Mass of aluminum, ma=80g=0.08Kg  
Mass of water, mw=150g=0.15Kg 
Mass of ice, mi=100g=0.1Kg 
Specific heat of water, sw=4182K/Kg/K 
Specific heat of aluminum,  sa=0.9J/gK=900J/Kg/K 
Specific heat of ice,  si=2.04KJ/Kg/K=2040J/Kg/K 
Latent heat of ice, Lf=3.33×105J/Kg 
Initial temperature, Ti=0∘C 
Final temperature, Tf=33∘C 
Heat generated by the coil in one minute, Q=1000×60J=60000J 
Let x amount of the ice that melt,
then assume that there is no loss of the heat,
(masa+mwsw+xsi)(Tf−Ti)+miLf=60000 
Putting values,
[(0.080×900)+(0.15×4182)+(x×2040)](33−0)+(0.1×3.33×105)=60000 
Solving it, we get, x=0.05Kg=50g 
Amount of ice left, M=100−50=50g 
                             
Comments
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