| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2005 |
| Session | June |
| 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 question requiring formation of a differential equation from a word problem, solving it with given initial conditions, and applying the solution. While it involves multiple steps (8 marks for part ii), the techniques are routine for C4: separating variables, integrating, applying boundary conditions, and substituting values. The algebraic manipulation is straightforward and follows a well-practiced template that students would have seen in textbooks. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| \(k = \text{const of proportionality}\), \(-\theta = \text{falling}\), \(\frac{d\theta}{dt} = \text{rate of change}\), \(\theta - 20 = \) diff betw obj & surrond temp | B2 2 | All 4 items (first two may be linked) S.R. Award B1 for any 2 items |
| (ii) \(\frac{1}{(\theta - 20)} d\theta = -k \int dt\) | M1, A1A1 | For separating variables; For integ each side (c not essential); Dep on 'c' being involved |
| \(\ln(\theta - 20) = -kt + c\) | M1 | |
| Subst \((θ,t) = (100,0)\) or \((68.5)\) | A1 | |
| k ] AG | ||
| \(c = \ln 80\) | A1, M1, A1 8 | |
| \(k = \frac{1}{5}\ln\frac{5}{3}\) | ||
| \(\theta = 20 + 80e^{-\ln(5/3)}\) | Subst into AEF of given eqn & solve; Accept \(15.7\) or \(15.8\); f.t. only if \(\theta =\) their \((68 - 32)\) or \(32\) | |
| (iii) Substitute \(\theta = 68 - 32\) | M1, A1 3 | |
| \(t = 15.75\) | ||
| Extra time \(= 10.75\), \their \(15.75 - 5\) | B1 3 |
$k = \text{const of proportionality}$, $-\theta = \text{falling}$, $\frac{d\theta}{dt} = \text{rate of change}$, $\theta - 20 = $ diff betw obj & surrond temp | B2 2 | All 4 items (first two may be linked) **S.R.** Award B1 for any 2 items
(ii) $\frac{1}{(\theta - 20)} d\theta = -k \int dt$ | M1, A1A1 | For separating variables; For integ each side (c not essential); Dep on 'c' being involved
$\ln(\theta - 20) = -kt + c$ | | M1 | or M2 for limits $(100,0)$ $(68.5) + $ A1 for
Subst $(θ,t) = (100,0)$ or $(68.5)$ | | A1 | for M2 for limits $(100,0)$ $(68.5) + $ A1
| | | k ] **AG**
$c = \ln 80$ | A1, M1, A1 8 |
$k = \frac{1}{5}\ln\frac{5}{3}$ | |
$\theta = 20 + 80e^{-\ln(5/3)}$ | | Subst into **AEF** of given eqn & solve; Accept $15.7$ or $15.8$; f.t. only if $\theta =$ their $(68 - 32)$ or $32$
(iii) Substitute $\theta = 68 - 32$ | | M1, A1 3 | **13**
$t = 15.75$ | |
Extra time $= 10.75$, \their $15.75 - 5$ | | B1 3 |Newton's law of cooling states that the rate at which the temperature of an object is falling at any instant is proportional to the difference between the temperature of the object and the temperature of its surroundings at that instant. A container of hot liquid is placed in a room which has a constant temperature of $20°C$. At time $t$ minutes later, the temperature of the liquid is $\theta°C$.
\begin{enumerate}[label=(\roman*)]
\item Explain how the information above leads to the differential equation
$$\frac{d\theta}{dt} = -k(\theta - 20),$$
where $k$ is a positive constant. [2]
\item The liquid is initially at a temperature of $100°C$. It takes 5 minutes for the liquid to cool from $100°C$ to $68°C$. Show that
$$\theta = 20 + 80e^{-(\frac{k}{5} \ln \frac{5}{3})t}.$$ [8]
\item Calculate how much longer it takes for the liquid to cool by a further $32°C$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 2005 Q9 [13]}}