Answer
(a) No
(b) $0C^{\circ}; 0.4Kg$
Work Step by Step
(a)
No, the $2.5\times10^5$ J is not enough to melt the ice completely.
(b)
We know that
$Q_1=mc_{ice}\Delta T$
$\implies Q_1=(1.1Kg)(2090J/Kg.C^{\circ})(0C^{\circ}-(-5C^{\circ}))=11495J$
and $Q_2=mL_f$
$Q_2=(1.1Kg)(33.5\times 10^4J/Kg)$
$Q_2=36.85\times 10^4J$
$Q_1+Q_2=11495J+368500J=3.79995\times 10^5J\gt Q=2.6\times 10^5J$
Thus the complete ice will not be melted -- that is, some amount of ice will left and hence the final temperature will not increase but remain at $0C^{\circ}$.
We know that the total heat of the system is given as
$Q=Q_1+Q_3$
$\implies 2.6\times 10^5J=11495J+M(33.5\times 10^4J/Kg)$
This simplifies to:
$M=0.7418Kg$
Now the mass of the ice that remains will be
$=m-M=1.1Kg-0.7418Kg=0.4Kg$
