## Question 14(a):

When $t = 0$: $82 = \theta_0 e^0$ so $\theta_0 = 82$ | B1 | **3.3** |

$t = 5$: $27 = \theta_0 e^{-5k}$ | M1 | **3.3** | Forming an equation for $k$ and attempt to solve

$k = \left[-\frac{1}{5}\ln\left(\frac{27}{82}\right)\right] = 0.222$ to 3 sf | A1 | **1.1b** | Allow for exact value or evaluated to at least 2 s.f.

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## Question 14(b):

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** | Must imply the model tends to zero and this does not match the real situation.

**[1]**

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## Question 14(c):

$\ln\theta = \ln\left(\theta_0 e^{-kt}\right) = \ln\theta_0 + \ln\left(e^{-kt}\right)$ | M1 | **2.1** | 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.

$\ln\theta = \ln 82 - 0.222t = [4.41 - 0.222t]$ | A1 | **2.1** | FT their values for $\theta_0$ and $k$. Accept as part of equation or $a$ and $b$ clearly stated

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## Question 14(d):

When $t = 0$, $\ln\theta = 3.4$ giving $\theta = 29.96$ so $30.0°$ C to 3 sf | B1 | **3.4** | Accept $30°$ www. Must be evaluated

$\theta = 29.96e^{-0.08t}$

$\frac{d\theta}{dt} = 29.96 \times -0.08e^{-0.08t}$ | M1 | **3.4** | Attempt to differentiate their exponential expression for $\theta$

| A1 | **3.4** | Any form e.g. $e^{3.4} \times -0.08e^{-0.08t}$ or $-0.08e^{3.4-0.08t}$

When $t = 0$: $\frac{d\theta}{dt} = -2.3968$ [object is cooling by $2.4°$ per minute] | A1 | **3.4** | Allow for correct negative value for $\frac{d\theta}{dt}$ or a clear statement that the rate of cooling is $2.4°$ per minute. Accept $= -0.08e^{3.4}$

**Alternative method:**

When $t = 0$, $\ln\theta = 3.4$ giving $\theta = 29.96$ so $30.0°$ C to 3 sf | B1 | **3.4** | Accept $30°$ www

Differentiate $\ln\theta = 3.4 - 0.08t$ w.r.t $t$:

$\frac{1}{\theta}\frac{d\theta}{dt} = -0.08$ | M1 | | Uses implicit differentiation w.r.t $t$

$\frac{d\theta}{dt} = -0.08\theta$ | A1 | | Correct derivative

When $t = 0$, $\theta = 29.96$: $\frac{d\theta}{dt} = -2.3968$, object is cooling by $2.4°$ per minute | A1 | | Allow for correct negative value for $\frac{d\theta}{dt}$ or a clear statement that the rate of cooling is $2.4°$ per minute

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## Question 14(e):

Solve simultaneously $\ln\theta = 3.4 - 0.08t$ and $\ln\theta = \ln 82 - 0.222t$ | M1 | **3.1b** | Attempting to find the intersection of their (c) and the given line. Could be BC

Gives $t = 7.089$, $t = 7.1$ [7 minutes and 5 seconds] | A1 | **3.4** | Accept awrt 7.0, 7.1 or 7.2

$\ln\theta = 2.8328$ gives $\theta = 17°$ C | A1 | **3.4** | Must be the value for $\theta$

**Alternative method:**

$82e^{-0.222t} = 30e^{-0.08t}$

$\frac{82}{30} = e^{0.142t}$ | M1 | | Equate their expressions for temperature and attempts to solve for $t$

$t = 7.08$ [7 minutes and 5 seconds] | A1 | | Accept awrt 7.0, 7.1 or 7.2

$\theta = 17°$ C | A1 | | Cao

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