Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 16 - Exercises and Problems: 48

Answer

a) Aluminium b) 83.68 Watts

Work Step by Step

We know: $\Delta Q = m_wS_w\Delta T_w$ $\Delta Q = m_xS_x\Delta T_x$ Setting these equal to each other and simplifying gives: $ S_x = \frac{\Delta T_w S_w m_w}{\Delta T_x m_x} \\ S_x = \frac{100\times10^{-3}\times 4184\times12}{100\times10^{-3}\times 56} = 900 \frac{J}{kg K }$ The material is $\fbox{Aluminum.}$ We now find the value of the heat flow rate: $= \frac{\Delta Q}{\Delta t} = \frac{5020.8}{60}=\fbox{83.68 W}$
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