Applications - Newton's Law of Cooling

You will learn how to set up and solve differential equations based on Newton’s Law of Cooling, work with initial value problems, and interpret the results in practical situations.

Newton’s Law of Cooling states that the rate at which an object changes temperature is proportional to the difference between its temperature and the ambient (surrounding) temperature.

Let $T(t)$ be the temperature of the object at time $t$, and let $T_a$ be the constant ambient temperature. Then:

$$\frac{dT}{dt} = -k(T - T_a)$$

where $k \gt 0$ is the cooling constant. This is a first-order linear (and also separable) differential equation.

The solution is:

$$T(t) = T_a + (T_0 - T_a)e^{-kt}$$

where $T_0$ is the initial temperature.

The visual shows two temperature curves: a hot object cooling down (red) and a cold object warming up (blue), both approaching the ambient temperature (dashed line) exponentially.

Example 1: Basic Cooling

A cup of coffee is 90°C when poured. The room is 22°C. After 10 minutes the coffee is 65°C. Find the temperature after 30 minutes.

Solution: $T_a = 22$, $T(0) = 90$, $T(10) = 65$.

First find $k$:

$$65 = 22 + (90 - 22)e^{-10k} \implies 43 = 68e^{-10k}$$ $$k = -\frac{1}{10}\ln\left(\frac{43}{68}\right) \approx 0.046$$

Then at $t = 30$:

$$T(30) = 22 + 68e^{-0.046 \cdot 30} \approx 38.9°\text{C}$$

Example 2: Warming Up

A cold can of soda at 4°C is placed in a 25°C room. After 20 minutes it reaches 12°C. When will it reach 20°C?

Solution: $T_a = 25$, $T_0 = 4$. Find $k$ from $T(20) = 12$:

$$12 = 25 + (4 - 25)e^{-20k} \implies -13 = -21e^{-20k}$$ $$k = -\frac{1}{20}\ln\left(\frac{13}{21}\right) \approx 0.024$$

Solve $T(t) = 20$:

$$20 = 25 - 21e^{-0.024t} \implies t = -\frac{1}{0.024}\ln\left(\frac{5}{21}\right) \approx 60 \text{ minutes}$$

Example 3: Crime Scene Application

A murder victim is found at 9:00 PM with body temperature 28°C. The room is 20°C. One hour later the body is 26.5°C. Estimate the time of death (normal body temp: 37°C).

Solution: From the two readings, find $k \approx 0.022$. Then solve for $t_0$ when $T = 37$:

$$37 = 20 + 17e^{-0.022 \cdot 0} \implies \text{need to work backwards}$$

Setting up the equation from time of death gives approximately 1.75 hours before 9:00 PM, so time of death was around 7:15 PM.

Newton’s Law of Cooling is used in forensics to estimate time of death, in food science to determine safe cooling times for cooked food, and in engineering to design cooling systems for electronics and engines. It also helps meteorologists understand how objects heat up in sunlight or how lakes cool at night. Even your morning coffee or evening tea follows this exact model.