October 2021 Paper 1 Q12
12 A cake is cooling so that, \(t\) minutes after it is removed from an oven, its temperature is \(\theta\,{}^\circ\mathrm{C}\).
When the cake is removed from the oven, its temperature is \(160\,{}^\circ\mathrm{C}\). After 10 minutes its temperature has fallen to \(125\,{}^\circ\mathrm{C}\).
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = -k\) | B1 | 3.3 |
| [1] | ||
| (ii) \(\theta = -3.5t + c\) | M1 | 3.4 |
| \(\theta = 160 - 3.5t\) | A1 | 1.1 |
| [2] | ||
| (iii) The model would predict that the temperature would fall below room temperature, and eventually below freezing point | B1 | 3.5b |
| [1] |
Notes
(a)(i) B1: Allow \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = k\) or \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = -3.5\)
Both sides of differential equation required
(a)(ii)
M1: Obtain equation of the form \(\theta = \pm 3.5t + c\), where \(c\) could already be numerical and possibly incorrect
Not dependent on correct differential equation in (i)
A1: Obtain correct equation
Alt method
For M1, integrate to get \(\theta = kt + c\), then use \((0, 160)\) and \((10, 125)\) to attempt \(c\) and hence \(k\)
(a)(iii) B1: Any sensible comment
Cooling rate unlikely to be linear
Identify that limit (ie room temperature) will be reached
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = -k(\theta - 20)\) | B1 | 3.3 |
| [1] | ||
| (ii) \(\displaystyle\int \frac{1}{\theta - 20}\,\mathrm{d}\theta = \int -k\,\mathrm{d}t\) | M1 | 3.1a |
| \(\ln|\theta - 20| = -kt + c\) | A1 | 1.1 |
| \(\ln 140 = c\) | M1 | 3.4 |
| \(\ln 105 = -10k + \ln 140\) \(k = -0.1\ln 0.75\) | M1 | 1.1a |
| \(\ln|\theta - 20| = (0.1\ln 0.75)t + \ln 140\) \(\theta - 20 = \mathrm{e}^{(0.1\ln 0.75)t + \ln 140} = 140\mathrm{e}^{(0.1\ln 0.75)t}\) | M1 | 1.1 |
| \(\theta = 20 + 140\mathrm{e}^{(0.1\ln 0.75)t}\) | A1 | 1.1 |
| [6] |
Notes
(b)(i) B1: Allow \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = k(\theta - 20)\)
Both sides of differential equation required
ISW if \(k = -3.5\) used once correct equation seen (but B0 if only ever seen with \(-3.5\))
(b)(ii)
M1: Separate variables (or invert each side) and attempt integration
Allow M1 for integration of a differential equation not of this form eg \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = \dfrac{-k}{(\theta - 20)}\), as long as \(t\) and/or \(\theta\) are involved – must be attempt at correct rearrangement of their diff eqn
A1: Obtain correct integral
Or \(\ln|\theta - 20| = kt + c\)
Condone brackets not modulus
M1: Use \(t = 0\), \(\theta = 160\) in an equation involving both \(k\) and \(c\)
Equation must be from integration attempt, but could follow M0
As far as numerical \(c\) or \(k\)
Using both pairs of values as limits in a definite integral is M2
M1: Use \(t = 10\), \(\theta = 125\) in an equation involving both \(k\) and \(c\) (\(c\) possibly now numerical)
As far as numerical \(c\) and \(k\)
M1: Attempt to make \(\theta\) the subject
As far as correctly removing logs
Equation must now be of the correct form ie \(\ln|a\theta + b| = ct + d\)
Could still be in terms of \(c\) and \(k\) to give eg \(\theta = A\mathrm{e}^{kt} + 20\)
A1: Obtain correct equation
Allow \(-0.0288\) (or better) for \(0.1\ln 0.75\) and/or 4.94 (or better) for \(\ln 140\)
Allow \(\theta = 20 + \mathrm{e}^{(0.1\ln 0.75)t + \ln 140}\)
Could see \(\theta = 140(0.75)^{0.1t} + 20\)
| Scheme | Marks | AO |
|---|---|---|
| \(25 = 160 - 3.5t \Rightarrow t = 38.6\) mins \(\ln 5 = (0.1\ln 0.75)t + \ln 140 \Rightarrow t = 115.8\) mins | M1 | 3.4 |
| 77 minutes | A1 | 3.4 |
| [2] |
Notes
M1: Use \(\theta = 25\) in both of their equations to find values for \(t\)
As far as two numerical values for \(t\)
A1: Obtain 77 minutes
Accept any answer rounding to 77, with no errors seen