Answer
Please see the work below.
Work Step by Step
We know that
$Q=Pt$
$Q=500\times 1200=600KJ$
We also know that
$Q=mc\Delta T+m^{\prime}L_v$
We plug in the known values to obtain:
$600KJ=(0.3Kg)(4.184\frac{KJ}{KgK})(80K)+m^{\prime}(2257\frac{KJ}{Kg})$
This simplifies to:
$m^{\prime}=\frac{500KJ}{2257\frac{KJ}{Kg}}=0.22Kg$
Now $mass\space of\space water \space left=0.3Kg-0.22Kg=0.08Kg$
