Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 6 - Sections 6.1-6.10 - Exercises - Problems by Topic - Page 288: 50c

Answer

$38g$

Work Step by Step

To find the mass we are going to use the equation $q= m\times C_{s}\times\Delta T$ where $q$ is the heat in joules, $m$ is the mass in grams, $C_{s}$ is the specific heat (found on table on pg. 257), and $\Delta T$ is the change in temperature in $^{\circ}C$. To find $\Delta T$ we must subtract the final temperature from the initial temperature: $44.2-23.0=21.2^{\circ}C$ Now we plug every thing into the equation. The specific heat of ethanol is $2.42$. $1.95\times10^{3}J= m\times 2.42\frac{J}{g \times^{\circ}C}\times\ 21.2^{\circ}C$ Rearranging we get: $\frac{1.95\times10^{3}}{21.2\times 2.42}g=m$ so the mass of the substance is $38g$
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