Answer
Please see the work below.
Work Step by Step
We know that
$Specific \space heat=\frac{q}{m\Delta t}$
We plug in the known values to obtain:
$Specific heat=\frac{325J}{121.6g(35.5C^{\circ}-20.4C^{\circ})}=0.12798\frac{J}{g.C}$
General Chemistry 10th Edition
by Ebbing, Darrell; Gammon, Steven D.
Published by
Cengage Learning
ISBN 10:
1-28505-137-8
ISBN 13:
978-1-28505-137-6
Chapter 6 - Thermochemistry - Questions and Problems - Page 264: 6.105
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