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A 0.180-kg cube of ice (frozen water) is floating in

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 57P Chapter 12

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 57P

A 0.180-kg cube of ice (frozen water) is floating in glycerine. The gylcerine is in a tall cylinder that has inside radius 3.50 cm. The level of the glycerine is well below the top of the cylinder. If the ice completely melts, by what distance does the height of liquid in the cylinder change? Does the level of liquid rise or fall? That is, is the surface of the water above or below the original level of the glycerine before the ice melted?

Step-by-Step Solution:

Solution 57P Introduction We have to calculate the change in the height of the liquid in the tube after ice melts. Step 1 The mass of the ice is m = 0.180 kg = 180 g . Hence the mass of the displaced glycerine is m = 180 g 3 The density of glycerine is g= 1.26 g/cm Hence the volume of the displaced glycerine is m 180 g 3 V =g g= 1.26 g/cm 142.9 cm Step 2 3 The density of water is w= 1 g/cm . 3 Hence the volume of the water after ice melts is V w = 180 cm Step 2 So the change in the volume is 3 3 3 V = 180 cm 142.9 cm = 37.1 cm Now the cross sectional area of the tube is A = r = (3.50 cm) = 38.5 cm 2 Hence the change in height is (37.1 cm ) h = A = (38.5 cm ) 0.964 cm Hence the change in the height will be 0.964 cm.

Step 3 of 3

Chapter 12, Problem 57P is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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