Question

How much heat is required to convert 5.23 g of ice at − 12.0 ∘C to water at 30.0 ∘C ? (The heat capacity of ice is 2.09 J/g∘C, ΔHvap(H2O)=40.7kJ/mol, ΔHfus(H2O)=6.02kJ/mol)

Answer 1

Mass of ice = 5.23 g

molar mass of water = 18 g/mol

moles of ice = mass/molar mass = 5.23/18 = 0.29 mol

From -12 to 0 ^oC, the state of the sample is solid, i.e. ice.

So, in this temperature range, heat = Q₁ = mcT

m = mass of ice

c = heat capacity = 2.09 J/g ^oC

T = 0-(-12) = 12 ^oC

Q₁ = mcT = 5.23*2.09*12 = 131.17 J

At 0 ^oC, Q₂ = number of moles*H_fus = 0.29*6.02 = 1.746 kJ/mol = 1746 J/mol

From 0 to 30 ^oC, the state of the sample is liquid, i.e. water.

So, in this temperature range, heat = Q₃ = mcT

m = mass of water = mass of ice = 5.23 g

c = heat capacity = 4.18 J/g ^oC

T = 30-0 = 30 ^oC

Q₃ = mcT = 5.23*4.18*30 = 655 J

Total heat required = Q₁ + Q₂ + Q₃ = 131.17+1746+655 = 2532 J = 2.532 kJ