A 0.160-kg cube of ice (frozen water) is floating in glycerine. The gylcerine is

Question

A 0.160-kg cube of ice (frozen water) is floating in glycerine. The gylcerine is in a tall cylinder that has inside radius 3.10cm . The level of the glycerine is well below the top of the cylinder.

Part 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.

Part B

Does the level of liquid rise or fall? That is, is the surface of the water above or below the original level of the gylcerine before the ice melted?

Does the level of liquid rise or fall? That is, is the surface of the water above or below the original level of the gylcerine before the ice melted?

Explanation / Answer

the weight of the displaced liquid equals the weight of the cube (or the mass of the displaced liquid equals the mass of the cube). And
mass = density*volume
m = ?*V
0.16 kg = 1.26 kg/L * V
V = 0.16 kg/1.26 kg/L = 0.1269 L (= dm^3 = displaced liquid)
Since the initial volume vi of the glycerol without the cube stays the same after the cube is added, the new volume (up to the surface only) in the cylinder after addition of the cube is the initial volume + the immersed volume
i.e., V = Vi + 0.1269 dm^3.

Now let the cube melt: 0.16 kg ice give 0.16 dm^3 water.
The volume in the cylinder is now vi + 0.16 dm^3.

The volume difference before, and after the melting, is then 0.16 - 0.1269 = 0.0331 dm^3
now find the height of 0.0331 dm^3 in the cylinder:
V = pir^2*h
h = V/(pir^2) with r in dm
h = 0.0331/(pi*0.31^2)
h = 0.1096 dm =1.096 cm > the liquid level is 1.096 cm higher after the melting.

(B )a.The level of liquid in the cylinder rises.

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