A student is conducting an experiment in a laboratory to investigate how quickly liquids cool to room temperature. A beaker containing a hot liquid at an initial temperature of 75°C cools so that the temperature, $ heta$, of the liquid at time t minutes can be modelled by the equation $$ heta = 5(4 + eta e^{-kt})$$ where $eta$ and $k$ are constants. After 2 minutes the temperature falls to 68°C. Find the temperature of the liquid after 15 minutes. Give your answer to three significant figures.

A-Level AQA Maths Pure Modelling with Functions: A student is conducting an exper

A student is conducting an experiment in a laboratory to investigate how quickly liquids cool to room temperature - AQA - A-Level Maths Pure - Question 8 - 2019 - Paper 3

Question 8

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A student is conducting an experiment in a laboratory to investigate how quickly liquids cool to room temperature. A beaker containing a hot liquid at an initial te... show full transcript

Worked Solution & Example Answer:A student is conducting an experiment in a laboratory to investigate how quickly liquids cool to room temperature - AQA - A-Level Maths Pure - Question 8 - 2019 - Paper 3

Step 1

Find the value of k using the equation with t = 2

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Answer

Substituting the values into the model equation:

$68 = 5(4 + eta e^{-2k})$

This simplifies to:

$68 = 20 + 5eta e^{-2k}$

Thus, we can rearrange to find:

$5eta e^{-2k} = 48$

This leads to:

$eta e^{-2k} = rac{48}{5} = 9.6$

To find the value of $k$ , we will need another equation, which we will obtain using the model again.

Step 2

Using the initial condition to find $eta$

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Answer

Next, using the initial condition when t = 0,

$75 = 5(4 + eta e^{-0})$

Thus:

$75 = 20 + 5eta$

Rearranging gives:

$5eta = 55$

So,

$eta = rac{55}{5} = 11$ .$$

Substituting this back into the equation from step 1:

$9.6 = 11 e^{-2k}$

So,

$e^{-2k} = rac{9.6}{11}$ .

Taking the natural logarithm:

$-2k = ext{ln} rac{9.6}{11}$ ,

yielding:

$k = - rac{1}{2} ext{ln} rac{9.6}{11} \\ k ext{ (to 5 significant figures)} = 0.06806.$

Step 3

Find the temperature after 15 minutes

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Answer

Using the found values of $eta$ and $k$ in the original equation:

$heta = 5(4 + 11e^{-15 imes 0.06806})$

Calculating this gives:

$e^{-15 imes 0.06806} ext{ (calculates to approximately 0.319)}$

So,

$heta = 5(4 + 11 imes 0.319)$

Continuing,

$heta = 5(4 + 3.509) = 5 imes 7.509 = 37.545.$

Rounding to three significant figures gives:

$heta = 38.0 °C.$

A student is conducting an experiment in a laboratory to investigate how quickly liquids cool to room temperature - AQA - A-Level Maths Pure - Question 8 - 2019 - Paper 3

Worked Solution & Example Answer:A student is conducting an experiment in a laboratory to investigate how quickly liquids cool to room temperature - AQA - A-Level Maths Pure - Question 8 - 2019 - Paper 3

Find the value of k using the equation with t = 2

Using the initial condition to find $eta$

Find the temperature after 15 minutes

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Using the initial condition to find $eta$