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: 83

Answer

Therefore the temperature change is $10.28^{\circ}$

Work Step by Step

We will use the equation $q=mc\Delta t$ q = heat transferred m = mass c = specific heat $\Delta t $ = change in temperature plug in the values. 1037.6 = (24) (4.2) ($\Delta t $) ***We multiply 24 times 4.2 and then divide 1037.6 by that answer *** solve using a calculator $ \Delta t = 10.28 ^{\circ} $ Therefore the temperature change is $10.28^{\circ}$
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