| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | January |
| Marks | 9 |
| 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 requiring separation of variables and applying initial conditions. Part (i) is routine integration, part (ii) involves straightforward substitution to find k and solving a logarithmic equation, and part (iii) tests conceptual understanding without calculation. Slightly above average due to the multi-part nature and the conceptual reasoning in part (iii), but all techniques are standard C4 material with no novel insights required. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Separating variables \(\int \frac{1}{\theta + 20}\,d\theta = \int -k\,dt\) | M1 | Or invert each side: \(\frac{dt}{d\theta} = -\frac{1}{k(\theta+20)}\); Must see \(\frac{1}{\theta+20}\); ignore posn 'k' |
| \(\ln(\theta + 20) = -kt\ (+c)\) or equivalent | A1 | "Eqn A" |
| \(\theta = Ae^{-kt} - 20\) oe (i.e. \(\theta = e^{-kt+c} - 20\)) | A1 | "Eqn B" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((-)3 = -k(40 + 20)\) | M1 | Using \(t=0,\ \theta=40,\ \frac{d\theta}{dt} = (-)3\) in given equation |
| \(k = \frac{1}{20}\) oe | *A1 | Not \(k = -\frac{1}{20}\) |
| Subst \(t=0,\ \theta=40\) & their \(k\) into Eqn A or Eqn B and solve for arbitrary constant | M1 | |
| Subst \(\theta = 0\) & their values of \(k\) and arbitrary constant into Eqn A or Eqn B | M1 | |
| \(t = 21.9722 = 22\) minutes cao | dep*A1 | www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(k\) is larger | B1 |
## Question 9:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Separating variables $\int \frac{1}{\theta + 20}\,d\theta = \int -k\,dt$ | M1 | Or invert each side: $\frac{dt}{d\theta} = -\frac{1}{k(\theta+20)}$; Must see $\frac{1}{\theta+20}$; ignore posn 'k' |
| $\ln(\theta + 20) = -kt\ (+c)$ or equivalent | A1 | "Eqn A" |
| $\theta = Ae^{-kt} - 20$ oe (i.e. $\theta = e^{-kt+c} - 20$) | A1 | "Eqn B" |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(-)3 = -k(40 + 20)$ | M1 | Using $t=0,\ \theta=40,\ \frac{d\theta}{dt} = (-)3$ in given equation |
| $k = \frac{1}{20}$ oe | *A1 | Not $k = -\frac{1}{20}$ |
| Subst $t=0,\ \theta=40$ & their $k$ into Eqn A or Eqn B and solve for arbitrary constant | M1 | |
| Subst $\theta = 0$ & their values of $k$ and arbitrary constant into Eqn A or Eqn B | M1 | |
| $t = 21.9722 = 22$ minutes cao | dep*A1 | www |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k$ is larger | B1 | |
---9 The temperature of a freezer is $- 20 ^ { \circ } \mathrm { C }$. A container of a liquid is placed in the freezer. The rate at which the temperature, $\theta ^ { \circ } \mathrm { C }$, of a liquid decreases is proportional to the difference in temperature between the liquid and its surroundings. The situation is modelled by the differential equation
$$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k ( \theta + 20 ) ,$$
where time $t$ is in minutes and $k$ is a positive constant.\\
(i) Express $\theta$ in terms of $t , k$ and an arbitrary constant.
Initially the temperature of the liquid in the container is $40 ^ { \circ } \mathrm { C }$ and, at this instant, the liquid is cooling at a rate of $3 ^ { \circ } \mathrm { C }$ per minute. The liquid freezes at $0 ^ { \circ } \mathrm { C }$.\\
(ii) Find the value of $k$ and find also the time it takes (to the nearest minute) for the liquid to freeze.
The procedure is repeated on another occasion with a different liquid. The initial temperature of this liquid is $90 ^ { \circ } \mathrm { C }$. After 19 minutes its temperature is $0 ^ { \circ } \mathrm { C }$.\\
(iii) Without any further calculation, explain what you can deduce about the value of $k$ in this case.
\hfill \mbox{\textit{OCR C4 2013 Q9 [9]}}