C3 June 2015 Q4
4. Water is being heated in an electric kettle. The temperature, \(\theta\,{}^\circ\mathrm{C}\), of the water \(t\) seconds after the kettle is switched on, is modelled by the equation\[\theta = 120 - 100\mathrm{e}^{-\lambda t}, \qquad 0 \leqslant t \leqslant T\]
Given that the temperature of the water in the kettle is \(70\,{}^\circ\mathrm{C}\) when \(t = 40\),
When \(t = T\), the temperature of the water reaches \(100\,{}^\circ\mathrm{C}\) and the kettle switches off.
| Scheme | Marks |
|---|---|
| \((\theta =)20\) | B1 |
| (1) |
Notes
B1: Sight of \((\theta =)20\)
| Scheme | Marks |
|---|---|
| Sub \(t = 40, \theta = 70 \Rightarrow 70 = 120 - 100\mathrm{e}^{-40\lambda}\) | |
| \(\Rightarrow \mathrm{e}^{-40\lambda} = 0.5\) | M1A1 |
| \(\Rightarrow \lambda = \dfrac{\ln 2}{40}\) | M1A1 |
| (4) |
Notes
M1: Sub \(t = 40, \theta = 70 \Rightarrow 70 = 120 - 100\mathrm{e}^{-40\lambda}\) and proceed to \(\mathrm{e}^{\pm 40\lambda} = A\) where \(A\) is a constant. Allow sign slips and copying errors.
A1: \(\mathrm{e}^{-40\lambda} = 0.5\), or \(\mathrm{e}^{40\lambda} = 2\) or exact equivalent
M1: For undoing the e's by taking ln's and proceeding to \(\lambda = ..\)
May be implied by the correct decimal answer awrt 0.017 or \(\lambda = \dfrac{\ln 0.5}{-40}\)
A1: cso \(\lambda = \dfrac{\ln 2}{40}\)
Accept equivalents in the form \(\dfrac{\ln a}{b}\), \(a, b \in \mathbb{Z}\) such as \(\lambda = \dfrac{\ln 4}{80}\)
Alt (b)
| Scheme | Marks |
|---|---|
| Sub \(t = 40, \theta = 70 \Rightarrow 100\mathrm{e}^{-40\lambda} = 50\) | |
| \(\Rightarrow \ln 100 - 40\lambda = \ln 50\) | M1A1 |
| \(\Rightarrow \lambda = \dfrac{\ln 100 - \ln 50}{40} = \dfrac{\ln 2}{40}\) | M1A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(\theta = 100 \Rightarrow T = \dfrac{\ln 0.2}{-\text{their '}\lambda\text{'}}\) | M1 |
| \(T = \text{awrt } 93\) | A1 |
| (2) | |
| (7 marks) |
Notes
M1: Substitutes \(\theta = 100\) and their numerical value of \(\lambda\) into \(\theta = 120 - 100\mathrm{e}^{-\lambda t}\) and proceed to \(T = \pm\dfrac{\ln 0.2}{\text{their '}\lambda\text{'}}\) or \(T = \pm\dfrac{\ln 5}{\text{their '}\lambda\text{'}}\) Allow inequalities here.
A1: awrt \(T = 93\)
Watch for candidates who lose the minus sign in (b) and use \(\lambda = \dfrac{\ln\frac{1}{2}}{40}\) in (c). Many then reach \(T = -93\) and ignore the minus. This is M1 A0