Answer
$44.2Kg$
Work Step by Step
We know that
$Q_{Al}=m_{Al}C_{Al}\Delta T$
$\implies Q_A=(155g(\frac{10^{-3}Kg}{1g}))(653J/Kg.C^{\circ})(196C^{\circ})$
$Q_{Al}=19838J$
Heat lost by water before it freezes
$Q_w=m_wC_w\Delta T$
$Q_w=(80g(\frac{10^{-3}Kg}{1g}))(4186J/Kg.C^{\circ})(15C^{\circ})$
$Q_w=5023.2J$
$Q_f=m_wL_f$
$\implies Q_f=(0.08Kg)(33.5\times 10^4J)=26,800J$
$Q_w+Q_f=5023.2J+26,800J=31823.2J$
Now $m_f=\frac{Q_{Al}-Q_w}{L_f}$
We plug in the known values to obtain:
$m_f=\frac{19838J-5023J}{33.5\times 10^4J/Kg}$
$\implies m_f=44.2Kg$
