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New Zealand
Level 8 - NCEA Level 3

Newton's law of cooling


Isaac Newton, in the early 1700s, observed that the rate of change of the temperature of a body was proportional to the difference between the temperatures of the body and its surroundings.

In mathematical symbols, we write $\frac{\mathrm{d}T}{\mathrm{d}t}=k\left(T-T_a\right)$dTdt=k(TTa) where $T$T is the temperature of the body at time $t$t and $T_a$Ta is the ambient temperature.

This is a differential equation that is almost in a form whose solution we recognise from previous work. To make it completely recognisable, we might do the following:

Put $y=T-T_a$y=TTa and hence, $\frac{\mathrm{d}y}{\mathrm{d}T}=1$dydT=1.

Now, using the chain rule,

$\frac{\mathrm{d}y}{\mathrm{d}t}$dydt $=$= $\frac{\mathrm{d}y}{\mathrm{d}T}.\frac{\mathrm{d}T}{\mathrm{d}t}$dydT.dTdt
  $=$= $1\times ky$1×ky
  $=$= $ky$ky

This differential equation has the solution $y=Ae^{rt}$y=Aert. So, $T-T_a=Ae^{rt}$TTa=Aert, or $T=T_a+Ae^{rt}$T=Ta+Aert.

We see that at time $t=0$t=0, the temperature of the body is  $T(0)=T_a+A$T(0)=Ta+A. We, therefore, set the parameter $A$A to $T(0)-T_a$T(0)Ta. So, the required function is:


The parameter $r$r will need to be determined from experimental results. If cooling is occurring, $r$r will be negative.


Example 1

A container of water is boiled and left to stand in a room with ambient temperature $20^\circ\text{C}$20°C. In $12$12 minutes, the water has cooled to $60^\circ\text{C}$60°C. How long will it take for the water to cool to $25^\circ\text{C}$25°C?

We can use the known solution to the differential equation and write 

$T$T $=$= $T_a+\left(T(0)-T_a\right)e^{rt}$Ta+(T(0)Ta)ert
  $=$= $20+80e^{rt}$20+80ert

From the given information, we can obtain the parameter $r$r.

$60$60 $=$= $20+80e^{12r}$20+80e12r
$\frac{40}{80}$4080 $=$= $e^{12r}$e12r
$\ln0.5$ln0.5 $=$= $12r$12r
$r$r $=$= $-0.0578$0.0578

Thus, with the parameter inserted, the formula is


We solve for $t$t when $T=25$T=25.

$25$25 $=$= $20+80e^{-0.0578t}$20+80e0.0578t
$\frac{5}{80}$580 $=$= $e^{-0.0578t}$e0.0578t
$\ln0.0625$ln0.0625 $=$= $-0.0578t$0.0578t
$t$t $=$= $-\frac{\ln0.0625}{0.0578}$ln0.06250.0578
  $=$= $47.97$47.97

The water cools to $25^\circ\text{C}$25°C in just under $48$48 minutes.





In the first part of this chapter, we solved a differential equation to obtain the explicit function $T=T_a+\left(T(0)-T_a\right)e^{rt}$T=Ta+(T(0)Ta)ert

Whenever a differential equation is solved, it is a good practice to verify that the proposed solution is correct by differentiating. In this case, we need to find $\frac{\mathrm{d}T}{\mathrm{d}t}$dTdt. Thus,

$T$T $=$= $T_a+\left(T(0)-T_a\right)e^{rt}$Ta+(T(0)Ta)ert
$\frac{\mathrm{d}T}{\mathrm{d}t}$dTdt $=$= $\left(T(0)-T_a\right)re^{rt}$(T(0)Ta)rert

But this is just $r(T-T_a)$r(TTa), because according to the function, $T-T_a=\left(T(0)-T_a\right)e^{rt}$TTa=(T(0)Ta)ert. So, the derivative is correct if we identify $r$r with $k$k.


finding the parameters

We found the general solution $T=T_a+Ae^{rt}$T=Ta+Aert for the differential equation $\frac{\mathrm{d}T}{\mathrm{d}t}=k(T-T_a)$dTdt=k(TTa).

We used the initial condition, at $t=0$t=0, $T=T(0)$T=T(0) to find that $A=T(0)-T(a)$A=T(0)T(a). That is, $A$A is the difference between the ambient temperature and the temperature at time $t=0$t=0.

In Example 1 we showed how the parameter $r$r is found from the experimental results. This number depends on the experimental conditions and is related to the heat-transferring properties of the materials involved. The units for the time and temperature measurements will also affect $r$r


solving for time

The function $T(t)=T_a+\left(T(0)-T_a\right)e^{rt}$T(t)=Ta+(T(0)Ta)ert enables us to calculate the temperature $T$T at time $t$t provided that we have determined the various constants in the formula. 

In Example 1 we had $T=20+80e^{-0.0578t}$T=20+80e0.0578t so that it is a simple matter to calculate $T$T for any given $t$t.

We may also wish to discover at what time $t$t the temperature $T$T will have reached a certain value. This can be done by trial-and-error, adjusting $t$t until the correct $T$T is found. However, in this case, the inverse function exists so that it is possible to write $t$t as a function of $T$T. This is illustrated in Example 1 but in general, we have

$T$T $=$= $T_a+\left(T(0)-T_a\right)e^{rt}$Ta+(T(0)Ta)ert
$T-T_a$TTa $=$=


$\frac{T-T_a}{T(0)-T_a}$TTaT(0)Ta $=$= $e^{rt}$ert
$\log\left(\frac{T-T_a}{T(0)-T_a}\right)$log(TTaT(0)Ta) $=$= $rt$rt
$t$t $=$= $\frac{1}{r}\log\left(\frac{T-T_a}{T(0)-T_a}\right)$1rlog(TTaT(0)Ta)



We began with the differential equation $\frac{\mathrm{d}T}{\mathrm{d}t}=r\left(T-T_a\right)$dTdt=r(TTa). We found when verifying that we had obtained the correct solution in terms of $T$T, that this was equivalent to $\frac{\mathrm{d}T}{\mathrm{d}t}=\left(T(0)-T_a\right)re^{rt}$dTdt=(T(0)Ta)rert

In this form, we can find the rate of change of $T$T at a time $t$t. In the original form, we can find the rate of change of $T$T when $T$T reaches some particular value.


Example 2

In Example 1, we had $T(0)=100^\circ$T(0)=100°, $T_a=20^\circ$Ta=20° and $r=-0.0578$r=0.0578.

What is $\frac{\mathrm{d}T}{\mathrm{d}t}$dTdt when $T=50^\circ$T=50°?
What is $\frac{\mathrm{d}T}{\mathrm{d}t}$dTdt when $t=15$t=15 minutes?

According to the original differential equation, we have $\frac{\mathrm{d}T}{\mathrm{d}t}=-0.0578(100-50)=-2.89$dTdt=0.0578(10050)=2.89 degrees per minute when $T=50^\circ$T=50°.

If the function $T=20+80e^{-0.0578t}$T=20+80e0.0578t is differentiated, we obtain the other form of the derivative, $\frac{\mathrm{d}T}{\mathrm{d}t}=-4.624e^{-0.0578t}$dTdt=4.624e0.0578t and when $t=15$t=15 this gives $\frac{\mathrm{d}T}{\mathrm{d}t}=-1.94$dTdt=1.94 degrees per minute.


Worked Examples

Question 1

A substance is placed in a room of constant temperature $25$25$°C$°C, and cools from $81$81$°C$°C to $53$53$°C$°C after $17$17 minutes according to Newton’s Law of Cooling.

The rate at which a substance cools is given by the equation $\frac{dT}{dt}=-k\left(T-25\right)$dTdt=k(T25) where $T$T is the temperature of the substance in degrees Celsius, $t$t is the elapsed time in minutes and $k$k is the cooling constant.

  1. Without finding $B$B or $k$k, use differentiation to show that $T=25+Be^{-kt}$T=25+Bekt is a solution to the equation $\frac{dT}{dt}=-k\left(T-25\right)$dTdt=k(T25), where $B$B is a constant.

  2. Solve for the number of minutes $t$t it takes the substance to cool down to $48$48$°C$°C.

    Round your answer to the nearest whole minute.

Question 2

The rate at which a substance cools is proportional to the difference between its temperature $T$T and the constant temperature $A$A of the surrounding air. This can be expressed by the equation $\frac{dT}{dt}=k\left(T-A\right)$dTdt=k(TA) where $t$t is the time in hours and $k$k is a constant.

  1. Using differentiation, show that $T=A+Be^{kt}$T=A+Bekt is a solution of the equation.

  2. Some sugar is heated to a temperature of $150$150$°C$°C, and when it is placed in a $18$18$°C$°C room, it takes $6$6 minutes to cool down to $50$50$°C$°C.

    Determine the value of $B$B.

  3. Considering how the sugar in part (b) cools, solve for the value of $k$k. State your answer in exact form.

  4. Find the temperature of the sugar after $12$12 minutes. Round your answer to the nearest degree.

Question 3

A cake cools at a rate proportional to the difference between the cake temperature $T$T and the room temperature $26$26$°C$°C. Initially the cake is $88$88$°C$°C and after $18$18 minutes the cake is $57$57$°C$°C.

  1. Show that $T=26+62e^{-kt}$T=26+62ekt satisfies the cooling equation $\frac{dT}{dt}=-k\left(T-26\right)$dTdt=k(T26).

  2. Find the cake temperature after $54$54 minutes. Round your answer to the nearest degree.

  3. Select the graph of the function $T=26+62e^{-\frac{\ln2}{18}t}$T=26+62eln218t.

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  4. Find when the cake temperature reaches $37$37$°C$°C. Round your answer to the nearest minute.

  5. Find the limiting temperature of the cake.

  6. What is the rate of change of temperature after exactly $18$18 minutes? Round your answer to two decimal places.



Form differential equations and interpret the solutions


Apply differentiation methods in solving problems

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