In: Chemistry
At 1 atm, how much energy is required to heat 89.0 g of H2O(s) at –12.0 °C to H2O(g) at 173.0 °C?
Data.
P = 1atm (standard condition)
mass H2O (s) = 89 g
To = -12°C
Tf = 173°C = 446 K
s = 4.184 J/g.°C
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Answer.
The equation to calculate the heat change is given by;
q = msΔT
where;
m = mass of sample
s = specific heat of water
ΔT = temperature change
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This procces must face two transitions:
From solid water (ice) to liquid
H2O (s) ==> H2O (l)
and, from liquid to vapor
H2O (l) ==> H2O (g)
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In this process we'll find the energy required by using heat change formula:
Step 1. (From ice to liquid)
q = (89 g)(4.184 J/g.°C)[0°C - (-12°C)]
q = 4.468 KJ
Step 1. (From ice to liquid)
q = (89 g)(4.184 J/g.°C)[0°C - (-12°C)]
q = 4.468 KJ
Step 2. (From liquid to vapor)
q = (89 g)(4.184 J/g.°C)[100°C - 0°C)]
q = 37.273 KJ
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q = (89 g)(4.184 J/g.°C)[173°C - 100°C)]
q = 27.183 KJ
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Additionally to this you have to sum up the latent heat of fusion for ice and vaporization for water on 89 g of water.
Cice = m Lf = 89 g x 334 J/g = 29726 J = 29.726 KJ
Cwater = m Lv = 89 g x 2257 J/g = 200873 J = 200.873 KJ
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Total energy required to Heat 89 g of water from -12°C to 173°C is
qt = 4.468 KJ + 37.273 KJ + 27.183 KJ = 68.92 KJ + 29.726 KJ + 200.873 KJ
qt = 299.523 KJ.
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