将一片面包放在烤箱中烤至 $154^\circ\text{C}$。随后将其取出并自然冷却，使其在从烤箱中取出后 $t$ 分钟的温度 $T^\circ\text{C}$ 由以下公式给出： $T = 26 + Ae^{-kt}$，其中 $A$ 和 $k$ 是常数， $t$ 是从烤箱中取出后的时间（以分钟为单位）。

1. Substitute the initial condition into the equation

When $t = 0$, $T = 154$.
$154 = 26 + Ae^{-k(0)}$

2. Solve for A

$154 = 26 + A(1)$
$A = 154 - 26$
$A = 128$

1. Interpret the constant term in the exponential decay model

As $t \to \infty$, $e^{-kt} \to 0$, so $T \to 26$. This means the temperature of the bread approaches $26^\circ\text{C}$ over a long time.

1. Substitute the given values and the value of $A$ from part (i) into the equation

From part (i), $A = 128$.
Substitute $t = 2$, $T = 116$ and $A = 128$ into $T = 26 + Ae^{-kt}$:
$116 = 26 + 128e^{-2k}$

2. Solve for $k$ using natural logarithms

$116 - 26 = 128e^{-2k}$
$90 = 128e^{-2k}$
$e^{-2k} = \frac{90}{128} = \frac{45}{64}$
$-2k = \ln\left(\frac{45}{64}\right)$
$k = -\frac{1}{2} \ln\left(\frac{45}{64}\right)$
$k \approx 0.17639 \approx 0.176 \text{ (to 3 s.f.)}$

1. Substitute the known values of $T$, $A$, and $k$ into the model equation

From earlier parts, $A = 128$ and $k = -\frac{1}{2}\ln\left(\frac{45}{64}\right) \approx 0.1761$.
Substitute $T = 45$ into the equation:
$45 = 26 + 128e^{-0.1761t}$

2. Solve the exponential equation for $t$

$19 = 128e^{-0.1761t}$
$e^{-0.1761t} = \frac{19}{128}$
$-0.1761t = \ln\left(\frac{19}{128}\right)$
$t = \frac{\ln(19/128)}{-0.1761} \approx 10.830$