Question

Part A

Calculate the enthalpy change, ΔH, for the process in which 45.1 g of water is converted from liquid at 17.6 ∘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).

Part B

How many grams of ice at -16.1 ∘C can be completely converted to liquid at 13.3 ∘C if the available heat for this process is 5.83×10³ kJ ?

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

Answer 1

a)

to heat the water from 17.6ºC to 25ºC, heat = mass x specific heat x deltaT
to heat the water to steam at 25ºC, heat = moles water x heat of vaporization

heat = 45.1g x 4.184J/g-ºC x 7.4ºC = 1396.36J =1.39kJ
heat = 45.1/18g/mole x 44kJ/mole = 110.24kJ
total heat = 1.39kJ + 110.24kJ = 111.63kJ

b)

The ice will do three things:

1) go up in temp by 16.1 C (the ice warming from -16.1 to 0)
2) melt at 0 C
3) go up in temp by 13.3 C (the liquid warming from 0 to 13.3)

Here's each set up:

Let’s assume x is the mass of ice completely converted

heat1 = mass x specific heat x deltaT = (x) (2.01) (16.1)
heat2 = moles ice x molar heat of fusion =(x / 18.015) (6010)
heat3 = mass x specific heat water x deltaT water = (x) (4.18) (13.3)


I used molar heat of fusion 6010 in heat 2 part so that the answer from that part would be in Joules, not kilojoules.

Given that The total energy involved is 5.83×10³ kJ= 5.83×10³ ×10³ J

= 5830 ×10³ J

Solve this equation:

5830 ×10³ J = [(x) (2.01) (16.1) + [(x / 18.015) (6010)] + [(x) (4.18) (13.3)]

5830×10³ J = 421.56 x

x= 13.829 ×10³ g

So the mass of ice completely converted = 13.829×10³ g