Question

A block of ice at 0

Answer 1

The water and aluminum calorimeter are at 0 deg C by the time 2 grams of ice are left.
The water went down 12 deg and had a mass of 210 grams and water has a specific heat capacity of 4.18
aluminum has a specific heat of .920 j/g/c went down 12 degrees and has a mass of 150 grams.

Q = M*Cv*dT
where Q= amount of heat
M = 210 gm
dT= temperature difference =12 deg
Cv = 4.18

Q= 210*4.18*12 = 10533.6 J

now for aluminum
Q= 150*0.92*12 = 1656 J
total lost = 10533.6 + 1656 = 12189.6

this means that ice gained 12.189 kJ

latent heat of melting for ice is 334 J/gm

ice melted = 12189/334 = 36.49

total ice = 36.49 + 2 = 38.49gm