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Subsection 3.3.2 Newton's Law of Cooling

Recall Newton's law of cooling from the start of this section:

The rate of change of temperature of an object is proportional to the difference in temperature between the object and its surroundings.

The temperature of the surroundings is sometimes called the ambient temperature. We then translated this statement into the following differential equation

where \(T(t)\) is the temperature of the object at time \(t\text{,}\) \(A\) is the temperature of its surroundings, and \(K\) is a constant of proportionality. This mathematical model of temperature change works well when studying a small object in a large, fixed temperature, environment. For example, a hot cup of coffee in a large room  7 It does not work so well when the object is of a similar size to its surroundings since the temperature of the surroundings will rise as the object cools. It also fails when there are phase transitions involved — for example, an ice-cube melting in a warm room does not obey Newton's law of cooling..

Before we worry about solving this equation, let's think a little about the sign of the constant of proportionality. At any time \(t\text{,}\) there are three possibilities.

  • If \(T(t) \gt A\text{,}\) that is, if the body is warmer than its surroundings, we would expect heat to flow from the body into its surroundings and so we would expect the body to cool off so that \(\diff{T}{t}(t) \lt 0\text{.}\) For this expectation to be consistent with equation 3.3.7, we need \(K \lt 0\text{.}\)
  • If \(T(t) \lt A\text{,}\) that is the body is cooler than its surroundings, we would expect heat to flow from the surroundings into the body and so we would expect the body to warm up so that \(\diff{T}{t}(t) \gt 0\text{.}\) For this expectation to be consistent with equation 3.3.7, we again need \(K \lt 0\text{.}\)
  • Finally if \(T(t)=A\text{,}\) that is the body and its environment have the same temperature, we would not expect any heat to flow between the two and so we would expect that \(\diff{T}{t}(t)=0\text{.}\) This does not impose any condition on \(K\text{.}\)

In conclusion, we would expect \(K \lt 0\text{.}\) Of course, we could have chosen to call the constant of proportionality \(-k\text{,}\) rather than \(K\text{.}\) Then the differential equation would be \(\diff{T}{t} = -k\big(T-A\big)\) and we would expect \(k \gt 0\text{.}\)

Now to find the general solution to equation 3.3.7. Since this equation is so similar in form to equation 3.3.1, we might expect a similar solution. Start by trying \(T(t) = Ce^{Kt}\) and let's see what goes wrong. Substitute it into the equation:

\begin{align*} \diff{T}{t} &= K( T(t)- A)\\ K C e^{Kt} &= KCe^{KT} - KA\\ ?0 & = -KA? & \text{the constant $A$ causes problems!} \end{align*}

Let's try something a little different — recall that the derivative of a constant is zero. So we can add or subtract a constant from \(T(t)\) without changing its derivative. Set \(Q(t) = T(t)+B\text{,}\) then

\begin{align*} \diff{Q}{t}(t) &= \diff{T}{t}(t) & \text{by Newton's law of cooling}\\ & = K(T(t)-A) = K(Q(t)-B-A) \end{align*}

So if we choose \(B=-A\) then we will have

\begin{align*} \diff{Q}{t}(t) &= K Q(t) \end{align*}

which is exactly the same form as equation 3.3.1, but with \(K=-k\text{.}\) So by Theorem 3.3.2

\begin{align*} Q(t) &= Q(0) e^{Kt} \end{align*}

We can translate back to \(T(t)\text{,}\) since \(Q(t)=T(t)-A\) and \(Q(0)=T(0)-A\text{.}\) This gives us the solution.

Just before we put this into action, we remind the reader that \(\log x = \log_e x = \ln x\text{.}\)

The temperature of a glass of iced tea is initially \(5^\circ\text{.}\) After 5 minutes, the tea has heated to \(10^\circ\) in a room where the air temperature is \(30^\circ\text{.}\)

  1. Determine the temperature as a function of time.
  2. What is the temperature after 10 minutes?
  3. Determine when the tea will reach a temperature of \(20^\circ\text{.}\)

Solution Part (a)

  • Denote by \(T(t)\) the temperature of the tea \(t\) minutes after it was removed from the fridge, and let \(A=30\) be the ambient temperature.
  • By Newton's law of cooling,
    \begin{gather*} \diff{T}{t}=K(T-A) = K(T-30) \end{gather*}
    for some, as yet unknown, constant of proportionality \(K\text{.}\)
  • By Corollary 3.3.8,
    \begin{gather*} T(t) = [T(0)-30]\,e^{Kt} + 30 =30-25 e^{Kt} \end{gather*}
    since the initial temperature \(T(0)=5\text{.}\)
  • This solution is not complete because it still contains an unknown constant, namely \(K\text{.}\) We have not yet used the given data that \(T(5)=10\text{.}\) We can use it to determine \(K\text{.}\) At \(t=5\text{,}\)
    \begin{align*} T(5) &=30-25 e^{5K}=10 & \text{rearrange}\\ e^{5K} &=\frac{20}{25}\\ 5K &=\log\frac{20}{25} & \text{and so}\\ K &=\frac{1}{5}\log\frac{4}{5}=-0.044629 & \text{ to 6 digits} \end{align*}

Part (b)

  • To find the temperature at 10 minutes we can just use the solution we have determined above.
    \begin{align*} T(10)&=30-25 e^{10K}\\ &=30-25 e^{10\times\frac{1}{5}\log\frac{4}{5}}\\ &=30-25 e^{2\log\frac{4}{5}} = 30-25 e^{\log\frac{16}{25}}\\ &=30-16=\text{$14^\circ$} \end{align*}

Part (c)

  • We can find when the temperature is \(20^\circ\) by solving \(T(t)=20\text{:}\)
    \begin{align*} 20 &= 30-25 e^{Kt} & \text{rearrange}\\ e^{Kt} &=\frac{10}{25} = \frac{2}{5}\\ K t &= \log \frac{2}{5}\\ t &= \frac{\log \frac{2}{5}}{K}\\ &= \text{20.5 minutes} & \text{ to 1 decimal place} \end{align*}

A slightly more gruesome example.

A dead body is discovered at 3:45pm in a room where the temperature is 20\(^\circ\)C. At that time the temperature of the body is 27\(^\circ\)C. Two hours later, at 5:45pm, the temperature of the body is 25.3 \(^\circ\)C. What was the time of death? Note that the normal (adult human) body temperature is \(37^\circ\text{.}\)

Solution We will assume  8 We don't know any other method! that the body's temperature obeys Newton's law of cooling.

  • Denote by \(T(t)\) the temperature of the body at time \(t\text{,}\) with \(t=0\) corresponding to 3:45pm. We wish to find the time of death — call it \(t_d\text{.}\)

  • There is a lot of data in the statement of the problem; we are told that

    • the ambient temperature: \(A=20\)
    • the temperature of the body when discovered: \(T(0)=27\)
    • the temperature of the body 2 hours later: \(T(2)=25.3\)
    • assuming the person was a healthy adult right up until he died, the temperature at the time of death: \(T(t_d)=37\text{.}\)
  • Since we assume the temperature of the body obeys Newton's law of cooling, we use Corollary 3.3.8 to find,
    \begin{gather*} T(t) = [T(0)-A]\,e^{Kt} + A =20+7 e^{Kt} \end{gather*}
    Two unknowns remain, \(K\) and \(t_d\text{.}\)
  • We can find the constant \(K\) by using \(T(2)=25.3\text{:}\)
    \begin{align*} 25.3=T(2)&= 20+7 e^{2K} & \text{rearrange}\\ 7 e^{2K}&=5.3 & \text{rearrange a bit more}\\ 2K &= \log\big(\tfrac{5.3}{7}\big)\\ K &= \tfrac{1}{2} \log\big(\tfrac{5.3}{7}\big) = -0.139 & \text{to 3 decimal places} \end{align*}
  • Since we know  9 Actually, we are assuming again. that \(t_d\) is determined by \(T(t_d)=37\text{,}\) we have
    \begin{align*} 37 = T(t_d) &= 20+7 e^{-0.139 t_d} & \text{rearrange}\\ e^{-0.139 t_d} &= \tfrac{17}{7}\\ -0.139 t_d &=\log\big(\tfrac{17}{7}\big)\\ t_d &= -\tfrac{1}{0.139}\log\big(\tfrac{17}{7}\big)\\ & = - 6.38 &\text{to 2 decimal places} \end{align*}
    Now \(6.38\) hours is \(6\) hours and \(0.38\times 60 = 23\) minutes. So the time of death was \(6\) hours and \(23\) minutes before 3:45pm, which is 9:22am.

A slightly tricky example — we need to determine the ambient temperature from three measurements at different times.

A glass of room-temperature water is carried out onto a balcony from an apartment where the temperature is \(22^\circ\)C. After one minute the water has temperature \(26^\circ\)C and after two minutes it has temperature \(28^\circ\)C. What is the outdoor temperature?

Solution We will assume that the temperature of the thermometer obeys Newton's law of cooling.

  • Let \(A\) be the outdoor temperature and \(T(t)\) be the temperature of the water \(t\) minutes after it is taken outside.
  • By Newton's law of cooling,
    \begin{gather*} T(t)=A+\big(T(0)-A\big)e^{Kt} \end{gather*}
    by Corollary 3.3.8. Notice there are 3 unknowns here — \(A\text{,}\) \(T(0)\) and \(K\) — so we need three pieces of information to find them all.
  • We are told \(T(0)=22\text{,}\) so
    \begin{align*} T(t) &=A+\big(22-A\big)e^{Kt}. \end{align*}
  • We are also told \(T(1)=26\text{,}\) which gives
    \begin{align*} 26 &=A+\big(22-A\big)e^{K} & \text{rearrange things}\\ e^K&=\frac{26-A}{22-A} \end{align*}
  • Finally, \(T(2)=28\text{,}\) so
    \begin{align*} 28&=A+\big(22-A\big)e^{2K} & \text{rearrange}\\ e^{2K} &= \frac{28-A}{22-A} & \text{but $e^K=\frac{26-A}{22-A}$, so}\\ \left(\frac{26-A}{22-A}\right)^2 &=\frac{28-A}{22-A} & \text{multiply through by $(22-A)^2$}\\ (26-A)^2 &= (28-A)(22-A) \end{align*}
    We can expand out both sides and collect up terms to get
    \begin{align*} \underbrace{26^2}_{=676}-52A+A^2 &= \underbrace{28\times22}_{=616}-50A+A^2\\ 60 &= 2A\\ 30 &= A \end{align*}
    So the temperature outside is \(30^\circ\text{.}\)
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