June 2022 Paper 1 Q14
14 Alex places a hot object into iced water and records the temperature \(\theta\,{}^\circ\mathrm{C}\) of the object every minute. The temperature of an object \(t\) minutes after being placed in iced water is modelled by \(\theta = \theta_0\mathrm{e}^{-kt}\) where \(\theta_0\) and \(k\) are constants whose values depend on the characteristics of the object.
The temperature of Alex’s object is \(82\,{}^\circ\mathrm{C}\) when it is placed into the water. After 5 minutes the temperature is \(27\,{}^\circ\mathrm{C}\).
Ben places a different object into iced water at the same time as Alex. The model for Ben’s object is \(\ln\theta = 3.4 - 0.08t\).
- the initial temperature of Ben’s object
- the rate at which Ben’s object is cooling initially.
Find this time and the corresponding temperature. [3]
| Scheme | Marks | AO |
|---|---|---|
| When \(t = 0\), \(82 = \theta_0\mathrm{e}^0\) so \(\theta_0 = 82\) | B1 | 3.3 |
| \(t = 5\), \(27 = \theta_0\mathrm{e}^{-5k}\) | M1 | 3.3 |
| giving \(k = \left[-\dfrac{1}{5}\ln\left(\dfrac{27}{82}\right)\right] = 0.222\) to 3 sf | A1 | 1.1b |
| [3] |
Notes
M1: Forming an equation for \(k\) and attempt to solve
A1: Allow for exact value or evaluated to at least 2 s.f.
| Scheme | Marks | AO |
|---|---|---|
| The model predicts that temperature tends to zero but if the quantity of water is small the water will warm up so it will not cool the object to zero. | E1 | 3.5b |
| [1] |
Notes
E1: Must imply to the model tends to zero and this does not match the real situation.
| Scheme | Marks | AO |
|---|---|---|
| \(\ln\theta = \ln\left(\theta_0\mathrm{e}^{-kt}\right) = \ln\theta_0 + \ln\left(\mathrm{e}^{-kt}\right)\) | M1 | 2.1 |
| \(\ln\theta = \ln 82 - 0.222t = [4.41 - 0.222t]\) | A1 | 2.1 |
| [2] |
Notes
M1: Taking logs and attempting to use laws of logs
Do not award for values of \(a\) and \(b\) obtained directly from the data and the natural log form of the model.
A1: FT their values for \(\theta_0\) and \(k\)
Accept as part of equation or \(a\) and \(b\) clearly stated
| Scheme | Marks | AO |
|---|---|---|
| When \(t = 0\), \(\ln\theta = 3.4\) giving \(\theta = 29.96\) so \(30.0^\circ\) C to 3 sf | B1 | 3.4 |
| \(\theta = 29.96\mathrm{e}^{-0.08t}\) \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = 29.96 \times -0.08\mathrm{e}^{-0.08t}\) | M1 A1 | 3.4 3.4 |
| When \(t = 0\), \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = -2.3968\) [object is cooling by \(2.4^\circ\) per minute] | A1 | 3.4 |
| [4] |
Notes
B1: Accept \(30^\circ\) www Must be evaluated
M1: Attempt to differentiate their exponential expression for \(\theta\)
A1: Any form eg \(\mathrm{e}^{3.4} \times -0.08\mathrm{e}^{-0.08t}\) or \(-0.08\mathrm{e}^{3.4-0.08t}\)
A1: Allow for correct negative value for \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\) or a clear statement that the rate of cooling is \(2.4^\circ\) per minute. Accept \(= -0.08\mathrm{e}^{3.4}\)
Alternative method
| Scheme | Marks |
|---|---|
| When \(t = 0\), \(\ln\theta = 3.4\) giving \(\theta = 29.96\) so \(30.0^\circ\) C to 3 sf | B1 |
| Differentiate \(\ln\theta = 3.4 - 0.08t\) w.r.t \(t\) \(\dfrac{1}{\theta}\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = -0.08\) | M1 |
| \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = -0.08\theta\) | A1 |
| When \(t = 0\), \(\theta = 29.96\) so \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = -2.3968\) object is cooling by \(2.4^\circ\) per minute | A1 |
B1: Accept \(30^\circ\) www
M1: Uses implicit differentiation w.r.t \(t\)
A1: Correct derivative
A1: Allow for correct negative value for \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\) or a clear statement that the rate of cooling is \(2.4^\circ\) per minute
| Scheme | Marks | AO |
|---|---|---|
| Solve simultaneously \(\ln\theta = 3.4 - 0.08t\) \(\ln\theta = \ln 82 - 0.222t\) | M1 | 3.1b |
| gives \(t = 7.089\), \(t = 7.1\) [7 minutes and 5 seconds] | A1 | 3.4 |
| \(\ln\theta = 2.8328\) gives \(\theta = 17^\circ\) C | A1 | 3.4 |
| [3] |
Notes
M1: Attempting to find the intersection of their (c) and the given line
Could be BC
A1: Accept awrt 7.0, 7.1 or 7.2
A1: Must be the value for \(\theta\)
Alternative method
| Scheme | Marks |
|---|---|
| \(82\mathrm{e}^{-0.222t} = 30\mathrm{e}^{-0.08t}\) \(\dfrac{82}{30} = \mathrm{e}^{0.142t}\) | M1 |
| \(t = 7.08\) [7 minutes and 5 seconds] | A1 |
| \(\theta = 17^\circ\) C | A1 |
M1: Equate their expressions for temperature and attempts to solve for \(t\)
A1: Accept awrt 7.0, 7.1 or 7.2
A1: Cao