One. (20 points) Place a rectangular iron block of size $a \mathrm{~cm} \times b \mathrm{~cm} \times c \mathrm{~cm}$ (where $a<b<c$) (as shown in Figure 3(a)) into a rectangular water tank (as shown in Figure 3(b)). The iron block does not touch the four walls of the water tank. Now, water is uniformly injected into the water tank until it is full. Because the iron block can be placed in three different ways in the water tank, the relationship between the water depth $h(\mathrm{~cm})$ and the injection time $t(\mathrm{~s})$ is represented by three different graphs (as shown in Figures 3(c), (d), and (e)) (the injection speed is the same for all three cases).

(1) Fill in the blanks based on the images:
(i) The depth of the water tank is $\qquad$ $\mathrm{cm}, a=$ $\qquad$ ,$b=$ ; $\qquad$
(ii) The relationship between $t_{1}$ and $t_{2}$ is $t_{1} \ldots t_{2}$.

(2) Find the bottom area of the water tank, the injection speed, and the values of $c$, $t_{1}$, and $t_{2}$.
(i) From the three images, the maximum value of $h$ is 10, so the depth of the water tank is $10 \, \text{cm}$.

From Figure 3(c), after $21 \, \text{s}$ of water injection, the water depth is $6 \, \text{cm}$. At this point, the water surface is at the same level as the top surface of the iron block or just covers the top surface of the iron block, so one of $a, b, c$ is 6.

From Figure 3(d), after $45 \, \text{s}$ of water injection, the water depth is $9 \, \text{cm}$. At this point, the water surface is at the same level as the top surface of the iron block or just covers the top surface of the iron block, so one of $a, b, c$ is 9.

From Figure 3(e), after $62 \, \text{s}$ (when the tank is full), the water depth is $10 \, \text{cm}$. At this point, the water has not submerged the iron block, so one of $a, b, c$ is not less than 10.
Since $a < b < c$, we have $a = 6, b = 9, c \geqslant 10$.
(ii) From Figures 3(c) and (d), the water level submerges the iron block before the tank is full, so the amount of water injected in these two cases is the same, which is the difference between the tank's volume and the iron block's volume; since the water injection speed is the same, the injection time for both cases is the same.
(2) Let the bottom area of the tank be $S \, \text{cm}^2$, and the water injection speed be $V \, \text{cm}^3/\text{s}$.
In Figure 3(c), when $t = 21 \, \text{s}$, we have
$$
6S - 6 \times 9 \times c = 21V,
$$
which simplifies to $6S - 54c = 21V$.
In Figure 3(d), when $t = 45 \, \text{s}$, we have
$$
9S - 6 \times 9 \times c = 45V,
$$
which simplifies to $9S - 54c = 45V$.
In Figure 3(e), when $t = 62 \, \text{s}$, we have
$$
10S - 6 \times 9 \times 10 = 62V,
$$
which simplifies to $10S - 540 = 62V$.
Solving the system of equations (1), (2), and (3), we get
$$
V = 30, S = 240, c = 15.
$$

In Figure 3(c), when $t = t_1$, we have
$$
10S - 6 \times 9 \times c = t_1 V.
$$
Solving for $t_1$, we get $t_1 = 53 \, \text{s}$. Since $t_1 = t_2$, we have $t_2 = 53 \, \text{s}$.
Therefore, the bottom area of the tank is $240 \, \text{cm}^2$, the water injection speed is $30 \, \text{cm}^3/\text{s}$, $c$ is $15$, $t_1$ is $53 \, \text{s}$, and $t_2$ is $53 \, \text{s}$.