To solve this problem, we apply the principle of buoyancy and Archimedes' principle, which states that the buoyant force on a submerged object is equal to the weight of the displaced fluid. Let's proceed step by step:

---

Step 1: Relationship between submerged and total volumes

Let:

• $h$ = total height of the ice block (in cm),
• $h_{\text{above}} = 25 \, \text{cm}$ (height of ice above the surface),
• $h_{\text{below}}$ = $h - h_{\text{above}}$ (height of ice below the surface).

The fraction of the ice block submerged ($\frac{h_{\text{below}}}{h}$) equals the ratio of the specific gravity of ice to that of seawater because the buoyant force equals the weight of the ice.

$\frac{h_{\text{below}}}{h} = \frac{\text{SG}_{\text{ice}}}{\text{SG}_{\text{seawater}}}$

Given: $\text{SG}_{\text{ice}} = 0.92, \quad \text{SG}_{\text{seawater}} = 1.025$

Substitute: $\frac{h_{\text{below}}}{h} = \frac{0.92}{1.025}$

---

Step 2: Express $h_{\text{below}}$ in terms of $h$

Rewriting the above equation: $h_{\text{below}} = h \cdot \frac{0.92}{1.025}$

---

Step 3: Express $h$ in terms of $h_{\text{above}}$ and solve

The total height of the block $h$ is the sum of the submerged height and the height above the surface: $h = h_{\text{below}} + h_{\text{above}}$

Substitute $h_{\text{below}} = h \cdot \frac{0.92}{1.025}$: $h = h \cdot \frac{0.92}{1.025} + h_{\text{above}}$

Rearrange to solve for $h$: $h - h \cdot \frac{0.92}{1.025} = h_{\text{above}}$

Factor out $h$: $h \left( 1 - \frac{0.92}{1.025} \right) = h_{\text{above}}$

Simplify: $h \cdot \frac{1.025 - 0.92}{1.025} = h_{\text{above}}$

$h = \frac{h_{\text{above}} \cdot 1.025}{1.025 - 0.92}$

Substitute $h_{\text{above}} = 25 \, \text{cm}$: $h = \frac{25 \cdot 1.025}{1.025 - 0.92}$

$h = \frac{25 \cdot 1.025}{0.105}$

$h \approx 243.81 \, \text{cm}$

---

Step 4: Calculate $h_{\text{below}}$

$h_{\text{below}} = h - h_{\text{above}}$

$h_{\text{below}} = 243.81 - 25 \approx 218.81 \, \text{cm}$

---

Final Answer:

The height of the ice block below the surface is approximately 218.81 cm.

---

Would you like further details or have any questions?

Here are 5 related questions for deeper understanding:

1. How does Archimedes' principle apply to different shapes, not just cubes?
2. What happens if the ice block is in freshwater instead of seawater?
3. How would you calculate the submerged height for an irregular ice shape?
4. How does temperature affect the specific gravity of seawater and ice?
5. What if the ice block were hollow? How would the calculation change?

Tip: Always remember that specific gravity is the ratio of density of the material to that of water, making it a dimensionless quantity helpful in buoyancy problems.