A Styrofoam bucket of negligible mass contains 1.60 kg of water and 0.425 kg of ice. More ice, from a refrigerator at -14.8 degree 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 0.768 kg.
Assuming no heat exchange with the surroundings, what mass of ice was added?
Physics help!! - Elena, Monday, April 16, 2012 at 3:19pm
The mixture of 1.6 kg of water and of 0.425 kg of ice must be at 0oC.
Mass m(new ice) at -14.8oC is added.
Q(of ice) = Q(freeze water)
m(new ice) • c•4180•(0 – (-14.8)) = m(freeze water) •r,
where r = 335000 J/kg is the heat of fusion.
c = 4180 J/kg•oC is heat capacity.
m(new ice) • 61864 = m(freeze water) •335000,
m(freeze water) = 0.185• m(new ice),
Mass of ice at the end = 0.768 kg.
0.768 kg = 0.425 kg + m(new ice) + m(freeze water),
0.768 kg = 0.425 kg + m(new ice) + 0.185• m(new ice),
(0.768 - 0.425) kg = 1.185• m(new ice),
m(new ice) = (0.768 - 0.425)/1.185 = 0.289 kg.