Introductory Chemistry (5th Edition)

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

Chapter 3 - Matter and Energy - Exercises - Cumulative Problems - Page 90: 97


$29756.608 kJ $

Work Step by Step

Using the equation $q=mc\Delta T$, we can solve for q which in this case is the heat that is required to heat water from $85^{\circ}F $ to $212^{\circ}F$. The change in temperature can be calculated by taking the final temperature and subtracting it from the initial. Therefore, its $212-85 = 127^{\circ}F$. The heat capacity ("c") of water is $4.184 \frac{J}{g\times^{\circ}C}$ Finally, the mass of the water is 56 kg or 56000 grams since the density of water is 1g/L and there is 56 L of water. Therefore, $q = 56,000 g \times 4.184 \frac{J}{g\times^{\circ}C}\times127 = 29756608 J $ $ 29756608 J \div 1000 = 29756.608 kJ $
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