Multi-step Problems
Calculate the energy to heat 12.0 g of water from
-100.0°C to 200.0°C.
q= ∆H x n
q=mcΔT
q=mcΔT
q= ∆H x n
q=mcΔT
We have to use both equations! This is a 5 step problem!
• Calculate the energy to heat 12.0 g of water from
-100.0°C to 200.0°C.
• Given:
C ice = 2.108 J/g°C
Cwater = 4.18 J/g°C
Csteam = 1.996 J/g°C
∆Hfus = 6.02 kJ/mol
∆Hvap = 40.7 kJ/mol
• Step 1: Calculate energy needed to heat ice to
the melting point. (q = Cm∆T)
• Step 2: Calculate energy needed to melt ice. (∆Hfus x n)
• Step 3: Calculate energy needed to heat liquid water to boiling point.
(q= Cm ∆T)
• Step 4: Calculate energy needed to boil water. (∆Hvap x n)
• Step 5: Calculate energy needed to heat steam to 200.0°C. (q = Cm ∆T)
• Finally, add all of these energy values together. Make sure that they are
all in the same units!!!!
• Calculate the energy to heat 12.0 g of water
from -100.0°C to 200.0°C.
• Given:
C ice = 2.108 J/g°C
Cwater = 4.18 J/g°C
Csteam = 1.996 J/g°C
∆Hfus = 6.02 kJ/mol
∆Hvap = 40.7 kJ/mol
• Step 1: q = Cm∆T
q = 2.108 J/g°C x 12.0 g x 100.0°C
q = 2530 J
• Step 2: ∆Hfus
xn
12.0 g x 1 mol x 6.02 kJ = 4.01 kJ
18.02g 1 mol
Now convert all values to kJ and add
together!!
• Step 3: q = Cm∆T
q = 4.18 J/g°C x 12.0 g x 100.0°C
q = 5020 J
• Step 4: ∆Hvap x n
12.0 g x 1 mol x 40.7kJ = 27.1 kJ
18.02g 1 mol
• Step 5: q = Cm∆T
q = 1.996 J/g°C x 12.0 g x
100.0°C
q = 2.40 x 103 J
How much heat is required to heat 12.0 g of water
from -100°C to 200°C?
• TOTAL energy required =
41.1 kJ
Helpful Hints!
• Look at heat curve to figure out what equations to use.
• Remember to keep an eye on what phase your specific heat
capacity is for and if you are using Hf or Hv .
• Make sure units are all the same when you add them!
Your turn!
Calculate the energy required to heat 45.0 g of water from 37.0°C to
103.0°C.
• Given:
– Specific Heat of ice = 2.108 J/g°C
– Specific Heat of water = 4.187 J/g°C
– Specific Heat of steam = 1.996 J/ g °C
– Heat of fusion for ice = 6.02 kJ/mol
– Heat of Vaporization for water = 40.7 kJ/mol
1. Draw a heating curve
2. Determine the number of steps
needed
3. Determine the equation needed
for each step
4. Find the energy needed for each
step
5. Add all energy values together.
(make sure they are all in the same
units!!)
Calculate the energy required to heat 45.0 g of water from
37.0°C to 103.0°C.
• Given:
– Specific Heat of ice = 2.108 J/g°C
– Specific Heat of water = 4.18 J/g°C
– Specific Heat of steam = 1.996 J/ g °C
– Heat of fusion for ice = 6.02 kJ/mol
– Heat of Vaporization for water = 40.7 kJ/mol
Total energy needed = 221 kJ
• 3 steps needed
– Heat the water to 100.0°C
(q = Cm∆T)
– Boil the water
(∆Hvap x n)
– Heat the steam to 103.0°C
(q = Cm∆T)