Question

A Styrofoam bucket of negligible mass contains 1.55 kg of water and 0.475 kg of ice. More ice, from a refrigerator at -15.4 ∘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 0.784 kg .

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

Answer 1

specific heat of water is 4.186 kJ/kgC
specific heat of ice is 2.06 kJ/kgC
heat of fusion of ice is 334 kJ/kg

initially, you have an ice water mixture at 0ºC
the added ice at –15.4`º will take energy to warm to 0º and more to melt.
the final temp of the mixture is still 0º
M is the mass of added ice
the amount of additional ice frozen is 0.784 – 0.475 = 0.309 kg
the energy needed to freeze that is
E1 = 334 kJ/kg x 0.309 kg = 103.20 kJ

Energy absorbed from added mass M is
E2 = [ 2.06 kJ/kgC x M x 15.4C ] + [ 334 kJ/kg x M ]

set them equal and solve for M

E1 = E2
31.724 M + 334 M = 103.20
365.72 M = 103.20
M = 0.282 kg