Respuesta :
Answer:
[tex]8.3\,\%[/tex] of that piece of ice would be above the freshwater. Assumptions:
- the density of the ice is [tex]\rho(\text{ice}) = 917\; \rm kg \cdot m^{-3}[/tex], and
- the density of freshwater is [tex]\rho(\text{water}) = 1.00 \times 10^3\; \rm kg \cdot m^{-3}[/tex] .
Explanation:
The volume of that chunk of ice can be split into two halves: volume above water [tex]V(\text{above})[/tex], and volume under water [tex]V(\text{under})[/tex]. The mass of the whole chunk of ice would be:
[tex]m(\text{ice}) = \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under}))[/tex].
Let [tex]g[/tex] be the acceleration due to gravity. The gravity on the entire chunk of ice would be
[tex]\begin{aligned}&W(\text{ice}) \\ &= m({\text{ice}}) \cdot g \\ &= \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) \cdot g\end{aligned}[/tex].
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: [tex]F(\text{bouyancy}) = W(\text{water displaced})[/tex].
Recall that [tex]V(\text{above})[/tex] is the volume of the ice above the water, and [tex]V(\text{under})[/tex] is the volume of the ice under the water.
The mass of water displaced would be equal to:
[tex]\begin{aligned}& m(\text{water displaced}) \\ &= \rho(\text{water}) \cdot V(\text{water displaced}) \\ &= \rho(\text{water}) \cdot V(\text{under})\end{aligned}[/tex].
The weight of that much water would be
[tex]\begin{aligned} &W(\text{water displaced}) \\ &= m(\text{water displaced}) \cdot g \\ &= \rho(\text{water}) \cdot V(\text{under}) \cdot g \end{aligned}[/tex].
Apply the equation [tex]F(\text{bouyancy}) = W(\text{water displaced})[/tex]. The bouyant force on this chunk of ice would be equal to [tex]\begin{aligned} &W(\text{water displaced}) = \rho(\text{water}) \cdot V(\text{under}) \cdot g \end{aligned}[/tex].
Since the ice is floating, the forces on it need to be balanced. In other words, [tex]\begin{aligned}W(\text{ice}) &= F(\text{bouyancy}) \\ &= \rho(\text{water}) \cdot V(\text{under}) \cdot g\end{aligned}[/tex].
On the other hand, recall that
[tex]\begin{aligned}&W(\text{ice}) = \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) \cdot g\end{aligned}[/tex].
Combine the two halves to obtain:
[tex]\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}[/tex].
[tex]\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}[/tex].
Divide both sides by [tex]g[/tex] (assume that [tex]g \ne 0[/tex]) to obtain:
[tex]\begin{aligned}& \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) = \rho(\text{water}) \cdot V(\text{under})\end{aligned}[/tex].
Rearrange to obtain:
[tex]\begin{aligned}& \frac{V(\text{under})}{V(\text{above}) + V(\text{under})} = \frac{\rho(\text{water})}{\rho(\text{ice})}\end{aligned}[/tex].
However, the question is asking for [tex]\displaystyle \frac{V(\text{above})}{V(\text{above}) + V(\text{under})}[/tex], the fraction of the volume above water. Note that
[tex]\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}[/tex].
Therefore,
[tex]\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}[/tex].
That's equivalent to [tex]8.3\,\%[/tex].
