Question

a) Calculate the enthalpy change, ΔHΔHDeltaH, for the process in which 10.8 gg of water is converted from liquid at 17.3 ∘C∘C to vapor at 25.0 ∘C∘C . For water, ΔHvapΔHvapH = 44.0 kJ/molkJ/mol at 25.0 ∘C∘C and CsCsC_s = 4.18  J/(g⋅∘C) J/(g⋅∘C) for H2O(l)H2O(l).

b)How many grams of ice at -19.6 ∘C∘C can be completely converted to liquid at 8.9 ∘C∘C if the available heat for this process is 4.83×10³ kJkJ ?

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

Answer 1

(a) enthalpy change = 26.75 kJ

(b) mass of ice = 1.18 x 10⁴ g

Explanation

(a) This process involves two steps

Step 1 : heating of water from initial temperature of 17.3 ^oC to final temperature of 25.0 ^oC

energy required, H1 = (mass of water) * (specific heat of water) * (final temp. - initial temp.)

H1 = (10.8 g) * (4.18 J/g.^oC) * (25.0 ^oC - 17.3 ^oC)

H1 = 347.6 J

H1 = 0.35 kJ

Step 2 : converting all liquid water to vapor at 25 ^oC

energy required, H2 = (mass of water / molar mass of water) * (enthalpy of vaporization)

H2 = (10.8 g / 18.0 g/mol) * (44.0 kJ/mol)

H2 = (0.6 mol) * (44.0 kJ/mol)

H2 = 26.4 kJ

enthalpy change = H1 + H2

enthalpy change = 26.4 kJ + 0.35 kJ

enthalpy change = 26.75 kJ