College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 11 - Heat - Learning Path Questions and Exercises - Exercises - Page 412: 10


(a) Less than (b) $1.283Kg$

Work Step by Step

(a) We know that $Q=mc\Delta T$ This can be rearranged as: $m=\frac{q}{c\Delta T}$ As $Q$ and $\Delta T$ are same for aluminum and copper $\implies m\propto \frac{1}{c}$ We know that $c_{Al}=0.900J/gC^{\circ}$ and $c_{Cu}=0.385J/gC^{\circ}$ which means $c_{Al}\gt c_{Cu}$ therefore $m_{Al}\lt m_{Cu}$ We conclude that the mass of the aluminum block is less than that of copper. (b) We know that $m_{Al}c_{Al}=m_{Cu}c_{Cu}$ This can be rearranged as: $m_{Al}=(\frac{c_{Cu}}{c_{Al}})m_{Cu}$ We plug in the known values to obtain: $m_{Al}=(\frac{0.385}{0.900})(3)=1.283Kg$
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