 minutes the gap halves, from 64 °C to 32 °C. Drag the first mark anywhere along the curve and watch the time to halve.

   Question: From 32 °C, how long does the gap take to halve again, to 16 °C?

   6 minutes, and 6 more to 8 °C. Half the gap closes at half the speed, so each half takes as long as the last, from wherever you start the clock. The half-life sets $k$: $e^{-6k} = \tfrac12$, so $k = \tfrac{\log 2}{6} \approx 0.116$ per minute.

   $$D(t) = 64\,e^{-kt}, \qquad k = \frac{\log 2}{6}$$$D(t)$ $=64e^{-kt},$$k$ $=\frac{\log 2}{6}$

   The figure: The gap between the soup and the room over 30 minutes, from 64 °C down towards 0. At 6.0 minutes it is 32.0 °C. Steps of 6 minutes from 6.0 minutes each halve the gap.

3. 3. The temperature does not halve

   The soup's temperature is the room plus the gap, $T = 18 + D$: the same curve lifted by 18 °C. After 6 minutes it reads 50 °C.

   Question: Lift it. Does the temperature halve every 6 minutes too?

   No. Every 6 minutes halves the gap above the dashed room line, so the temperature goes 82, 50, 34, 26 °C, closing in on 18 °C and never passing it. Only the part above the room decays exponentially.

   $$T(t) = 18 + 64\,e^{-kt}$$$T(t)$ $=18+64e^{-kt}$

   The figure: The soup's temperature over 30 minutes, from 82 °C down towards the room's 18 °C, drawn dashed. At 0.0 minutes it is 82.0 °C. Steps of 6 minutes from 0.0 minutes each halve the gap.

4. 4. Decay the gap, not the temperature

   A common slip decays the temperature itself, $T = 82e^{-ct}$, with $c$ chosen so that it too reads 50 °C at 6 minutes. In amber, it agrees with the soup at the start and at 6 minutes. Drag the point to compare them later.

   Question: What does the slip’s model say the soup reads after 24 minutes?

   About 11.3 °C, colder than the room it sits in, and still falling towards 0 °C. The soup itself reads 22 °C, 4 °C above the room. The room's temperature is where the soup ends up, so it belongs in the model: decay $T - 18$, never $T$.

   The figure: The soup's temperature over 30 minutes, from 82 °C down towards the room's 18 °C, drawn dashed. At 24.0 minutes it is 22.0 °C. In amber, the slip's model 82e^(−ct), which agrees at 0 and 6 minutes, crosses the room's temperature and reads 11.3 °C at 24.0 minutes.

5. 5. Read the time off the gap

   How long until the soup reads 26 °C? Work with the gap: it has to fall from 64 °C to 8 °C, a factor of 8.

   Drag the point to when the soup reads 26 °C.

   18 minutes: a factor of $8 = 2^3$ is three half-lives. On paper, set the gap equal to its target and take logs: $64e^{-kt} = 8$ gives $t = \frac{\log 8}{k} = 18$.

   The figure: The soup's temperature over 30 minutes, from 82 °C down towards the room's 18 °C, drawn dashed. At 18.0 minutes it is 26.0 °C.

The film, as said, 1:44

1. 0:00

   Soup at 82 °C cools in a room at 18 °C. Newton's law of cooling is about the gap between them, $D = T - 18$.

2. 0:07

   The gap shrinks at a rate proportional to itself, $D' = -kD$. At 64 °C it closes at about 7.4 °C a minute.

3. 0:14

   Follow it down. At 32 °C it closes at about 3.7 °C a minute: half the gap, half the rate.

4. 0:24

   Mark steps of 6 minutes. Each one halves the gap: 64, 32, 16, 8 °C.

5. 0:30

   Start the clock anywhere: the gap still halves every 6 minutes, because half the gap closes at half the speed.

6. 0:40

   That fixed half-life sets $k$: $e^{-6k} = \tfrac12$, so $k = \tfrac{\log 2}{6} \approx 0.116$ per minute.

7. 0:44

   The temperature is the room plus the gap, $T = 18 + D$: the same curve, lifted by 18 °C.

8. 0:51

   It reads 82, 50, 34, 26 °C. The gap above the room halves; the temperature does not.

9. 0:57

   A common slip decays the temperature itself, $T = 82e^{-ct}$, fitted to read 50 °C at 6 minutes too. It is in amber.

10. 1:04

    Run the clock. The amber model sinks below the room and reads about 11.3 °C at 24 minutes, against the soup’s 22.

11. 1:15

    The room is where the soup ends up, so it belongs in the model: decay $T - 18$, never $T$.

12. 1:21

    When does the soup read 26 °C? Then the gap is 8 °C, down from 64 by a factor of 8.

13. 1:30

    A factor of $8 = 2^3$ is three half-lives: 18 minutes.

14. 1:36

    What shrinks at a rate proportional to itself halves on a steady clock. In cooling, that is the gap to the room.