Answer
$\dot{Q}_{removed}=16.47\ kW$
Work Step by Step
The energy balance for a single ball:
$Q=\Delta U=mc(T_2-T_1)$
The mass is given by:
$m=\rho \frac{\pi}{6}D^3$
With $\rho=8522\ kg/m³,\ D=0.05\ m$
we get $m=0.558\ kg$
Since $c=0.385\ kJ/kg.K, T_2=74°C, T_1=120°C$
$Q=-9.89\ kJ$
The heat removal rate from the water must be equal to the heat rate wich is added to the water by the balls.
With the frequency of $\dot{n}=100 balls/min$:
$\dot{Q}_{removed}=-\dot{n}Q$
$\dot{Q}_{removed}=16.47\ kW$
