June 2022 Paper 3 Q6
6 A hot drink is cooling. The temperature of the drink at time \(t\) minutes is \(T\,{}^\circ\mathrm{C}\).
The rate of decrease in temperature of the drink is proportional to \((T - 20)\).
Determine how long it takes for the drink to cool from \(90\,{}^\circ\mathrm{C}\) to \(40\,{}^\circ\mathrm{C}\). [6]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}T}{\mathrm{d}t} = k(T - 20)\) [where \(k\) is a constant] | B2 | 3.3 1.1 |
| [2] |
Notes
B2: May be \(-k\) and/or \((20 - T)\)
B1 if no constant
If additional terms seen B0
| Scheme | Marks | AO |
|---|---|---|
| \(k = +0.07\) OR \(-0.07\) oe | B1 | 1.1 |
| \(\displaystyle\int \frac{1}{(T-20)}\,\mathrm{d}T = \int k\,\mathrm{d}t\) | M1 | 3.1a |
| \(\ln(T - 20) = kt\ [+c]\) | M1 | 1.1 |
| \(\ln 70 = c\) | M1 | 1.1 |
| \(\ln(T - 20) = -0.07t + \ln 70\) \(\ln(20) = -0.07t + \ln 70\) | M1 | 3.4 |
| \(t \approx 17.9\) [minutes] | A1 | 3.2a |
| [6] |
Notes
M1: Appropriate separation of variables for their DE soi by correct integration or answer
Condone missing \(\mathrm{d}t\) and/or \(\mathrm{d}T\)
M1: o.e. For correct integration to include a ln term
Or \(T - 20 = e^{kt\ [+c]}\)
M1: To find \(c\) from their equation using \(t = 0\), \(T = 90\)
Or \(e^c = 70\)
M1: Clear substitution of \(T = 40\) into their equation in \(T\) and \(t\) coming from integration
\(\ln\left(\dfrac{2}{7}\right) = -0.07t\)
A1: 17.8966…
Any accuracy \(\geqslant\) 2sf
17.9 nfww gets 6
If there is any doubt about the rigour of an otherwise correct solution withhold the A1.