Question

the density of ice is 917.what fraction of the volume of a piece of ice will be above the liquid when floating in fresh water

Answer 1

Answer:

$8.3\,\%$ of that piece of ice would be above the freshwater.  Assumptions:

• the density of the ice is $\rho(\text{ice}) = 917\; \rm kg \cdot m^{-3}$, and
• the density of freshwater is $\rho(\text{water}) = 1.00 \times 10^3\; \rm kg \cdot m^{-3}$ .

Explanation:

The volume of that chunk of ice can be split into two halves: volume above water $V(\text{above})$, and volume under water $V(\text{under})$. The mass of the whole chunk of ice would be:

$m(\text{ice}) = \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under}))$.

Let $g$ be the acceleration due to gravity. The gravity on the entire chunk of ice would be

$\begin{aligned}&W(\text{ice}) \\ &= m({\text{ice}}) \cdot g \\ &= \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) \cdot g\end{aligned}$.

On the other hand, the size of buoyant force on an object is equal to the weight of the liquid that it displaces. That is: $F(\text{bouyancy}) = W(\text{water displaced})$.

Recall that $V(\text{above})$ is the volume of the ice above the water, and $V(\text{under})$ is the volume of the ice under the water.

The mass of water displaced would be equal to:

$\begin{aligned}& m(\text{water displaced}) \\ &= \rho(\text{water}) \cdot V(\text{water displaced}) \\ &= \rho(\text{water}) \cdot V(\text{under})\end{aligned}$.

The weight of that much water would be

$\begin{aligned} &W(\text{water displaced}) \\ &= m(\text{water displaced}) \cdot g \\ &= \rho(\text{water}) \cdot V(\text{under}) \cdot g \end{aligned}$.

Apply the equation $F(\text{bouyancy}) = W(\text{water displaced})$. The bouyant force on this chunk of ice would be equal to $\begin{aligned} &W(\text{water displaced}) = \rho(\text{water}) \cdot V(\text{under}) \cdot g \end{aligned}$.

Since the ice is floating, the forces on it need to be balanced. In other words, $\begin{aligned}W(\text{ice}) &= F(\text{bouyancy}) \\ &= \rho(\text{water}) \cdot V(\text{under}) \cdot g\end{aligned}$.

On the other hand, recall that

$\begin{aligned}&W(\text{ice}) = \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) \cdot g\end{aligned}$.

Combine the two halves to obtain:

$\begin{aligned}& \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) \cdot g \\ &= W(\text{ice}) = \rho(\text{water}) \cdot V(\text{under}) \cdot g\end{aligned}$.

$\begin{aligned}& \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) \cdot g = \rho(\text{water}) \cdot V(\text{under}) \cdot g\end{aligned}$.

Divide both sides by $g$ (assume that $g \ne 0$) to obtain:

$\begin{aligned}& \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) = \rho(\text{water}) \cdot V(\text{under})\end{aligned}$.

Rearrange to obtain:

$\begin{aligned}& \frac{V(\text{under})}{V(\text{above}) + V(\text{under})} = \frac{\rho(\text{water})}{\rho(\text{ice})}\end{aligned}$.

However, the question is asking for $\displaystyle \frac{V(\text{above})}{V(\text{above}) + V(\text{under})}$, the fraction of the volume above water. Note that

$\begin{aligned}& \frac{V(\text{under})}{V(\text{above}) + V(\text{under})} + \frac{V(\text{above})}{V(\text{above}) + V(\text{under})} = 1\end{aligned}$.

Therefore,

$\begin{aligned} &\frac{V(\text{above})}{V(\text{above}) + V(\text{under})} \\ &= 1 - \frac{V(\text{under})}{V(\text{above}) + V(\text{under})} \\ &= 1 - \frac{\rho(\text{water})}{\rho(\text{ice})} = 1 - \frac{917}{10^3} = 0.083\end{aligned}$.

That's equivalent to $8.3\,\%$.