## 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?

Next page - History and applications - Logarithmic plots