A heated metal block at $160^\circ$C is left to cool in a laboratory with an ambient temperature of $30^\circ$C. The temperature, $T$ in $^\circ$C, of the metal block $t$ minutes after it is left in the laboratory is model 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 $160^\circ$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 $100^\circ$C.

1. Formulate the differential equation using Newton's Law of Cooling.

The rate of cooling is proportional to the difference between the object's temperature ($T$) and the ambient temperature ($30^\circ\text{C}$).
$\frac{\mathrm{d}T}{\mathrm{d}t} = -k(T - 30)$

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

$\begin{aligned} \int \frac{1}{T-30} \, \mathrm{d}T &= \int -k \, \mathrm{d}t \\ \ln(T-30) &= -kt + C \\ T - 30 &= \mathrm{e}^{-kt+C} \\ T &= 30 + A\mathrm{e}^{-kt} \quad \text{where } A = \mathrm{e}^C \end{aligned}$

3. Use the initial condition to find the particular expression for $T(t)$.

The metal block is initially at $160^\circ\text{C}$, so at $t = 0$, $T = 160$.
$\begin{aligned} 160 &= 30 + A\mathrm{e}^{0} \\ A &= 130 \\ \therefore T(t) &= 30 + 130\mathrm{e}^{-kt} \end{aligned}$

1. State the value of $a$ by determining the required time shift.

The initial model starts cooling at $t = 0$. In the actual experiment, the cooling is delayed and starts 5 minutes later at $t = 5$. This represents a translation of 5 units in the positive $t$-direction.
Therefore, $t$ must be replaced by $t - 5$.
Comparing this with $t + a$, we get:
$a = -5$

1. Concisely derive the expression for $T(t)$ and the value of $a$ from parts (a) and (b).

From part (a), the differential equation for cooling is $\frac{\mathrm{d}T}{\mathrm{d}t} = -k(T - 30)$.
Integrating gives $\ln(T - 30) = -kt + c$, which simplifies to $T(t) = 30 + Ae^{-kt}$.
Since the initial temperature of the un-delayed block is $160^\circ\text{C}$ at $t = 0$, $A = 160 - 30 = 130$, giving $T(t) = 30 + 130e^{-kt}$.

From part (b), cooling begins after a 5-minute delay, so the effective duration of cooling is $t - 5$. Replacing $t$ with $t - 5$ yields $a = -5$.

2. Use the temperature at 10:15 am to find the cooling constant $k$.

At 10:15 am, $t = 15$. The measured temperature is $100^\circ\text{C}$.
Using the second piece of the piecewise function ($t \ge 5$):
$T_p(15) = T(15 - 5) = T(10) = 100$
Substitute $t = 10$ into $T(t)$:
$30 + 130e^{-10k} = 100$
$130e^{-10k} = 70 \implies e^{-10k} = \frac{7}{13}$
Taking the natural logarithm:
$-10k = \ln\left(\frac{7}{13}\right) \implies k = -\frac{1}{10}\ln\left(\frac{7}{13}\right) \approx 0.061904$

3. Substitute $a$ and $k$ into the expression for $T(t+a)$ to show the final result.

Substitute $a = -5$ and $k \approx 0.061904$ into $T(t+a)$:
$T(t - 5) = 30 + 130e^{-0.061904(t - 5)}$
$T(t - 5) = 30 + 130e^{5(0.061904)} e^{-0.061904t}$
$T(t - 5) = 30 + 130(1.3628) e^{-0.061904t}$
$T(t - 5) = 30 + 177.16 e^{-0.061904t}$
Rounding to 3 significant figures, we get:
$T(t - 5) \approx 30 + 177e^{-0.0619t} \text{ (shown)}$

1. Equate the temperature model from part (c) to $60^\circ\text{C}$.

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

2. Solve the exponential equation for $t$.

Taking the natural logarithm on both sides:
$-0.0619t = \ln\left(\frac{10}{59}\right)$
$t = \frac{\ln\left(\frac{10}{59}\right)}{-0.0619}$
$t \approx 28.674$
$t \approx 28.7 \text{ minutes (to 3 s.f.)}$

1. Analyse the behaviour of the graph for $0 \le t < 5$.

From the given piecewise function, $T_p(t) = 160$ for $0 \le t < 5$.

This is represented by a horizontal line segment from $t=0$ to $t=5$ at a height of $160$. The $y$-intercept is $(0, 160)$.

2. Analyse the behaviour of the graph for $t \ge 5$ and identify asymptotes.

For $t \ge 5$, $T_p(t) = 30 + 177e^{-0.0619t}$.

At $t=5$, $T_p(5) \approx 30 + 177e^{-0.0619(5)} \approx 160$, so the graph is continuous and connects at $(5, 160)$.

As $t \to \infty$, the exponential term $e^{-0.0619t} \to 0$.
$\lim_{t \to \infty} T_p(t) = 30 + 0 = 30$

Hence, for $t \ge 5$, the graph is an exponential decay curve that decreases from $(5, 160)$ and approaches the horizontal asymptote $T_p(t) = 30$.