Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 15 - Thermodynamics - Problems - Page 415: 7

Answer

(a) $-5.03\times10^{5}J$ (b) $120.16\space Calories$

Work Step by Step

(a) Here we use equations 15.1 $Q=\Delta U+W$ and 12.5 $Q=mL_{v}$ to find the change in internal energy of the weight lifter. $Q=\Delta U+W=>\Delta U=Q-W=-mL_{v}-W$ Let's plug known values into this equation. $\Delta U=-(150\space kg)(2.42\times10^{6}J/kg)-(1.4\times10^{5}J)$ $\Delta U=-5.03\times10^{5}J$ (change in internal energy of the weight lifter) (b) The number of nutritional = $(5.03\times10^{5}J)(\frac{1\space Calorie}{4186\space J})$ calories The number of nutritional$=120.16\space Calories$ calories
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