Question

Part A

What mass of steam at 100∘C must be added to 1.90 kg of ice at 0∘C to yield liquid water at 18 ∘C? The heat of fusion for water is 333 kJ/kg , the specific heat is 4186 J/kg⋅C∘ , the heat of vaporization is 2260 kJ/kg .

Express your answer to two significant figures and include the appropriate units.

m =

Answer 1

specific heat of water is 4.186 kJ/kgC

heat of fusion of ice is 333 kJ/kg

heat of vaporization of water is 2260 kJ/kg

Call M the unknown mass of steam.

The energy from condensing the steam and cooling it to 19 degree has to equal the energy required to melt the ice

and warm it to 19 degree

E1 = (2260 kJ/kg * M) + (4.186 kJ/kgC * M * (100-18)C)

E2 = (333 kJ/kg * 1.90 kg) + (4.186 kJ/kgC * 1.90kg * (18-0)C)

set them equal and solve for M

2260M + 343.25M = 632.7 + 143.16

2603.25M = 775.86

M = 775.86/2603.25

M = 0.298 kg