Ice Floats: Calculating Water Level Percent

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SUMMARY

The discussion focuses on calculating the percentage of ice that remains above water when floating, using the densities of water (1000 kg/m³) and ice (920 kg/m³). The solution reveals that 8% of the ice volume is above the water level. By considering a volume of ice (100 m³), the mass is calculated as 92,000 kg, which displaces 92 m³ of water. This leads to the conclusion that the remaining volume of ice above the water line is 8 m³, resulting in the final percentage of 8% above the water level.

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Miri
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Homework Statement


If ice floats in water, how many percent of it show above the water level?
density of water=1000kg/m^3
density of ice=920kg/m^3

Homework Equations


I really don't know how to start and what I have to do with those densities...just give me a hint. solution would be 8%
 
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Consider a volume of ice. How much does it weigh?

Then how much water needs to be displaced to support that?

Subtract that from 1 to get the % above the water line.
 
So I take for example 100m^3 for the volume of ice. So the mass is (920kg/m^3)*100m^3=92000kg. Then I divide 92000kg by the density of water and I get 92m^3. And then??
 
No, no, that won't get you anywhere. Try making a free body diagram of a block of ice, mass m, suspended in water.
What are the forces acting on it? What is the sum of these forces, and what does that say about the volume of ice submerged relative to its total volume? (The %)
 
Miri said:
So I take for example 100m^3 for the volume of ice. So the mass is (920kg/m^3)*100m^3=92000kg. Then I divide 92000kg by the density of water and I get 92m^3. And then??

So that means that only 92 m3 of water are needed to support 100 m3 of ice. What's the difference? Isn't that what's left over above the water line? What's the percentage since that's the form they want the answer in?
 
Ok, thanks, I got it...
 

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