## History and applications

### Newton's law of cooling

The temperature of many objects can be modelled using a differential equation. **Newton's law of cooling** (or heating) states that the temperature of a body changes at a rate proportional to the difference in temperature between the body and its surroundings. It is a reasonably accurate approximation in some circumstances.

More precisely, let \(T\) denote the temperature of an object and \(T_0\) the ambient temperature. If \(t\) denotes time, then Newton's law states that

\[ \dfrac{dT}{dt} = -k (T - T_0), \]where \(k\) is a positive constant. Thus, if the object is much hotter than its surroundings, then \(T - T_0\) is large and positive, so \(\dfrac{dT}{dt}\) is large and negative, so the object cools quickly. If the object is only slightly hotter than its surroundings, then \(T - T_0\) is small positive, and the object cools slowly. So a cup of hot coffee will cool more quickly if you put it in the refrigerator!

This differential equation is of the same type as ones seen previously in this module.

#### Example

You take an ice-cream out of the freezer, kept at \(-18\) \(^\circ\)C. Outside it is 32 \(^\circ\)C. After one minute, the ice-cream has warmed to \(-8\) \(^\circ\)C. What is the temperature of the ice-cream after five minutes?

#### Solution

Let \(T\) be the temperature of the ice-cream (in \(^\circ\)C) after \(t\) minutes out of the freezer. Then Newton's law gives

\[ \dfrac{dT}{dt} = - k(T - 32) = -kT + 32k. \]Remembering that \(k\) is a constant, solving this differential equation gives a general solution of

\[ T = Ce^{-kt} + 32. \]Since \(T=-18\) when \(t=0\), we obtain \(-18 = C + 32\), so \(C = -50\). Since \(T = -8\) when \(t=1\), we have \(-8 = -50 \, e^{-k} + 32\), which gives \(k = \log_e \dfrac{5}{4}\). We obtain

\begin{align*} T &= - 50 \, e^{- (\log_e \dfrac{5}{4}) \, t} + 32 \\ &= -50 \, \Bigl( \dfrac{4}{5} \Bigr)^t + 32. \end{align*} (Here we used some index laws.) Hence, when \(t=5\), we have \[ T = - 50 \, \Bigl( \dfrac{4}{5} \Bigr)^5 + 32 = \dfrac{1952}{125} \approx 15.6\ ^\circ\text{C}. \]Your ice-cream has well and truly melted! Note that we have effectively assumed that the ice-cream is a block of a single temperature — not very realistic.

Exercise 14

A cup of coffee is made with boiling water at a temperature of 100 \(^\circ\)C, in a room at temperature 20 \(^\circ\)C. After two minutes it has cooled to 80 \(^\circ\)C. What is its temperature after five minutes? When will the coffee drop below 40 \(^\circ\)C and taste cold?

Exercise 15

You are given a very hot sample of metal, and wish to know its temperature. You have a thermometer, but it only measures up to 200 \(^\circ\)C and the metal is hotter than that!

You leave the metal in a room kept at 20 \(^\circ\)C. After six minutes it has cooled sufficiently that you can measure its temperature; it is 80 \(^\circ\)C. After another two minutes it is 50 \(^\circ\)C. What was the initial temperature of the metal?