Thermodynamics: An Engineering Approach 8th Edition

Published by McGraw-Hill Education
ISBN 10: 0-07339-817-9
ISBN 13: 978-0-07339-817-4

Chapter 6 - The Second Law of Thermodynamics - Problems - Page 323: 6-138

Answer

$\Delta t=3.95\ min$

Work Step by Step

$m=\rho V,\ \rho=1.2\ kg/m³,\ V=12\times2.3\times3.5=96.6\ m³$ $m=116\ kg$ $Q=mc_p\Delta T,\ c_p=1.0\ kJ/kg.°C,\ \Delta T=(25-5)°C$ $Q_{pc}=2,320\ kJ$ $\dot{Q}_i=UA.\Delta T,\ UA=120\ W/°C,\ \Delta T=(25-15)°C$ $\dot{Q}_i=1.2\ kJ/s$ $\dot{Q}_r=\dot{Q}_i+Q_{pc}/\Delta t,\ \dot{Q}_r=11\ kW$ $\Delta t=237s=3.95\ min$
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