Question

A Styrofoam bucket of negligible mass contains 1.90 kg of water and 0.460 kg of ice. More ice, from a refrigerator at -16.0 ∘C, is added to the mixture in the bucket, and when thermal equilibrium has been reached, the total mass of ice in the bucket is 0.788 kg .

Assuming no heat exchange with the surroundings, what mass of ice was added? (kg)

Answer 1

Specific Heat of Water, Cw = 4.186 J/gm k
Specific Heat of ice, Ci = 2.108 J/gm k
Latent Heat of ice, Li = 333.55 J/gm
Mass of ice , mi = 0.460 kg = 460 gm
Mass of water, = 1.90 kg = 1900 gm
Let the mass of ice added b m

New ice which is frozen = 0.788 - 0.460 = 0.328 kg = 328 gm
Energy needed to Freeze this much ice = 328 * 333.55 J
Energy needed to Freeze this much ice = 109404.4 J

Now,
m*Ci*16 + m*Li = 109404.4
m*2.108 * 16 + 333.55 * m = 109404.4
m = 297.9 gm
Mass of ice added, m = 0.298 kg