We are given the temperature function:

$T(t) = 85 - 3\sqrt{25 - t} \quad \text{for} \quad 0 \leq t \leq 25.$

We are tasked with solving the following:

Part (a): Find the room's temperature at $t = 0$, $t = 16$, and $t = 25$.

Let's substitute each value of $t$ into the temperature function:

1. When $t = 0$: $T(0) = 85 - 3\sqrt{25 - 0} = 85 - 3\sqrt{25} = 85 - 3(5) = 85 - 15 = 70.$ So, the temperature at $t = 0$ is $70^\circ F$.

2. When $t = 16$: $T(16) = 85 - 3\sqrt{25 - 16} = 85 - 3\sqrt{9} = 85 - 3(3) = 85 - 9 = 76.$ So, the temperature at $t = 16$ is $76^\circ F$.

3. When $t = 25$: $T(25) = 85 - 3\sqrt{25 - 25} = 85 - 3\sqrt{0} = 85 - 0 = 85.$ So, the temperature at $t = 25$ is $85^\circ F$.

Part (b): Find the room's average temperature for $0 \leq t \leq 25$.

The average temperature is found by computing the average value of the function over the interval from $0$ to $25$. The formula for the average value of a continuous function is:

$\text{Average temperature} = \frac{1}{b - a} \int_{a}^{b} T(t) \, dt.$

Here, $a = 0$, $b = 25$, and the temperature function is $T(t) = 85 - 3\sqrt{25 - t}$. Thus, we need to compute:

$\text{Average temperature} = \frac{1}{25 - 0} \int_0^{25} \left( 85 - 3\sqrt{25 - t} \right) dt.$

This integral can be split:

$\int_0^{25} \left( 85 - 3\sqrt{25 - t} \right) dt = \int_0^{25} 85 \, dt - 3 \int_0^{25} \sqrt{25 - t} \, dt.$

Now, we compute each integral:

1. For $\int_0^{25} 85 \, dt$: $\int_0^{25} 85 \, dt = 85 \times 25 = 2125.$

2. For $\int_0^{25} \sqrt{25 - t} \, dt$: We can use a substitution. Let $u = 25 - t$, then $du = -dt$, and when $t = 0$, $u = 25$; when $t = 25$, $u = 0$. So the integral becomes: $\int_0^{25} \sqrt{25 - t} \, dt = \int_{25}^{0} \sqrt{u} (-du) = \int_0^{25} \sqrt{u} \, du.$ The integral of $\sqrt{u}$ is: $\int \sqrt{u} \, du = \frac{2}{3} u^{3/2},$ evaluated from 0 to 25: $\frac{2}{3} (25^{3/2} - 0^{3/2}) = \frac{2}{3} \times 125 = \frac{250}{3}.$

So, the second integral is: $3 \times \frac{250}{3} = 250.$

Thus, the total integral is: $2125 - 250 = 1875.$

Finally, the average temperature is: $\frac{1}{25} \times 1875 = 75.$

So, the average temperature over the interval $0 \leq t \leq 25$ is $75^\circ F$.

Would you like further details or explanations? Here are some related questions you could explore:

1. How does the temperature change with respect to time?
2. What is the derivative of the temperature function, and what does it represent physically?
3. How would the average temperature change if the upper limit was different, say $t = 20$?
4. Can the temperature function be modeled for times beyond $t = 25$?
5. How would the result change if the square root function had a different coefficient?

Tip: When computing integrals involving square roots, substitution can simplify the process significantly, especially for expressions of the form $\sqrt{a - t}$.