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(T−Ta) 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=T−Ta 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, $T-T_a=Ae^{rt}$T−Ta=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:
$T=T_a+\left(T(0)-T_a\right)e^{rt}$T=Ta+(T(0)−Ta)ert
The parameter $r$r will need to be determined from experimental results. If cooling is occurring, $r$r will be negative.
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
$T=20+80e^{-0.0578t}$T=20+80e−0.0578t.
We solve for $t$t when $T=25$T=25.
$25$25 | $=$= | $20+80e^{-0.0578t}$20+80e−0.0578t |
$\frac{5}{80}$580 | $=$= | $e^{-0.0578t}$e−0.0578t |
$\ln0.0625$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(T−Ta), because according to the function, $T-T_a=\left(T(0)-T_a\right)e^{rt}$T−Ta=(T(0)−Ta)ert. So, the derivative is correct if we identify $r$r with $k$k.
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(T−Ta).
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.
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+80e−0.0578t 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 |
$T-T_a$T−Ta | $=$= |
$\left(T(0)-T_a\right)e^{rt}$(T(0)−Ta)ert |
$\frac{T-T_a}{T(0)-T_a}$T−TaT(0)−Ta | $=$= | $e^{rt}$ert |
$\log\left(\frac{T-T_a}{T(0)-T_a}\right)$log(T−TaT(0)−Ta) | $=$= | $rt$rt |
$t$t | $=$= | $\frac{1}{r}\log\left(\frac{T-T_a}{T(0)-T_a}\right)$1rlog(T−TaT(0)−Ta) |
We began with the differential equation $\frac{\mathrm{d}T}{\mathrm{d}t}=r\left(T-T_a\right)$dTdt=r(T−Ta). 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.
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(100−50)=−2.89 degrees per minute when $T=50^\circ$T=50°.
If the function $T=20+80e^{-0.0578t}$T=20+80e−0.0578t is differentiated, we obtain the other form of the derivative, $\frac{\mathrm{d}T}{\mathrm{d}t}=-4.624e^{-0.0578t}$dTdt=−4.624e−0.0578t and when $t=15$t=15 this gives $\frac{\mathrm{d}T}{\mathrm{d}t}=-1.94$dTdt=−1.94 degrees per minute.
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(T−25) 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.
Without finding $B$B or $k$k, use differentiation to show that $T=25+Be^{-kt}$T=25+Be−kt is a solution to the equation $\frac{dT}{dt}=-k\left(T-25\right)$dTdt=−k(T−25), where $B$B is a constant.
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.
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(T−A) where $t$t is the time in hours and $k$k is a constant.
Using differentiation, show that $T=A+Be^{kt}$T=A+Bekt is a solution of the equation.
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.
Considering how the sugar in part (b) cools, solve for the value of $k$k. State your answer in exact form.
Find the temperature of the sugar after $12$12 minutes. Round your answer to the nearest degree.
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.
Show that $T=26+62e^{-kt}$T=26+62e−kt satisfies the cooling equation $\frac{dT}{dt}=-k\left(T-26\right)$dTdt=−k(T−26).
Find the cake temperature after $54$54 minutes. Round your answer to the nearest degree.
Select the graph of the function $T=26+62e^{-\frac{\ln2}{18}t}$T=26+62e−ln218t.
Find when the cake temperature reaches $37$37$°C$°C. Round your answer to the nearest minute.
Find the limiting temperature of the cake.
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