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.


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?


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?

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