Question

At 1 atm, how much energy is required to heat 85.0 g of H2O(s) at –10.0 °C to H2O(g) at 161.0 °C?

Answer 1

constants that you must use

Heat of fusion of water: ΔHf = 334 J/g
Heat of vaporization of water: ΔHv = 2257 J/g
Specific heat of ice: c = 2.09 J/g∘C;
Specific heat of water: c = 4.18 J/g∘C;
Specific heat of steam: c = 2.09 J/g∘C;

Notice that you must go from ice to vapor, which means that you must go through two phase changes - solid to liquid and liquid to vapor.

The steps that describe your specific transition are

1. Going from ice at -10∘C to ice at 0∘C

q1=m⋅cice⋅ΔT1=85.0 g⋅2.09Jg⋅∘C⋅(0−(−10))∘C

q1= 1776.5J

2. Going from ice at 0∘C to water at 0∘C

q2=m⋅ΔHf=85.0 g⋅334Jg= 28390J

3. Going from water at 0∘C to water at 100∘C

q3=m⋅cwater⋅ΔT3

q3=85.0 g⋅4.18Jg⋅∘C⋅(100−0)∘C= 35530J

4. Going from water at 100∘C to steam at 100∘C

q4=m⋅ΔHv=85.0 g⋅2257Jg= 191845J

5. Going from steam at 100∘C to steam at 161∘C

q5=m⋅csteam⋅ΔT5

q5=85.0 g⋅2.09Jg⋅∘C⋅(161−100)∘C= 10836.65J

The total amount of energy required to go from ice at -10∘C to steam at 161∘C will be the sum of all the calculated energies

qtotal=q1+q2+q3+q4+q5

qtotal= 1776.5J + 28390J + 35530J + 191845J + 10836.65J = 268378.15J

In kJ and rounded to three sig figs, the answer will be

qtotal= 268.378kJ= 268kJ

Energy required= 268 kJ