| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Newton's law of cooling |
| Difficulty | Moderate -0.3 This is a straightforward application of Newton's law of cooling with clear scaffolding. Part (i) is trivial arithmetic (20/k₁), part (ii) requires writing dθ/dt = -k₂(θ-20), and part (iii) involves separating variables and integrating ln(θ-20), which is standard C4 technique. The context is slightly more complex than a basic textbook example, but the mathematical steps are routine and well-signposted. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{20}{k_1}\) (seconds) | B1 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d\theta}{dt}=-k_2(\theta-20)\) | B1 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Separate variables or invert each side | M1 | Correct eqn or very similar |
| Correct int of each side \((+c)\) | A1,A1 | for each integration |
| Subst \(\theta=60\) when \(t=0\) into eqn containing '\(c\)' | M1 | or \(\theta=60\) when \(t=\) their (i) |
| \(c\) (or \(-c\)) \(=\ln 40\) or \(\frac{1}{k_2}\ln 40\) or \(\frac{1}{k_2}\ln 40k_2\) | A1 | Check carefully their '\(c\)' |
| Subst their value of \(c\) and \(\theta=40\) back into equation | M1 | Use scheme on LHS |
| \(t=\frac{1}{k_2}\ln 2\) | A1 | Ignore scheme on LHS |
| Total time \(=\frac{1}{k_2}\ln 2+\text{their (i)}\) (seconds) | \(\sqrt{}\)A1 8 |
# Question 9:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{20}{k_1}$ (seconds) | B1 **1** | |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d\theta}{dt}=-k_2(\theta-20)$ | B1 **1** | |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Separate variables or invert each side | M1 | Correct eqn or very similar |
| Correct int of each side $(+c)$ | A1,A1 | for each integration |
| Subst $\theta=60$ when $t=0$ into eqn containing '$c$' | M1 | or $\theta=60$ when $t=$ their (i) |
| $c$ (or $-c$) $=\ln 40$ or $\frac{1}{k_2}\ln 40$ or $\frac{1}{k_2}\ln 40k_2$ | A1 | Check carefully their '$c$' |
| Subst their value of $c$ and $\theta=40$ back into equation | M1 | Use scheme on LHS |
| $t=\frac{1}{k_2}\ln 2$ | A1 | Ignore scheme on LHS |
| Total time $=\frac{1}{k_2}\ln 2+\text{their (i)}$ (seconds) | $\sqrt{}$A1 **8** | |9 A tank contains water which is heated by an electric water heater working under the action of a thermostat. The temperature of the water, $\theta ^ { \circ } \mathrm { C }$, may be modelled as follows. When the water heater is first switched on, $\theta = 40$. The heater causes the temperature to increase at a rate $k _ { 1 } { } ^ { \circ } \mathrm { C }$ per second, where $k _ { 1 }$ is a constant, until $\theta = 60$. The heater then switches off.\\
(i) Write down, in terms of $k _ { 1 }$, how long it takes for the temperature to increase from $40 ^ { \circ } \mathrm { C }$ to $60 ^ { \circ } \mathrm { C }$.
The temperature of the water then immediately starts to decrease at a variable rate $k _ { 2 } ( \theta - 20 ) ^ { \circ } \mathrm { C }$ per second, where $k _ { 2 }$ is a constant, until $\theta = 40$.\\
(ii) Write down a differential equation to represent the situation as the temperature is decreasing.\\
(iii) Find the total length of time for the temperature to increase from $40 ^ { \circ } \mathrm { C }$ to $60 ^ { \circ } \mathrm { C }$ and then decrease to $40 ^ { \circ } \mathrm { C }$. Give your answer in terms of $k _ { 1 }$ and $k _ { 2 }$.
4
\hfill \mbox{\textit{OCR C4 2009 Q9 [10]}}