| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Linear transformation to find constants |
| Difficulty | Moderate -0.3 This is a standard exponential modeling question requiring data transformation (subtracting room temperature), taking logarithms to linearize, plotting points, and reading gradient/intercept from a graph. While multi-step, each component is routine C2 material with clear scaffolding through parts (i)-(iii). The physical interpretation and algebraic manipulation are straightforward, making this slightly easier than average. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form1.06i Exponential growth/decay: in modelling context |
| Time (minutes) | 10 | 20 | 30 | 40 | 50 |
| Temperature \(\left( { } ^ { \circ } \mathrm { C } \right)\) | 68 | 53 | 42 | 36 | 31 |
| Answer | Marks |
|---|---|
| At \(t = 0\) oe | B1, B1 |
| 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \log z_0 - k\log 10\) | B1, B1 | nb AG |
| 2 marks | ||
| iii \(Z=\) | If \(z=\) 68, 53...P1, L1,M1, M1, M1 available | |
| \(46\) | T1 | |
| \(31\) | P1 | ft their values, within 2mm |
| \(20\) | L1 | Ruled, using their points |
| \(14\) | ||
| \(9\) | ||
| \(\log z :\) \(1.66\) \(1.49\) \(1.30\) \(1.15\) \(0.95\) | G2 | M1 for attempting +/- gradient |
| \(k = 0.017\) to \(0.019\) or \(0.02\) | B2 | M1 for \((\log) z_0= 1.82\) to \(1.86\) |
| \(z_0= 66\) to \(73\) | C2 | M1 \(3\) to \(5\) or their \(69 × 10^{-70}\) x their \(k\) |
| temp of drink = 25 to 27 | [13] | |
| 9 marks |
**i** Excess temperature
At $t = 0$ oe | B1, B1 |
| | 2 marks |
**ii** $\log z = \log z_0 + \log(10^{kt})$
$= \log z_0 - k\log 10$ | B1, B1 | nb AG
| | 2 marks |
**iii** $Z=$ | | If $z=$ 68, 53...P1, L1,M1, M1, M1 available
$46$ | T1 |
$31$ | P1 | ft their values, within 2mm
$20$ | L1 | Ruled, using their points
$14$ | |
$9$ | |
$\log z :$ $1.66$ $1.49$ $1.30$ $1.15$ $0.95$ | G2 | M1 for attempting +/- gradient
$k = 0.017$ to $0.019$ or $0.02$ | B2 | M1 for $(\log) z_0= 1.82$ to $1.86$
$z_0= 66$ to $73$ | C2 | M1 $3$ to $5$ or their $69 × 10^{-70}$ x their $k$
temp of drink = 25 to 27 | | [13]
| | 9 marks |11 Answer part (iii) of this question on the insert provided.\\
A hot drink is made and left to cool. The table shows its temperature at ten-minute intervals after it is made.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Time (minutes) & 10 & 20 & 30 & 40 & 50 \\
\hline
Temperature $\left( { } ^ { \circ } \mathrm { C } \right)$ & 68 & 53 & 42 & 36 & 31 \\
\hline
\end{tabular}
\end{center}
The room temperature is $22 ^ { \circ } \mathrm { C }$. The difference between the temperature of the drink and room temperature at time $t$ minutes is $z ^ { \circ } \mathrm { C }$. The relationship between $z$ and $t$ is modelled by
$$z = z _ { 0 } 10 ^ { - k t }$$
where $z _ { 0 }$ and $k$ are positive constants.\\
(i) Give a physical interpretation for the constant $z _ { 0 }$.\\
(ii) Show that $\log _ { 10 } z = - k t + \log _ { 10 } z _ { 0 }$.\\
(iii) On the insert, complete the table and draw the graph of $\log _ { 10 } z$ against $t$.
Use your graph to estimate the values of $k$ and $z _ { 0 }$.\\
Hence estimate the temperature of the drink 70 minutes after it is made.
\hfill \mbox{\textit{OCR MEI C2 2005 Q11 [13]}}