Question

# A Styrofoam bucket of negligible mass contains 1.55 kg of water and 0.475 kg of ice....

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?

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