Growth, Decay and Newton's Law of Cooling

When a quantity changes at a rate proportional to its size, $A' = kA$, the solution is $A = A_0e^{kt}$, the exponential model you already know. Newton's law of cooling is the same idea with one change: the rate is proportional to the difference between an object's temperature and its surroundings.

Idea clave

Growth and decay: $A' = kA$ gives $A = A_0e^{kt}$; doubling time or half-life $= \frac{\ln 2}{|k|}$.
Newton's law of cooling: $T' = -k(T - T_s)$ with $k > 0$ gives $$T(t) = T_s + (T_0 - T_s)e^{-kt}.$$ The temperature approaches the surrounding temperature $T_s$, not 0.

Por qué funciona

Let $D = T - T_s$, the gap to room temperature. Since $T_s$ is constant, $D' = T' = -kD$: the gap decays exponentially, $D = D_0e^{-kt}$. Exponential decay means equal time steps multiply the gap by the same factor, so you can often use a ratio from the data instead of computing $k$. At $T = T_s$ the rate is 0: the room temperature is the equilibrium.

Where do the half-life and doubling-time formulas come from? Set $A_0e^{kt} = 2A_0$ and take logarithms: $kt = \ln 2$, so the doubling time is $\frac{\ln 2}{k}$, the same no matter when you start counting. For decay ($k < 0$) the same step with $\frac{1}{2}$ gives the half-life $\frac{\ln 2}{|k|}$. In cooling problems you rarely need $k$ itself: two temperature readings tell you the factor by which the gap shrinks over a known time, and that factor repeats.

Ejemplo resuelto: Half-life

A sample of 160 mg decays with a half-life of 8 years. How much remains after 24 years, and what is $k$?

24 years is 3 half-lives: $160 \cdot \left(\frac{1}{2}\right)^{3} = 20$.
$e^{-8k} = \frac{1}{2}$ gives $k = \frac{\ln 2}{8} \approx 0.0866$ per year.

Respuesta: 20 mg; $k \approx 0.0866$ per year

Ejemplo resuelto: Cooling coffee

Coffee at 90°C sits in a 20°C room. After 10 minutes it is 60°C. What is it after 20 minutes?

The gap starts at 70 and is 40 after 10 minutes: a factor $\frac{40}{70} = \frac{4}{7}$ per 10 minutes.
After 20 minutes the gap is $70\left(\frac{4}{7}\right)^{2} = \frac{160}{7} \approx 22.9$.
$T = 20 + 22.9$.

Respuesta: about 42.9°C

Ejemplo resuelto: When will it reach a temperature?

For the same coffee, when is it 30°C?

The gap must shrink from 70 to 10: $70\left(\frac{4}{7}\right)^{t/10} = 10$.
$\frac{t}{10}\ln\frac{4}{7} = \ln\frac{1}{7}$, so $t = 10\cdot\frac{\ln 7}{\ln(7/4)}$.

Respuesta: $t \approx 34.8$ minutes

Error común

Writing $T' = -kT$ forgets the room: it predicts the coffee cools toward 0°C. The rate must be zero when $T = T_s$, so the model needs $T - T_s$. A quick check: plug $T = T_s$ into your equation; you should get $T' = 0$. Also note that $T_s$ may be larger than $T_0$: a cold drink warms toward room temperature by the same law.

The direction field of $T' = -0.3(T - 20)$ ($t$ in minutes across, temperature up). Every solution, hot or cold, flattens out at the room temperature 20°C.

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