| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | January |
| Marks | 11 |
| 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 (i) requires translating the given statement into dθ/dt = k(160-θ), which is direct recall. Part (ii) involves separating variables, integrating, applying two initial conditions to find k, then evaluating at t=10. While it requires multiple steps, each is routine for C4 level with no novel insight needed—slightly easier than average due to the guided structure. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d\theta}{dt} = \ldots\) | B1 | |
| \(k(160-\theta)\) | B1 (2) | The 2 @ 'B1' are independent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Separate variables with \((160-\theta)\) in denom; or invert | \*M1 | \(\int\frac{1}{160-\theta}\,d\theta = \int k, \frac{1}{k}, 1\, dt\) |
| Indication that LHS \(= \ln f(\theta)\) | A1 | If wrong ln, final \(3@A=0\) |
| RHS \(= kt\) or \(\frac{1}{k}t\) or \(t\) \(\quad (+c)\) | A1 | |
| Subst. \(t=0, \theta=20\) into equation containing \(c\) | dep\*M1 | |
| Subst \(t=5, \theta=65\) into equation containing \(c\) & \(k\) | dep\*M1 | |
| \(c = -\ln 140 \quad (-4.94)\) ISW | A1 | |
| \(k = \frac{1}{5}\ln\frac{140}{95} \quad (\approx 0.077 \text{ or } 0.078)\) ISW | A1 | |
| Using their \(c\) & \(k\), subst \(t=10\) & evaluate \(\theta\) | dep\*M1 | |
| \(\theta = 96\ (95.535714) \quad \left(95\tfrac{15}{28}\right)\) | A1 (9) | |
| Total: 11 |
# Question 9(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d\theta}{dt} = \ldots$ | B1 | |
| $k(160-\theta)$ | B1 **(2)** | The 2 @ 'B1' are independent |
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# Question 9(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Separate variables with $(160-\theta)$ in denom; or invert | \*M1 | $\int\frac{1}{160-\theta}\,d\theta = \int k, \frac{1}{k}, 1\, dt$ |
| Indication that LHS $= \ln f(\theta)$ | A1 | If wrong ln, final $3@A=0$ |
| RHS $= kt$ or $\frac{1}{k}t$ or $t$ $\quad (+c)$ | A1 | |
| Subst. $t=0, \theta=20$ into equation containing $c$ | dep\*M1 | |
| Subst $t=5, \theta=65$ into equation containing $c$ & $k$ | dep\*M1 | |
| $c = -\ln 140 \quad (-4.94)$ ISW | A1 | |
| $k = \frac{1}{5}\ln\frac{140}{95} \quad (\approx 0.077 \text{ or } 0.078)$ ISW | A1 | |
| Using their $c$ & $k$, subst $t=10$ & evaluate $\theta$ | dep\*M1 | |
| $\theta = 96\ (95.535714) \quad \left(95\tfrac{15}{28}\right)$ | A1 **(9)** | |
| **Total: 11** | | |9 A liquid is being heated in an oven maintained at a constant temperature of $160 ^ { \circ } \mathrm { C }$. It may be assumed that the rate of increase of the temperature of the liquid at any particular time $t$ minutes is proportional to $160 - \theta$, where $\theta ^ { \circ } \mathrm { C }$ is the temperature of the liquid at that time.\\
(i) Write down a differential equation connecting $\theta$ and $t$.
When the liquid was placed in the oven, its temperature was $20 ^ { \circ } \mathrm { C }$ and 5 minutes later its temperature had risen to $65 ^ { \circ } \mathrm { C }$.\\
(ii) Find the temperature of the liquid, correct to the nearest degree, after another 5 minutes.
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\hfill \mbox{\textit{OCR C4 2009 Q9 [11]}}