A heated metal block at $160^\circ\text{C}$ is left to cool in a laboratory with an ambient temperature of $30^\circ\text{C}$. The temperature, $T$ in $^\circ\text{C}$, of the metal block $t$ minutes after it is left in the laboratory is modelled as a function $T(t)$. The rate of cooling is proportional to the difference between the object's temperature and the ambient temperature, with a positive constant of proportionality $k$. An experiment to study the cooling of the heated metal block starts at 10:00 am. However, a protective thermal casing is used to delay the cooling of the metal block for 5 minutes, resulting in the temperature of the metal block remaining constant at $160^\circ\text{C}$ during this time. At 10:05 am, the casing is removed and cooling begins immediately. At 10:15 am, the temperature of the metal block is measured to be $100^\circ\text{C}$.

1. Formulate the differential equation based on the given rate of cooling.

Let $T$ be the temperature of the metal block and $t$ be the time in minutes.
The rate of cooling is the negative rate of change of temperature, $-\frac{\mathrm{d}T}{\mathrm{d}t}$.
Since it is proportional to the difference between the object's temperature and the ambient temperature ($30^\circ\text{C}$) with a positive constant $k$:
$-\frac{\mathrm{d}T}{\mathrm{d}t} = k(T - 30)$
$\frac{\mathrm{d}T}{\mathrm{d}t} = -k(T - 30)$

2. Solve the differential equation by separating the variables.

$\int \frac{1}{T - 30} \, \mathrm{d}T = \int -k \, \mathrm{d}t$
$\ln |T - 30| = -kt + C$

3. Apply the initial condition to find the constant of integration and express $T(t)$.

From the first paragraph, the cooling begins at an initial temperature of $160^\circ\text{C}$.
When $t = 0$, $T = 160$:
$\ln (160 - 30) = -k(0) + C$
$C = \ln 130$

Substituting $C$ back into the equation:
$\ln(T - 30) = -kt + \ln 130$
$T - 30 = e^{-kt + \ln 130}$
$T - 30 = 130e^{-kt}$
$T(t) = 30 + 130e^{-kt}$

1. Determine the value of $a$ by applying a horizontal translation to account for the delayed cooling.

Since the cooling is delayed by 5 minutes, it effectively starts at $t = 5$ instead of $t = 0$.
To shift the original cooling function $T(t)$ 5 units to the right along the time axis, we must replace $t$ with $t - 5$.
Comparing $t - 5$ with $t + a$, we obtain:
$a = -5$

1. State the results from parts (a) and (b) and apply the given condition at 10:15 am to form an equation for $k$.

From parts (a) and (b), we have $T(t) = 30 + 130e^{-kt}$ and $a = -5$.
At 10:15 am, $t = 15$ minutes since 10:00 am.
The temperature is $100^\circ\text{C}$, so $T_p(15) = 100$.
Since $15 \ge 5$, we use the second branch of the piecewise function:
$T_p(15) = T(15 - 5) = T(10)$
$30 + 130e^{-10k} = 100$

2. Solve the exponential equation for the exact and approximate value of the cooling constant $k$.

$130e^{-10k} = 70$
$e^{-10k} = \frac{7}{13}$
Taking the natural logarithm on both sides:
$-10k = \ln\left(\frac{7}{13}\right)$
$k = -0.1\ln\left(\frac{7}{13}\right) \approx 0.061904$

3. Substitute $k$ and $a$ into $T(t+a)$ to obtain the required expression.

Substitute $k \approx 0.061904$ and $a = -5$ into $T(t+a)$:
$T(t+a) = T(t - 5) = 30 + 130e^{-0.061904(t - 5)}$
Using exponent laws to expand:
$T(t+a) = 30 + 130e^{-0.061904t} \cdot e^{0.061904 \times 5}$
$T(t+a) = 30 + 130e^{0.30952} e^{-0.061904t}$
$T(t+a) = 30 + 130(1.3628) e^{-0.061904t}$
$T(t+a) = 30 + 177.16 e^{-0.061904t}$
Rounding the constants to appropriate significant figures, we get:
$T(t+a) \approx 30 + 177e^{-0.0619t} \text{ (shown)}$

1. Substitute $T_p(t) = 60$ into the given equation for $t \ge 5$ and simplify.

$30 + 177e^{-0.0619t} = 60$
$177e^{-0.0619t} = 30$
$e^{-0.0619t} = \frac{30}{177} = \frac{10}{59}$

2. Use natural logarithms to solve for $t$.

$-0.0619t = \ln\left(\frac{10}{59}\right)$
$t = \frac{\ln\left(\frac{10}{59}\right)}{-0.0619}$
$t \approx 28.675$

1. Determine the shape and key points of the graph for $0 \le t < 5$.

For $0 \le t < 5$, the function is defined as $T_p(t) = 160$.
This represents a constant temperature. On the graph, it is a horizontal line segment starting from the y-intercept at $(0, 160)$ and ending at $(5, 160)$.

2. Determine the shape and asymptote of the graph for $t \ge 5$.

For $t \ge 5$, the function is $T_p(t) = 30 + 177e^{-0.0619t}$.
At $t = 5$, $T_p(5) \approx 30 + 177(0.7338) \approx 160$, ensuring the graph is continuous at $t = 5$.
As $t \to \infty$, the exponential term $e^{-0.0619t} \to 0$, which means $T_p(t) \to 30$.
This signifies an exponential decay curve that strictly decreases from $(5, 160)$ and approaches a horizontal asymptote at $T_p = 30$.

3. Combine the features into a comprehensive sketch.

Draw the axes $T_p(t)$ and $t$.
Plot the horizontal segment from $(0, 160)$ to $(5, 160)$.
From $(5, 160)$, draw a smooth decreasing curve that asymptotically flattens out along the dashed line $T_p = 30$.
(Marking the point $(15, 100)$ provides additional clarity to the scale of the cooling curve.)