Answer
$2.2137Kg$
Work Step by Step
We can find the mass of the block of the ice as follows:
$Q_1=mc_{ice}\Delta T$
$\implies Q_1=m(2090J/KgC^{\circ})(15C^{\circ})$
$Q_2=mL_f$
$Q_2=m(33.5\times 10^4J/Kg)$
and $Q_3=mc_{water}\Delta T$
$Q_3=m(4186J/KgC^{\circ})(15C^{\circ})$
Now $Q=Q_1+Q_2+Q_3$
We plug in the known values to obtain:
$9.5\times 10^5J=m(2090J/KgC^{\circ})(15C^{\circ})+m(33.5\times 10^4J/Kg)+m(4186J/KgC^{\circ})(15C^{\circ})$
This simplifies to:
$m=2.2137Kg$
