A freshly brewed cup of tea, initially at a temperature of $85^\circ\text{C}$, is left out to cool in a room with a room temperature of $25^\circ\text{C}$. After 20 minutes, the temperature of the tea is $55^\circ\text{C}$. It is known that the rate of change of the temperature of the tea is proportional to the difference between the temperature of the tea and the room temperature.

1. Formulate a differential equation representing the rate of change of the tea's temperature.

Let the temperature of the tea at time $t$ be $\theta$.
Given that the rate of change of the temperature is proportional to the difference between the tea temperature and the room temperature ($25^\circ\text{C}$):
$\frac{\text{d}\theta}{\text{d}t} \propto (\theta - 25)$
$\frac{\text{d}\theta}{\text{d}t} = -k(\theta - 25)$
where $k$ is a positive constant of proportionality (we use $-k$ because the tea is cooling, so the rate of change is negative).

2. Separate variables and integrate to find the general solution.

Separating the variables $\theta$ and $t$:
$\int \frac{1}{\theta - 25} \, \text{d}\theta = \int -k \, \text{d}t$
$\ln |\theta - 25| = -kt + C$
Since the initial temperature is $85^\circ\text{C}$ and it cools down, $\theta > 25$, so $|\theta - 25| = \theta - 25$.
$\theta - 25 = \text{e}^{-kt + C}$
$\theta = 25 + A\text{e}^{-kt}$
where $A = \text{e}^C$ is an arbitrary constant.

3. Use the initial condition to determine the constant $A$ and show the given equation.

At $t = 0$, the initial temperature $\theta = 85$.
$85 = 25 + A\text{e}^{0}$
$85 = 25 + A$
$A = 60$
Substituting $A = 60$ into the general solution:
$\theta = 25 + 60\text{e}^{-kt} \quad \text{(shown)}$

4. Substitute the condition at $t = 20$ to set up an equation for $k$.

We are given that when $t = 20$, $\theta = 55$.
$55 = 25 + 60\text{e}^{-20k}$
$30 = 60\text{e}^{-20k}$
$\text{e}^{-20k} = \frac{30}{60} = 0.5$

5. Solve the exponential equation to determine the exact and approximate value of $k$.

Taking the natural logarithm on both sides:
$-20k = \ln(0.5)$
$k = \frac{\ln(0.5)}{-20}$
$k = \frac{\ln(2^{-1})}{-20} = \frac{-\ln 2}{-20} = \frac{\ln 2}{20}$
$k \approx 0.034657 \dots \approx 0.0347 \text{ (to 3 s.f.)}$

1. Determine the initial value ($\theta$-intercept) and the horizontal asymptote.

At $t = 0$, the initial temperature is $\theta = 25 + 60\text{e}^{0} = 85$.
Since $k > 0$ (as the tea is cooling), as $t \to \infty$, $\text{e}^{-kt} \to 0$.
Therefore, the long-term temperature is $\lim_{t \to \infty} \theta = 25 + 60(0) = 25$, which gives a horizontal asymptote of $\theta = 25$.

2. Identify the domain and the shape of the graph to be sketched.

The domain is $t \ge 0$ since time cannot be negative.
The term $60\text{e}^{-kt}$ represents exponential decay, so the curve will start from the point $(0, 85)$ and decrease with a flattening gradient to approach the dotted horizontal line $\theta = 25$ from above.