Answer
(a) $0.469Kg$
(b) $0.320Kg$
(c) $0.171Kg$
Work Step by Step
(a) We know that
$Q=5.00\times 10^4J-2.30\times 10^4J=2.70\times 10^4J$
Now $m=\frac{Q}{L_f}$
We plug in the known values to obtain:
$m=\frac{2.70\times 10^4J}{33.5\times 10^4J/Kg}$
$m=0.081Kg$
$m_{ice}=0.55Kg-0.081Kg=0.469Kg$
(b) As $Q=1.00\times 10^4J-2.30\times 10^4J=0.77\times 10^4J$
and $m=\frac{Q}{L_f}$
$\implies m=\frac{0.77\times 10^4J}{33.5\times 10^4J/Kg}$
$m=0.230Kg$
$m_{ice}=0.55Kg-0.230Kg=0.320Kg$
(c) $Q=1.50\times 10^4J-2.30\times 10^4J=1.27\times 10^4J$
$m=\frac{Q}{L_f}$
$m=\frac{1.27\times 10^4J}{33.5\times 10^4J/Kg}=0.379Kg$
$m_{ice}=0.55Kg-0.379Kg=0.171Kg$
