Question 17
The temperature, $T$ °C, of a cup of coffee $t$ minutes after it is poured can be modelled by the differential equation
$$\frac{dT}{dt}=-k(T-A)$$
where $k$ is a positive constant and $A$ °C is the constant temperature of the room.
The coffee is poured at a temperature of 88 °C. Its temperature is 72 °C four minutes later and 60 °C eight minutes after it is poured.
Use a calculus approach to show that the temperature of the room is 24 °C.
[4 marks] b)Determine the time taken for the coffee to cool to 40 °C.
[2 marks]Three readings, one room temperature
The coffee cools towards the room, and the gap above the room shrinks exponentially. Watch the room temperature slide until the gaps at 0, 4 and 8 minutes shrink by the same factor.
The gap above the room is $T-A=(88-A)e^{-kt}$. The readings are 4 minutes apart, so the gaps must shrink by the same factor each time. Only $A=24$ makes that happen.
Find the room temperature, then the time
For each room temperature you pick, the curve $T=A+(88-A)e^{-kt}$ is fitted through 88 °C at the start and 72 °C at 4 minutes. Change $A$ until the curve also passes through 60 °C at 8 minutes. Then move the time until the coffee is at 40 °C.
With $A=0$ the model is the pure exponential $T=88e^{-kt}$. It predicts about 58.9 °C at $t=8$, not 60 °C, so it does not fit the data. The room is warmer than this.
Your curve passes through 88 °C and 72 °C, but at 8 minutes it predicts about °C. That is below the 60 °C reading, so the room is warmer than this.
Your curve passes through 88 °C and 72 °C, but at 8 minutes it predicts about °C. That is above the 60 °C reading, so the room is cooler than this.
All three readings fit. Here $(72-24)^2=2304$ and $(88-24)(60-24)=2304$, so the room is at 24 °C. Now move the time for part b).
Part b) uses the room temperature you found. Set it back to 24 °C.
At minutes the coffee is still at about °C. It has not cooled to 40 °C yet.
At minutes the coffee is already down to about °C. It reached 40 °C earlier.
The coffee is at 40 °C after about 19.3 minutes. Exactly, $t=\dfrac{\ln(4)}{k}\approx19.2754$. Now try $k=0.07$ to see what early rounding does.
With $k=0.07$ the coffee reaches 40 °C at about 19.8 minutes, not 19.3. Rounding $k$ too early loses the final mark. Keep $k$ unrounded in the calculator.
First, find the room temperature that fits all three readings.
The room is at 24 °C. Now find when the coffee is at 40 °C.
You have both answers: the room is at 24 °C, and the coffee takes about 19.3 minutes to cool to 40 °C.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
Divide by $T-A$, then integrate each side in its own variable.
$$\begin{aligned}\int\frac{1}{T-A}\,dT&=\int -k\,dt\\[4pt]\ln|T-A|&=-kt+c\end{aligned}$$
Marker: correctly separates the variables and integrates
Solve for $T$, then use $t=0$, $T=88$ to find the constant.
$$\begin{aligned}T&=A+Be^{-kt}\\88&=A+B\ \Rightarrow\ B=88-A\\T&=A+(88-A)e^{-kt}\end{aligned}$$
Marker: correctly determines $T$ in terms of $t$, $k$ and $A$
Substitute both readings, then make each exponential the subject.
$$\begin{aligned}72&=A+(88-A)e^{-4k}\\60&=A+(88-A)e^{-8k}\\[6pt]e^{-4k}&=\frac{72-A}{88-A}\\[4pt]e^{-8k}&=\frac{60-A}{88-A}\end{aligned}$$
$A$ and $k$ may be determined simultaneously with technology once the two equations are formed.
Marker: correctly forms two equations using the temperature readings
Square the expression for $e^{-4k}$, set it equal to $e^{-8k}$ and solve for $A$.
$$\begin{aligned}\left(\frac{72-A}{88-A}\right)^2&=\frac{60-A}{88-A}\\[4pt](72-A)^2&=(88-A)(60-A)\\A^2-144A+5184&=A^2-148A+5280\\4A&=96\\A&=24\end{aligned}$$
The readings are 4 minutes apart, so $\left(e^{-4k}\right)^2=e^{-8k}$. This removes $k$ and leaves one equation in $A$ alone.
Marker: correctly shows that the temperature of the room is 24 °C
Put $A=24$ into the equation for $e^{-4k}$ and solve for $k$.
$$\begin{aligned}e^{-4k}&=\frac{48}{64}=\frac{3}{4}\\[4pt]k&=\frac{1}{4}\ln\left(\frac{4}{3}\right)\approx0.0719\end{aligned}$$
Enter ln(4/3)/4 to get 0.07192051811. Store this value (or keep it as Ans) rather than retyping a rounded version.
Marker: correctly determines the value of $k$
Set $T=40$ in $T=24+64e^{-kt}$ and solve for $t$ with the unrounded $k$.
$$\begin{aligned}40&=24+64e^{-kt}\\[4pt]e^{-kt}&=\frac{1}{4}\\[4pt]t&=\frac{\ln(4)}{k}\approx19.3\end{aligned}$$
Straight after finding $k$, enter ln(4)/Ans to get 19.27536672.
$k$ must be carried to at least four decimal places. $k=0.07$ gives 19.8, which does not gain the final mark.
Marker: correctly determines the time
Each phrase of the question gave you something
If I had to guess where you might lose marks…
- Rounding $k$ too early. Using $k=0.07$ gives $t=\dfrac{\ln(4)}{0.07}\approx19.8$ minutes, which loses the final mark. Keep $k$ unrounded in the calculator.
- Doubling instead of squaring. Writing $e^{-8k}=2e^{-4k}$ gives $60-A=2(72-A)$, so $A=84$. The index law is $e^{-8k}=\left(e^{-4k}\right)^2$.
- Applying the initial condition as $B=88$. Writing $T=A+88e^{-kt}$ leads to $(72-A)^2=88(60-A)$, that is $A^2-56A-96=0$, so $A=28\pm4\sqrt{55}\approx57.6648$ or $-1.66479$. Neither is 24. At $t=0$, $T=A+B$, so $B=88-A$.
- Forgetting to subtract the room temperature in b). Solving $40=64e^{-kt}$ gives $t=\dfrac{\ln(1.6)}{k}\approx6.5$ minutes.
- Ignoring $A$ entirely. The pure exponential $T=88e^{-kt}$ gives $k=\frac{1}{4}\ln\left(\frac{88}{72}\right)\approx0.0501677$ and $t\approx15.7$ minutes. The model then predicts about 58.9 °C at $t=8$, not 60 °C, so it does not fit the data.
- A sign error when integrating. Writing $\ln|T-A|=kt+c$ leads to $k=\frac{1}{4}\ln\left(\frac{3}{4}\right)\approx-0.0719205$, which contradicts “$k$ is a positive constant”.
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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