A 0.180 kg cube of ice (frozen water) is floating in glycerin. The glycerin is in a tall cylinder that has inside radius 5.50 cm. The level of the glycerin is well below the top of the cylinder.

A)If the ice completely melts, by what distance does the height of liquid in the cylinder change?

Express your answer with the appropriate units. Enter positive value if the surface of the water is above the original level of the glycerin before the ice melted and negative value if the surface of the water is below the original level of the glycerin.

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sampsonola sampsonola · Answer 1 · 2020-01-03T11:38:20+01:00

Answer:

0.00389 m

Explanation:

Archimedes principle states that the buoyant force when a body is immersed in a liquid equals the weight of the liquid displaced

B = weight of the liquid displaced

density is defined as the mass divided by the volume of the substance and the units is in kg/m³

Density of glycerin = mass / volume

ρg × volume = mass of glycerin displaced

since the object was floating, the upthrust from the liquid equals the weight of the liquid

ρg × volume × g = mg

divide both side by g

ρg × volume = 0.180 where ρg (density of glycerin) = 1260 kg / m³

1260 × volume of glycerin displaced = 0.18 / 1260 = 0.000143 m³

volume of glycerin displaced = πr²h₁ where h₁ = of liquid displaced

πr²h₁ = 0.000143

h₁ = 0.000143 / ( 3.142 × 0.055² ) = 0.01505 m

when the ice completely melted, it will displaces liquid equal to it own volume

density of water = mass of water /volume

1000 = 0.18 / v

v = 0.18 / 1000 = 0.00018 m³

volume = πr²h₂ where h₂ = height of the melted water

πr²h₂ = 0.00018

h₂ = 0.00018 / ( 3.142 × 0.055²) = 0.01894 m

change in height of the liquid = h₂ - h₁ = 0.01894 m - 0.01505 m = 0.00389 m