Question

A) Four ice cubes at exactly 0 ∘C with a total mass of 53.5 g are combined with 140 g of water at 85 ∘Cin an insulated container. (ΔH∘fus=6.02 kJ/mol, cwater=4.18J/g⋅∘C)

If no heat is lost to the surroundings, what is the final temperature of the mixture?

B) A sample of steam with a mass of 0.510 g and at a temperature of 100 ∘C condenses into an insulated container holding 4.50 g of water at 2.0 ∘C.( ΔH∘vap=40.7 kJ/mol, Cwater=4.18 J/g⋅∘C)

Assuming that no heat is lost to the surroundings, what is the final temperature of the mixture?

Answer 1

A) Four ice cubes at exactly 0 ∘C with a total mass of 53.5 g are combined with 140 g of water at 85 ∘Cin an insulated container. (ΔH∘fus=6.02 kJ/mol, cwater=4.18J/g⋅∘C)

Hice + Hwater = 0

Hice = melting + cold water = mice*Latent heat + mice*Cwater*(Tf-0)

Hwater = hot water = mwater*Cwater*(Tf-85)

so

mice*Latent heat + mice*Cwater*(Tf-0) + mwater*Cwater*(Tf-85) = 0

53.5*334 + 53.5*4.184 * (Tf) + 140 * 4.184*(Tf-85) =

17869 + 53.5*4.184*Tf +585.76*(Tf-85) = 0

Tf(223.84 +585.76) = -17869 + 585.76*85

Tf = 31920.6 /(223.84 +585.76)

Tf = 39.42 °C