A sample of water has a temperature of 87.5 d/C. The volume of the sample is 430 mL. What mass of ice at 0.0 d/C will melt with this sample of water if the final temperature of water and melted ice is 0.0 d/?

Answer 1

Σ q = 0
(Mc∆T)water + (M∆Hfusion)ice + (Mc∆T)ice water = 0
[(430g)(1cal/g-oC)(0oC – 87.5oC)] + [(M-ice)(80cal/g)] + [(M-ice water)(1cal/goC)(0oC - 0oC)] = 0
[(430)(-87.5)] + [(M-ice)(80)] + [0] = 0
[(M-ice)(80)] = 37625 => M-ice = (37625/80)g-ice = 470g-ice

Answer 2

which equations did you use for this problem?