Question

A.

Calculate the enthalpy change, ΔH, for the process in which 49.3 g of water is converted from liquid at 18.5 ∘C to vapor at 25.0 ∘C .

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

B.

How many grams of ice at -6.4 ∘C can be completely converted to liquid at 12.3 ∘C if the available heat for this process is 5.33×10³ kJ ?

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

Answer 1

A. dH for overall reaction = dH for water going from liquid to vapor + vaporization

dH for liquid H2O moving from 18.5 oC to 25 oC

dH = mass H2O x specific heat H2O x (Tfinal-Tinitial)

      = 49.3 x 4.18 x (25-18.5)

      = 1.34 kJ

dH for vaporization at 25 oC

dH = mass H2O x Hvap

     = 49.3 x 44 x 1000

     = 2169.2 kJ

dH for reaction = 1.34 + 2169.2 = 2170.54 kJ

B. Let us calculate the indidivual heat needed for this transformation

q1 = heat to raise ice from -6.4 oC to 0 oC

q1 = mass of ice x specific heat of ice x (Tfinal-T(initial)

     = g x 2.01 x (6.4) = g x 12.864 J

q2 = heat to melt ice

q2 = mass ice x heat fusion

     = g x 6.01 x 1000

     = g x 6010 J

q3 = heat to raise temperature of liquid H2O from 0 oC to 12.3 oC

q3 = mass of water x specific heat H2O x (Tfinal-Tinitial)

     = g x 4.18 x (12.3 - (-6.4))

     = g x 78.17 J

Now we have total heat supplied for melting = 5.33 x 10^3 kJ = 5.33 x 10^6 J

5.33 x 10^6 = q1 + q2 + q3 = (12.864 + 6010 + 78.17)g

g = 873.62 g of ice

Thus, it would melt 873.62 g (0.87 kg) of ice to liquid water