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.
Key idea
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.
Why it works
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.
Worked example: Half-life
- 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.
Answer: 20 mg; $k \approx 0.0866$ per year
Worked example: Cooling coffee
- 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 $70\left(\frac{4}{7}\right)^{2} = \frac{160}{7} \approx 22.9$.
- $T = 20 + 22.9$.
Answer: about 42.9°C
Worked example: When will it reach a temperature?
- 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 $t = 10\cdot\frac{\ln 7}{\ln(7/4)}$.
Answer: $t \approx 34.8$ minutes
Common mistake