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 - General Problems - Page 411: 67

Answer

a. $3.6\times10^7J$ b. 63 minutes

Work Step by Step

Use equation 14–2. One liter of water has a mass of 1 kg. a. $Q=mc\Delta T=(245kg)(\frac{4186J}{kg\cdot C^{\circ}})(45^{\circ}C-10^{\circ}C)=3.6\times10^7J$ b. Power is the rate at which energy is used. $$P=9500\frac{J}{s}=\frac{3.6\times10^7J }{t}$$ Solve for t = 3778s, or about 63 minutes.
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