| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Newton's law of cooling |
| Difficulty | Standard +0.3 This is a standard Newton's law of cooling problem with straightforward setup and solution. Part (a) requires translating a word statement into a differential equation (routine for C4), part (b) involves separating variables and using two conditions to find constants, and part (c) applies the model. All steps are textbook-standard with no novel insight required, making it slightly easier than average. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| \((a)\) \(\frac{d\theta}{dt} = -k(\theta - 20)\) | B2 | |
| \((b)\) \(\int \frac{1}{\theta - 20} \, d\theta = \int -k \, dt\) | M1 | |
| \(\ln | \theta - 20 | = -kt + c\) |
| \(t = 0, \theta = 37 \Rightarrow c = \ln 17\) | M1 | |
| \(\ln\left | \frac{\theta - 20}{17}\right | = -kt, \theta = 20 + 17e^{-kt}\) |
| \(t = 4, \theta = 36 \Rightarrow 36 = 20 + 17e^{-4k}\) | M1 | |
| \(k = -\frac{1}{4}\ln\frac{16}{17} = 0.01516\) | A1 | |
| \(t = 10, \theta = 20 + 17e^{-0.01516 \times 10} = 34.6°C\) (3sf) | A1 | |
| \((c)\) \(33 = 20 + 17e^{-0.01516t}\) | M1 | |
| \(t = -\frac{1}{0.01516}\ln\frac{13}{17} = 17.70\) minutes \(= 17\) mins 42 secs | M1 A1 | (13 marks) |
$(a)$ $\frac{d\theta}{dt} = -k(\theta - 20)$ | B2 |
$(b)$ $\int \frac{1}{\theta - 20} \, d\theta = \int -k \, dt$ | M1 |
$\ln|\theta - 20| = -kt + c$ | M1 A1 |
$t = 0, \theta = 37 \Rightarrow c = \ln 17$ | M1 |
$\ln\left|\frac{\theta - 20}{17}\right| = -kt, \theta = 20 + 17e^{-kt}$ | A1 |
$t = 4, \theta = 36 \Rightarrow 36 = 20 + 17e^{-4k}$ | M1 |
$k = -\frac{1}{4}\ln\frac{16}{17} = 0.01516$ | A1 |
$t = 10, \theta = 20 + 17e^{-0.01516 \times 10} = 34.6°C$ (3sf) | A1 |
$(c)$ $33 = 20 + 17e^{-0.01516t}$ | M1 |
$t = -\frac{1}{0.01516}\ln\frac{13}{17} = 17.70$ minutes $= 17$ mins 42 secs | M1 A1 | (13 marks)5. A bath is filled with hot water which is allowed to cool. The temperature of the water is $\theta ^ { \circ } \mathrm { C }$ after cooling for $t$ minutes and the temperature of the room is assumed to remain constant at $20 ^ { \circ } \mathrm { C }$.
Given that the rate at which the temperature of the water decreases is proportional to the difference in temperature between the water and the room,
\begin{enumerate}[label=(\alph*)]
\item write down a differential equation connecting $\theta$ and $t$.
Given also that the temperature of the water is initially $37 ^ { \circ } \mathrm { C }$ and that it is $36 ^ { \circ } \mathrm { C }$ after cooling for four minutes,
\item find, to 3 significant figures, the temperature of the water after ten minutes.
Advice suggests that the temperature of the water should be allowed to cool to $33 ^ { \circ } \mathrm { C }$ before a child gets in.
\item Find, to the nearest second, how long a child should wait before getting into the bath.\\
5. continued
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q5 [13]}}