Answer
$\dot{Q}_T=27.22\ kW$
Work Step by Step
For the bottles:
$\dot{Q}+\dot{m}h_1=\dot{m}h_2$
$\dot{Q}=\dot{m}c_p(T_2-T_1)$
$\dot{m}=1.125\ kg/s,\ c_p=0.8\ kJ/kg.s,\ T_2=50°C,\ T_1=20°C$
$\dot{Q}=27\ kW$
For the water removed:
$\dot{m}=450\ bottles/min \times 0.2\ g/bottle=90\ g/min=0.0015\ kg/s$
$\dot{Q}=\dot{m}c_p(T_2-T_1)$
$c_p=4.18\ kJ/kg.°C,\ T_2=50°C,\ T_1=15°C$
$\dot{Q}=0.2195\ kW$
Total heat removed:
$\dot{Q}_T=\dot{Q}_B+\dot{Q}_w$
$\dot{Q}_T=27.22\ kW$
