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

The mixture of 1.55 kg of water and of 0.475 kg of ice must be at 0oC.
Mass m(new ice) at -15.4degree C is added.
Q(of ice) = Q(freeze water)
m(new ice) x4190 x (0 – (-15.4)) = m(freeze water) x r
where r = 335000 J/kg is the heat of fusion.
c = 4190 J/kg•oC is heat capacity.
m(new ice) x 64526 = m(freeze water) x 335000
m(freeze water) = 0.1926x m(new ice),
Mass of ice at the end = 0.784 kg.
So
0.784 kg = 0.475 kg + m(new ice) + m(freeze water),

0.784 kg = 0.475 kg + m(new ice) + 0.1926x m(new ice),
(0.784 - 0.475) kg = 0.309 kg = 1.1926 x m(new ice),
m(new ice) = 0.309/1.1926 = 0.259 kg .............................................ans