# Question 12:

## Part (a)(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{d\theta}{dt} = -k$ | B1 | Allow $\frac{d\theta}{dt} = k$ or $\frac{d\theta}{dt} = -3.5$. Both sides of differential equation required |

**[1]**

## Part (a)(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\theta = -3.5t + c$ | M1 | Obtain equation of the form $\theta = \pm3.5t + c$, where $c$ could already be numerical and possibly incorrect. Not dependent on correct differential equation in (i) |
| $\theta = 160 - 3.5t$ | 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 $k$ |

**[2]**

## Part (a)(iii):

| Answer | Mark | Guidance |
|--------|------|----------|
| The model would predict that the temperature would fall below room temperature, and eventually below freezing point | B1 | Any sensible comment. Cooling rate unlikely to be linear. Identify that limit (ie room temperature) will be reached |

**[1]**

## Part (b)(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{d\theta}{dt} = -k(\theta - 20)$ | B1 | Allow $\frac{d\theta}{dt} = k(\theta-20)$. Both sides of differential equation required. ISW if $k = -3.5$ used once correct equation seen (but B0 if only ever seen with $-3.5$) |

**[1]**

## Part (b)(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\int \frac{1}{\theta-20}\,d\theta = \int -k\,dt$ | M1 | Separate variables (or invert each side) and attempt integration. Allow M1 for integration of a differential equation not of this form eg $\frac{d\theta}{dt} = \frac{-k}{(\theta-20)}$, as long as $t$ and/or $\theta$ are involved – must be attempt at correct rearrangement of their diff eqn |
| $\ln|\theta - 20| = -kt + c$ | A1 | Obtain correct integral. Or $\ln|\theta-20| = kt + c$. Condone brackets not modulus |
| $\ln 140 = c$ | 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 |
| $\ln 105 = -10k + \ln 140$, $k = -0.1\ln 0.75$ | M1 | Use $t=10$, $\theta=125$ in an equation involving both $k$ and $c$ ($c$ possibly now numerical). As far as numerical $c$ and $k$ |
| $\ln|\theta-20| = (0.1\ln 0.75)t + \ln 140$ | 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 $\theta = Ae^{kt}+20$ |
| $\theta = 20 + 140e^{(0.1\ln 0.75)t}$ | A1 | Obtain correct equation. Allow $-0.0288$ (or better) for $0.1\ln0.75$ and/or $4.94$ (or better) for $\ln140$. Allow $\theta = 20 + e^{(0.1\ln0.75)t + \ln140}$. Could see $\theta = 140(0.75)^{0.1t}+20$ |

**[6]**

## Part (c):

| Answer | Mark | Guidance |
|--------|------|----------|
| $25 = 160 - 3.5t \Rightarrow t = 38.6$ mins; $\ln 5 = (0.1\ln0.75)t + \ln140 \Rightarrow t = 115.8$ mins | M1 | Use $\theta = 25$ in both of their equations to find values for $t$. As far as two numerical values for $t$ |
| $77$ minutes | A1 | Obtain 77 minutes. Accept any answer rounding to 77, with no errors seen |

**[2]**