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.

11 players on a field that has a maximum of 11 players

the cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.