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 410: 53

Answer

$6.6\times10^3kcal$

Work Step by Step

Eighty percent of the heat from the cyclist causes the water to evaporate. The water is at approximately room temperature, so use the heat of vaporization at room temperature as discussed on page 399. One liter of water has a mass of 1 kg. $$0.80Q_{cyclist}=mL_v$$ $$ Q_{cyclist}=\frac{mL_v }{0.8}=\frac{(9.0kg)(585kcal/kg)}{0.80}=6.6\times10^3kcal$$
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