As a pot of tea cools, the temperature of the tea is modeled by a differentiable function $H$ for $0 \leq t \leq 10$, where time $t$ is measured in minutes and temperature $H(t)$ is measured in degrees Celsius. Values of $H(t)$ at selected values of time $t$ are shown in the table below.

\begin{center}
\begin{tabular}{ | c | | c | c | c | c | c | }
\hline
\begin{tabular}{ c }
$t$ \\
(minutes) \\
\end{tabular} & 0 & 2 & 5 & 9 & 10 \\
\hline
\begin{tabular}{ c }
$H(t)$ \\
(degrees Celsius) \\
\end{tabular} & 66 & 60 & 52 & 44 & 43 \\
\hline
\end{tabular}
\end{center}

(a) Use the data in the table to approximate the rate at which the temperature of the tea is changing at time $t = 3.5$. Show the computations that lead to your answer.

(b) Using correct units, explain the meaning of $\frac{1}{10}\int_{0}^{10} H(t)\,dt$ in the context of this problem. Use a trapezoidal sum with the four subintervals indicated by the table to estimate $\frac{1}{10}\int_{0}^{10} H(t)\,dt$.

(c) Evaluate $\int_{0}^{10} H'(t)\,dt$. Using correct units, explain the meaning of the expression in the context of this problem.

(d) At time $t = 0$, biscuits with temperature $100^{\circ}\mathrm{C}$ were removed from an oven. The temperature of the biscuits at time $t$ is modeled by a differentiable function $B$ for which it is known that $B'(t) = -13.84e^{-0.173t}$. Using the given models, at time $t = 10$, how much cooler are the biscuits than the tea?