AQA Paper 3 2019 June — Question 8 12 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
Year2019
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
TypeFinding x from given y value
DifficultyStandard +0.3 This is a standard exponential decay modelling question requiring substitution to find constants, evaluation at a given time, and interpretation of limiting behaviour. The multi-step nature and 12 marks suggest moderate length, but all techniques (substituting initial conditions, solving for constants, finding limits) are routine A-level procedures with no novel insight required. Slightly easier than average due to straightforward structure.
Spec1.02z Models in context: use functions in modelling1.06a Exponential function: a^x and e^x graphs and properties1.06i Exponential growth/decay: in modelling context
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, \(\theta °C\), of the liquid at time \(t\) minutes can be modelled by the equation $$\theta = 5(4 + \lambda e^{-kt})$$ where \(\lambda\) and \(k\) are constants. After 2 minutes the temperature falls to \(68°C\).
  1. Find the temperature of the liquid after 15 minutes. Give your answer to three significant figures. [7 marks]
    1. Find the room temperature of the laboratory, giving a reason for your answer. [2 marks]
    2. Find the time taken in minutes for the liquid to cool to \(1°C\) above the room temperature of the laboratory. [2 marks]
  3. Explain why the model might need to be changed if the experiment was conducted in a different place. [1 mark]
Question 8:
AnswerMarks Guidance
8(a)Uses model with t = 0 and
θ = 75 to form an equation3.4 M1
75=5 4+λe0
λ=11
68=5 ( 4+11e−2k )
k =0.068066
θ=5 ( 4+11e−0.068066×15 )
=39.8C
AnswerMarks Guidance
Obtains correct λ1.1b A1
Uses model with t = 2, θ = 68
AnswerMarks Guidance
and their λ to form an equation3.4 M1
Solves their equation correctly to
AnswerMarks Guidance
find k1.1a M1
Obtains correct k
AWRT 0.07
AnswerMarks Guidance
OE1.1b A1
Uses model with their λand their
AnswerMarks Guidance
k and t = 153.4 M1
Obtains correct temperature
Condone missing units
AnswerMarks Guidance
AWRT 39.81.1b A1
(b)(i) ---
8
AnswerMarks
(b)(i)States correct room temperature
Condone missing units
AnswerMarks Guidance
CAO3.4 B1
As t gets large the temperature
predicted by the model will get
close to room temperature
Explains that the temperature
predicted by the model will
approach room temperature as t
increases.
AnswerMarks Guidance
OE2.4 E1
(b)(ii) ---
8
AnswerMarks
(b)(ii)Uses the model with their k and
their room temperature+1 to form
AnswerMarks Guidance
equation for t3.4 M1
t =58.87
Obtains the correct value of t
AWRT 59
AnswerMarks Guidance
ISW1.1b A1
AnswerMarks
8(c)Room temperature
change/higher/lower
Cooling rate change/higher/lower
or identifies a factor that may be
AnswerMarks Guidance
different in a different place.3.5a E1
change
AnswerMarks Guidance
Total12
QMarking instructions AO 
Question 8:
--- 8(a) ---
8(a) | Uses model with t = 0 and
θ = 75 to form an equation | 3.4 | M1 | ( )
75=5 4+λe0
λ=11
68=5 ( 4+11e−2k )
k =0.068066
θ=5 ( 4+11e−0.068066×15 )
=39.8C
Obtains correct λ | 1.1b | A1
Uses model with t = 2, θ = 68
and their λ to form an equation | 3.4 | M1
Solves their equation correctly to
find k | 1.1a | M1
Obtains correct k
AWRT 0.07
OE | 1.1b | A1
Uses model with their λand their
k and t = 15 | 3.4 | M1
Obtains correct temperature
Condone missing units
AWRT 39.8 | 1.1b | A1
--- 8
(b)(i) ---
8
(b)(i) | States correct room temperature
Condone missing units
CAO | 3.4 | B1 | 20C
As t gets large the temperature
predicted by the model will get
close to room temperature
Explains that the temperature
predicted by the model will
approach room temperature as t
increases.
OE | 2.4 | E1
--- 8
(b)(ii) ---
8
(b)(ii) | Uses the model with their k and
their room temperature+1 to form
equation for t | 3.4 | M1 | 5 ( 4+11e−0.068066t ) =21
t =58.87
Obtains the correct value of t
AWRT 59
ISW | 1.1b | A1
--- 8(c) ---
8(c) | Room temperature
change/higher/lower
Cooling rate change/higher/lower
or identifies a factor that may be
different in a different place. | 3.5a | E1 | The new room temperature might
change
Total | 12
Q | Marking instructions | AO | Mark | Typical solution
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, $\theta °C$, of the liquid at time $t$ minutes can be modelled by the equation

$$\theta = 5(4 + \lambda e^{-kt})$$

where $\lambda$ and $k$ are constants.

After 2 minutes the temperature falls to $68°C$.

\begin{enumerate}[label=(\alph*)]
\item Find the temperature of the liquid after 15 minutes.

Give your answer to three significant figures. [7 marks]

\item \begin{enumerate}[label=(\roman*)]
\item Find the room temperature of the laboratory, giving a reason for your answer. [2 marks]

\item Find the time taken in minutes for the liquid to cool to $1°C$ above the room temperature of the laboratory. [2 marks]
\end{enumerate}

\item Explain why the model might need to be changed if the experiment was conducted in a different place. [1 mark]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 3 2019 Q8 [12]}}
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