Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 14 - Heat - Problems - Page 408: 10

Answer

$1700\frac{J}{kg\cdot C^{\circ}}$

Work Step by Step

Use equation 14–2. $$Q=mc\Delta T$$ $$135\times10^3J =(4.1kg)c (37.2^{\circ}C-18.0^{\circ}C) $$ Solve for $c=1715\frac{J}{kg\cdot C^{\circ}}\approx1700\frac{J}{kg\cdot C^{\circ}}$
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