Chapter 5 - Mass and Energy Analysis of Control Volumes - Problems - Page 260: 5-98

$\dot{m}=0.0104\ kg/s$ $f=23.8\%$

From the energy balance: $\dot{Q}+\dot{W}_s+\dot{m}h_1=\dot{m}h_2$ $\dot{Q}+\dot{W}_s=\dot{m}c_p\Delta T$ and since $\dot{Q}=8\times10W=80W,\ \dot{W}_s=25W,\ c_p=1.005\ kJ/kg.C,\ \Delta T=10°C$: $\dot{m}=0.0104\ kg/s$ $f=\frac{\dot{W}_s}{\dot{Q}+\dot{W}_s}=0.238=23.8\%$