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
This is really helpful