A48 g piece of ice at 0.0 ∘c is added to a sample of water at 7.4 ∘c. all of the ice melts and the temperature of the water decreases to 0.0 ∘c. how many grams of water were in the sample?

mass of water = 519 grams

Explanation:

∑Q = Q(melting ice) + Q(warming ice water) + Q(cooling solvent water) = 0

m·ΔH(fusion ice) + m·c·ΔT(ice water) + m·c·ΔT(solvent water) = 0

48g(80cal/g) + 48g(1cal/g°C)(0°C) + m(solvent water)(1cal/g°C)(0°C - 7.4°C) = 0

3840cal = m(solvent water)(1cal/g°C)(7.4°C)

m(solvent water) = 3840cal/7.4cal/g = 519 g solvent water.

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