A metal cube is heated to a temperature of $212^\circ\text{C}$ before being dropped into a liquid.
As the cube cools, its temperature $T^\circ\text{C}$, $t$ minutes after it enters the liquid is given by
$T = P + 180e^{-kt}$, where $P$ and $k$ are constants.
It is recorded that when $t = 5$, $T = 185$.

1. Use the initial condition to find $P$.

When $t = 0$, $T = 212$.
$212 = P + 180e^{-k(0)}$
$212 = P + 180(1)$
$P = 212 - 180 = 32$

2. Use the second condition to form an equation for $k$.

When $t = 5$, $T = 185$.
$185 = 32 + 180e^{-5k}$
$153 = 180e^{-5k}$
$e^{-5k} = \frac{153}{180} = 0.85$

3. Solve for $k$ using logarithms.

$-5k = \ln(0.85)$
$k = -\frac{1}{5} \ln(0.85)$
$k \approx 0.0325037$
$k = 0.0325 \text{ (to 3 s.f.)}$

1. Sketch the graph of $T$ against $t$ using the values from part (a)

From part (a), $P = 32$ and $k > 0$. The equation is $T = 32 + 180e^{-kt}$.
When $t = 0$, $T = 32 + 180 = 212$.
As $t \to \infty$, $e^{-kt} \to 0$, so $T \to 32$.
The graph is an exponential decay curve starting at $(0, 212)$ and approaching the horizontal asymptote $T = 32$.

2. Explain why $T$ cannot be 30 based on the graph

From the sketch, the horizontal asymptote is $T = 32$ and the curve lies entirely above this line for all $t \ge 0$.
Therefore, $T > 32$ for all $t \ge 0$, which means $T$ can never reach 30.