Two 20.0g-g icecubes at -11.0 degrees Celcius are placed into 265g of water at 2

Question

Two 20.0g-g icecubes at -11.0 degrees Celcius are placed into 265g of water at 25.0 degrees Celcius. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts.

ndiana University of Pe x C Home I chegg.com shmad/ibis view.php?id 31643 Saplinglea ming-Com Sapling Learning Jump to... Sapling Learning l ndiana University of Pennsylvania CHEM 112 Spring17-ASHE l Activities and Due Dates HW1 Chapter 12) My Assignment Gradebook 2/7/2017 11:55 PM A 76.2/100 2/2/2017 07:17 PM Attempts Score Periodic Table 100 Question B of 25 Map 100 General Chemistry 4th Edition University Science Books presented by Leaming 100 Two 20.0-g ice cubes at-11.0 C are placed into 265 g of water at 25.0 'C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts, heat capacity of H20(s) 37.7 J/(mol. K) 100 Number heat capacity of H20() 75.3 J/(mol. K) enthalpy of fusion af H20 6.01 kJimol 100 100 10 100 100 12 c 2011-2017 Sapling Learning, Inc. about us careers partners privacy policy terms of us Search all the things Logout Resources Assignment Information Available From: /26/2017 09:15 AM 2/7/2017 11:55 PM Due Date. Points Possible: Grade Category: Graded Description CHEM 111 Fall 2016 Policies: You can check your answers. You can view solutions after the due date. You have ten attempts per question. There is no penalty for incorrect answers. For multiple-choice questions, the penalty depends af ch 30 PM 2/7/2017 521

Explanation / Answer

1) The ice does three things:

a) heat up from -11 to 0
b) melt at 0
c) heat up from 0 to unknown temperature

2) The three calculations are:

qa = (20 x 37.7 x 11 /18) = 460.77Joules

qb = (20 x 6010 /18) = 6677.77 Joules
qc = (X x 75.3 x 20/18) = 83.66 X

The heat lost by ice = The heat gained by water

460.77 + 6677.77 + 83.66 X + 83.66 = (25-X) 265 x 75.3 /18

7222.22 + 83.66 X = 27714.58 - 1108.58 X

1192.24 X = 20492.35

X = 17.188 degrees Celcius

Hence the final temperature of water is  17.188 degrees Celcius

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