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

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