Answer
$\dot{m}_w=1.56\ kg/s$
Work Step by Step
The energy balance for the chicken:
$\dot{Q}_c=\dot{m}_cc_{p,c}(T_1-T_2)_c$
Given $c_{p,c}=3.54\ kJ/kg.°C,\ T_{2,c}=15°C,\ T_{1,c}=3°C$
$\quad \quad\quad\dot{m}_c=500\ chicken/h\times2.2\ kg/chicken=0.3056\ kg/s$
$\dot{Q}_c=13.0\ kW$
The heat gain from the water with $\dot{Q}_{air}=0$:
$\dot{Q}_w=\dot{Q}_{air}+\dot{Q}_{chicken}$
$\dot{Q}_w=13.0\ kW$
From the energy balance for the water:
$\dot{Q}_w=\dot{m}_wc_{p,w}\Delta T_w$
Given $c_{p,w}=4.18\ kJ/kg.°C,\ \Delta T_w=2°C$:
$\dot{m}_w=1.56\ kg/s$
