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: 92

Answer

3.62°C

Work Step by Step

Given Mass of ice cube= 14.7g Mass of water= 324 g. The amount of heat required for melting 1 mol of ice is 6.02 kJ. To melt 8.5 g of ice is: q= mC$\Delta t$. q= ((14.7g x(1mol / 18.02 g )x(6.0 kJ/mol )x($10^3$J/ 1 kJ)) q= 4.91 x $10^3$ J The amount of heat produced by water is calculated using: q= mC$\Delta t$. here m= 324g C= 4.184 J/g $\Delta t$= ? $\Delta t$ = q/mC= ((4.91x $10^3$ J)/ (324g)x(4.184 J/g)) = 3.62°C Change in temperature is 3.62°C.
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