13. What fraction of ice is submerged when it floats in freshwater, given the density of water and ice at $$ 0^{\circ} \mathrm{C} $$ respectively are very close to 1000 and $$ 917 \mathrm{~kg} / \mathrm{m}^{3} $$ ? 14. A rock with a mass of 540 g in air is found to have an apparent mass of 342 g when submerged in water. (a) What mass of water is displaced? (b) What is the volume of the rock? (c) What is its average density?

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UpStudy AI · Accepted Answer

To solve these problems, we will use the principles of buoyancy and density.

### Problem 13: Fraction of Ice Submerged

When ice floats in water, the fraction of the ice submerged can be determined using the principle of buoyancy, which states that the weight of the water displaced by the ice is equal to the weight of the ice itself.

Let:
- $$ \rho_{\text{ice}} = 917 \, \text{kg/m}^3 $$ (density of ice)
- $$ \rho_{\text{water}} = 1000 \, \text{kg/m}^3 $$ (density of water)

Let $$ V $$ be the total volume of the ice, and $$ V_s $$ be the volume of the ice that is submerged. The weight of the ice is given by:

$$
\text{Weight of ice} = \rho_{\text{ice}} \cdot V \cdot g
$$

The weight of the water displaced by the submerged part of the ice is:

$$
\text{Weight of displaced water} = \rho_{\text{water}} \cdot V_s \cdot g
$$

According to Archimedes' principle, these two weights are equal:

$$
\rho_{\text{ice}} \cdot V \cdot g = \rho_{\text{water}} \cdot V_s \cdot g
$$

The $$ g $$ cancels out, and we can express the submerged volume as a fraction of the total volume:

$$
\rho_{\text{ice}} \cdot V = \rho_{\text{water}} \cdot V_s
$$

Since $$ V_s = \text{fraction submerged} \cdot V $$, we can write:

$$
\rho_{\text{ice}} = \rho_{\text{water}} \cdot \text{fraction submerged}
$$

Thus, the fraction submerged is:

$$
\text{fraction submerged} = \frac{\rho_{\text{ice}}}{\rho_{\text{water}}} = \frac{917}{1000} = 0.917
$$

So, approximately **91.7%** of the ice is submerged when it floats in freshwater.

### Problem 14: Rock in Water

Given:
- Mass of the rock in air, $$ m_{\text{air}} = 540 \, \text{g} $$
- Apparent mass of the rock when submerged in water, $$ m_{\text{apparent}} = 342 \, \text{g} $$

#### (a) Mass of Water Displaced

The mass of water displaced is equal to the loss of weight of the rock when submerged:

$$
\text{Mass of water displaced} = m_{\text{air}} - m_{\text{apparent}} = 540 \, \text{g} - 342 \, \text{g} = 198 \, \text{g}
$$

#### (b) Volume of the Rock

The volume of the rock can be found using the mass of the water displaced, since the volume of water displaced is equal to the volume of the rock:

Using the density of water ($$ \rho_{\text{water}} = 1000 \, \text{kg/m}^3 $$ or $$ 1 \, \text{g/cm}^3 $$):

$$
\text{Volume of rock} = \frac{\text{Mass of water displaced}}{\rho_{\text{water}}} = \frac{198 \, \text{g}}{1 \, \text{g/cm}^3} = 198 \, \text{cm}^3
$$

#### (c) Average Density of the Rock

The average density of the rock can be calculated using its mass and volume:

$$
\text{Density of rock} = \frac{m_{\text{air}}}{\text{Volume of rock}} = \frac{540 \, \text{g}}{198 \, \text{cm}^3} \approx 2.727 \, \text{g/cm}^3
$$

### Summary of Answers:
- **Fraction of ice submerged:** $$ 0.917 $$ (or $$ 91.7\% $$)
- **Mass of water displaced:** $$ 198 \, \text{g} $$
- **Volume of the rock:** $$ 198 \, \text{cm}^3 $$
- **Average density of the rock:** $$ \approx 2.727 \, \text{g/cm}^3 $$
- Fraction of ice submerged: 0.917 (or 91.7%) 	- Mass of water displaced: 198 g 	- Volume of the rock: 198 cm³ 	- Average density of the rock: ≈ 2.727 g/cm³

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