一杯热巧克力牛奶冷却至室温时的温度 $\theta^\circ C$ 可以用公式 $\theta = 27 + q b^{-t}$ 来建模，其中 $q$ 和 $b$ 为常数，$t$ 是冷却时间（单位：分钟）。$t$ 和 $\theta$ 的测量值如下表所示。

1. Rearrange the given equation to isolate the exponential term and take logarithms on both sides.

$\theta = 27 + q b^{-t}$
$\theta - 27 = q b^{-t}$
$\lg(\theta - 27) = \lg(q b^{-t})$
$\lg(\theta - 27) = \lg q - t \lg b$

2. Explain how this represents a straight line graph.

The equation $\lg(\theta - 27) = (-\lg b)t + \lg q$ is in the linear form $Y = mX + c$.
By plotting $Y = \lg(\theta - 27)$ against $X = t$, a straight line graph can be drawn with gradient $m = -\lg b$ and $Y$-intercept $c = \lg q$.

3. Calculate the values of $\lg(\theta - 27)$ for the given data points.

4. Draw the straight line graph.

Plot the points $(t, \lg(\theta - 27))$ on a grid and draw the line of best fit passing through them.

1. Relate the constants $q$ and $b$ to the features of the linear graph

From part (a), the linear form is $\lg(\theta - 27) = -t \lg b + \lg q$.
The $Y$-intercept is $\lg q$ and the gradient is $-\lg b$.

2. Estimate $q$ from the $Y$-intercept

From the graph, the $Y$-intercept is approximately $1.78$.
$\lg q = 1.78$
$q = 10^{1.78} \approx 60.3 \approx 60$

3. Estimate $b$ from the gradient

Using points $(2, 1.64)$ and $(10, 1.04)$ from the line of best fit:
Gradient $m = \frac{1.04 - 1.64}{10 - 2} = \frac{-0.60}{8} = -0.075$
$-\lg b = -0.075$
$\lg b = 0.075$
$b = 10^{0.075} \approx 1.188 \approx 1.2$

1. Read the value of the vertical axis variable from the graph at $t = 7$

From the graph of $\lg(\theta - 27)$ against $t$, when $t = 7$,
$\lg(\theta - 27) \approx 1.26$

2. Solve for the original variable $\theta$

$\theta - 27 = 10^{1.26}$
$\theta - 27 \approx 18.2$
$\theta \approx 45.2$

1. Identify the behaviour of the cooling term as time becomes very large

As $t \to \infty$, the term $b^{-t} \to 0$ (since $b > 1$ for cooling).

2. Determine the room temperature from the limit

Therefore, $\theta \to 27 + q(0) = 27$. The room temperature is $27^\circ\text{C}$.