| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Topic | Differential equations |
| Type | Newton's law of cooling |
| Difficulty | Moderate -0.3 This is a standard Newton's law of cooling problem with straightforward setup, separation of variables, and parameter finding. Part (i) tests understanding of the model, part (ii) is routine integration with given initial conditions, and part (iii) is simple substitution and logarithm manipulation. While it requires multiple techniques, each step follows a well-practiced template with no novel insight needed, making it slightly easier than average. |
| Spec | 1.02z Models in context: use functions in modelling1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| \[\ln | T-20 | = -kt+c\] |
| Answer | Marks | Guidance |
|---|---|---|
| - M1: Rearrange expression of form \(\pm\ln | T-20 | =\pm kt\pm c\) to given expression, including correct manipulation of logs and exponentials – allow if still in terms of \(c\). Must be sound algebra throughout |
**Question 6:**
**(i)**
- $\frac{dT}{dt}$ is the rate of change of $T$
- $T-20$ is difference between $T$ and the temp of the room
- $k$ is the constant of proportionality
- negative since the temperature is decreasing
- **B1**: At least two correct points
- **B1**: Fully correct explanation
**Total for (i): [2]**
**(ii)**
$$\int\frac{1}{T-20}\,dT = \int -k\,dt$$
$$\ln|T-20| = -kt+c$$
$$\ln 60 = c$$
$$T-20 = e^{-kt+\ln 60} = e^{-kt}e^{\ln 60} = 60e^{-kt}$$
$$T = 20+60e^{-kt}$$
- **M1**: Separate variables and attempt integration of both sides oe
- **A1**: Correct $\ln(T-20)$
- **A1**: Correct $-kt$ (allow no $+c$)
- **M1**: Attempt $c$ using $T=80$, $t=0$ (clear detail required if rearrangement is done first), e.g. $80=20+A$
- **M1**: Rearrange expression of form $\pm\ln|T-20|=\pm kt\pm c$ to given expression, including correct manipulation of logs and exponentials – allow if still in terms of $c$. Must be sound algebra throughout
- **A1**: Obtain $T=20+60e^{-kt}$, detail required and no errors seen. Must see $e^{-kt+\ln 60}=e^{-kt}e^{\ln 60}=60e^{-kt}$ (oe in terms of $c$)
**Total for (ii): [6]**
**(iii)**
$$\ln 40 = -2k+\ln 60 \quad \text{OR} \quad 60=20+60e^{-2k}$$
$$2k = \ln\frac{3}{2} \qquad e^{-2k}=\frac{2}{3}$$
$$k=\frac{1}{2}\ln\frac{3}{2} \qquad k=-\frac{1}{2}\ln\frac{2}{3}$$
- **M1**: Substitute $T=60$, $t=2$ into given expression, oe
- **M1**: Attempt to find $k$, allow one slip but must be using correct order of operations
- **A1**: Obtain $k=\frac{1}{2}\ln\frac{3}{2}$ oe, including 0.203
**Total for (iii): [3]**6 A cup of tea is served at $80 ^ { \circ } \mathrm { C }$ in a room which is kept at a constant $20 ^ { \circ } \mathrm { C }$. The temperature, $T ^ { \circ } \mathrm { C }$, of the tea after $t$ minutes can be modelled by assuming that the rate of change of $T$ is proportional to the difference in temperature between the tea and the room.\\
(i) Explain why the rate of change of the temperature in this model is given by $\frac { \mathrm { d } T } { \mathrm {~d} t } = - k ( T - 20 )$, where $k$ is a positive constant.\\
(ii) Show by integration that the temperature of the tea after $t$ minutes is given by $T = 20 + 60 \mathrm { e } ^ { - k t }$.\\
(iii) After 2 minutes the tea has cooled to $60 ^ { \circ } \mathrm { C }$. Find the value of $k$.
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2015 Q6 [11]}}