Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 19 - Exercises and Problems: 60

Answer

Please see the work below.

Work Step by Step

(a) We know that $m_1c_{copper}(T_1-T_c)=m_2c_{water}(T_c-T_2)$ We plug in the known values to obtain: $0.5\times 386\times (80-T_c)=1\times 4184\times (T_c-10)$ $Tc=13.1C^{\circ}$ (b) As $\Delta S_{copper}=0.5\times 386\times \ln(\frac{13.1+273}{80+273})=-40.5\frac{J}{K}$ $\Delta S_{water}=1\times 4184\times \ln(\frac{13.1+273}{10+273})=44.1\frac{J}{K}$ Now $\Delta S_{system}=\Delta S_{copper}+\Delta S_{water}=-40.5+44.1=3.7\frac{J}{K}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.