Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 12 - Liquids, Solids, and Intermolecular Forces - Exercises - Cumulative Problems: 93

Answer

1.1x $10^2$ g

Work Step by Step

Given Volume of water= 352 mL. density of water = 1.g /mL Mass of water = volume x density = 352 mL x 1 g/mL= 352 g. The amount of heat produced by water is calculated using: q= mC$\Delta t$. here m= 352g C= 4.184 J/g $\Delta t$ = (0 C-25C)= -25 C q= mC$\Delta t$ = (352 g)x( 4.184 J/g)x(-25C) q= - 36819 J= -36.8 kJ The amount of heat required for melting 1 mol of ice =6.02 kJ. So, with -37 kJ = 37kJx( 1mol/ 6.02 kJ)x( 18.02 g/ 1 mol h2O) = 1.1 x $10^2$ g. So with -37 kJ, we can melt 1.1x $10^2$ g of ice.
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