Physics: Principles with Applications (7th Edition)

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

Chapter 15 - The Laws of Thermodynamics - Problems - Page 440: 37

Answer

78 liters.

Work Step by Step

The coefficient of performance for a refrigerator is given by Eq. 15–6a, $Q_L/W$. $Q_L$ is the heat removed from the low-temperature area, and is the latent heat removed to freeze the ice. $$COP=\frac{Q_L}{W}=\frac{mL_f}{W}$$ The work done to remove the heat is 1.2 kW multiplied by the elapsed time. The mass of water frozen is its density multiplied by its volume. $$COP=\frac{Q_L}{W}=\frac{(\rho V) L_f}{Pt}$$ Solve for the volume. $$V=\frac{(COP)Pt}{\rho L_f}=\frac{(6.0)(1200W)(3600s)}{(1000kg/m^3)(3.33\times10^5 J/kg)}=0.0778m^3$$
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