Moderate -0.8 This is a straightforward exponential modelling question requiring only routine substitution and basic logarithm manipulation. Part (a) uses t=0 to find A, part (b) is direct substitution, and part (c) involves rearranging and taking logarithms—all standard textbook exercises with no problem-solving insight required.
9. A cup of tea is cooling down in a room.
The temperature of tea, \(\theta ^ { \circ } \mathrm { C }\), at time \(t\) minutes after the tea is made, is modelled by the equation
$$\theta = A + 70 e ^ { - 0.025 t }$$
where \(A\) is a positive constant.
Given that the initial temperature of the tea is \(85 ^ { \circ } \mathrm { C }\)
a. find the value of \(A\).
b. Find the temperature of the tea 20 minutes after it is made.
c. Find how long it will take the tea to cool down to \(43 ^ { \circ } \mathrm { C }\).
(4)
Find value of \(A\) where \(\theta = A + 70e^{-0.025t}\) and initial temperature is \(85°C\)
(1)
B1 Substitutes \(t = 0\), \(\theta = 85\) into equation to obtain value of \(A\)
Answer: \(A = 15\)
Part (b):
Answer
Marks
Guidance
Find temperature 20 minutes after tea is made
(2)
M1 Substitutes \(t = 20\) into given equation with their \(A\) to obtain value for \(\theta\); A1 Answer 57.5°C
Part (c):
Answer
Marks
Guidance
Find how long for tea to cool to \(43°C\)
(4)
M1 Substitutes \(\theta = 43\) into given equation with their \(A\) and proceeds to form \(P = Qe^{-0.025t}\) or \(M = Ne^{0.025t}\); A1 For \(e^{-0.025t} = \frac{28}{70}\) or \(e^{0.025t} = \frac{70}{28}\) or equivalent; M1 Takes ln's correctly to reach \(-0.025t = \ln\alpha\), \(\alpha > 0\); A1 Answer 36.7 minutes; e.g. \(t = \frac{\ln\frac{2}{5}}{-0.025} = 36.7\)
**Part (a):**
Find value of $A$ where $\theta = A + 70e^{-0.025t}$ and initial temperature is $85°C$ | (1) | B1 Substitutes $t = 0$, $\theta = 85$ into equation to obtain value of $A$
Answer: $A = 15$
**Part (b):**
Find temperature 20 minutes after tea is made | (2) | M1 Substitutes $t = 20$ into given equation with their $A$ to obtain value for $\theta$; A1 Answer 57.5°C
**Part (c):**
Find how long for tea to cool to $43°C$ | (4) | M1 Substitutes $\theta = 43$ into given equation with their $A$ and proceeds to form $P = Qe^{-0.025t}$ or $M = Ne^{0.025t}$; A1 For $e^{-0.025t} = \frac{28}{70}$ or $e^{0.025t} = \frac{70}{28}$ or equivalent; M1 Takes ln's correctly to reach $-0.025t = \ln\alpha$, $\alpha > 0$; A1 Answer 36.7 minutes; e.g. $t = \frac{\ln\frac{2}{5}}{-0.025} = 36.7$
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9. A cup of tea is cooling down in a room.
The temperature of tea, $\theta ^ { \circ } \mathrm { C }$, at time $t$ minutes after the tea is made, is modelled by the equation
$$\theta = A + 70 e ^ { - 0.025 t }$$
where $A$ is a positive constant.\\
Given that the initial temperature of the tea is $85 ^ { \circ } \mathrm { C }$\\
a. find the value of $A$.\\
b. Find the temperature of the tea 20 minutes after it is made.\\
c. Find how long it will take the tea to cool down to $43 ^ { \circ } \mathrm { C }$.\\
(4)\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q9 [7]}}