Question

Calculate the enthalpy change, ΔH, for the process in which 10.3 g of water is converted from liquid at 9.4 ∘C to vapor at 25.0 ∘C .

For water, ΔHvap = 44.0 kJ/mol at 25.0 ∘C and Cs = 4.18  J/(g⋅∘C) for H2O(l).

How many grams of ice at -24.5 ∘C can be completely converted to liquid at 9.8 ∘C if the available heat for this process is 5.03×10³ kJ ?

For ice, use a specific heat of 2.01 J/(g⋅∘C) and ΔHfus=6.01kJ/mol.

Substance	Specific heat [J/(g⋅∘C)]	ΔH (kJ/mol)
water	4.18	44.0
ice	2.01	6.01

Answer 1

1.

Convert ΔHvap = 44.0 kJ/mol = 44.0 kJ/mol /18 g H₂O/mol= 2.44 kJ/g = 2440 j/g

Heat from 9.4 ∘C to 25.0 ∘C (ΔH1) and allow to vaporize (ΔH2).

ΔH = ΔH1 + ΔH2 =

      = m x C_s x ΔT + m x ΔHvap =

       = m ( C_s x ΔT +   ΔHvap ) =

      = 10.3 g (4.18  J/(g⋅∘C) x (25.0-9.4)^oC + 2440 J/g ) = 2505 J = 2.50 kJ     

2.

Convert 6.01 kJ/mol = 6.01 kJ/mol / 18 g/mol = 0.334 kJ/g=334 J/g

You have 3 steps: Heat ice, fuse ice, heat water.

ΔH = ΔH1 + ΔH2 + ΔH3

5.03×10³ kJ = m ( 2.01 J/(g.^oC) x 24.5^oC + 334 J/g + 4.18 J/(g.^oC)x9.8^oC

m = 5030 J / 424.2 J/g = 11.86 g = 11.9 g