If we measure the heat change (q) of everything in a system except the calorimeter, we can use the first law of thermodynamics (∆Euniverse = 0) to calculate the heat change of the calorimeter (qcal).  In this reaction, we combine 5.59 g of 66.2 ºC water with 5.62 g of 16.1 ºC water in a calorimeter.  Since some of the heat is absorbed by the calorimeter, the final temperature of the water is 26.6ºC.  Since the hot and cold water are combined, the final temp of the water is the final temp for the hot water and the cold water.  Calculate the heat change of the calorimeter in J. Remember that all of the heat changes must add up to zero so you can calculate the heat change of the hot water (qhotwater) and the heat change of the cold water (qcoldwater) using for each where Cs = 4.186 J/gºC.  The first law of thermodynamics tells us that all of these changes sum up to zero so   Enter your answer to one decimal place (tenths

Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter4: Energy And Chemical Reactions
Section4.8: Measuring Reaction Enthalpies: Calorimetry
Problem 4.17E
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If we measure the heat change (q) of everything in a system except the calorimeter, we can use the first law of thermodynamics (∆Euniverse = 0) to calculate the heat change of the calorimeter (qcal).  In this reaction, we combine 5.59 g of 66.2 ºC water with 5.62 g of 16.1 ºC water in a calorimeter.  Since some of the heat is absorbed by the calorimeter, the final temperature of the water is 26.6ºC.  Since the hot and cold water are combined, the final temp of the water is the final temp for the hot water and the cold water.  Calculate the heat change of the calorimeter in J.

Remember that all of the heat changes must add up to zero so you can calculate the heat change of the hot water (qhotwater) and the heat change of the cold water (qcoldwater) using for each where C= 4.186 J/gºC.  The first law of thermodynamics tells us that all of these changes sum up to zero so  

Enter your answer to one decimal place (tenths

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