1. Multi-step Problems

2. 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!

3. Calculate the energy to heat 12.0 g of water from -100.0°C to 200.0°C.
Given:
C ice = J/g°C ∆Hfus = 6.02 kJ/mol
Cwater = 4.18 J/g°C ∆Hvap = 40.7 kJ/mol
Csteam = J/g°C
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!!!!

4. Step 1: q = Cm∆T Step 2: ∆Hfus x n Step 3: q = Cm∆T
Calculate the energy to heat 12.0 g of water from °C to 200.0°C.
Given:
C ice = J/g°C ∆Hfus = 6.02 kJ/mol
Cwater = 4.18 J/g°C ∆Hvap = 40.7 kJ/mol
Csteam = J/g°C
Step 1: q = Cm∆T
q = J/g°C x 12.0 g x 100.0°C
q = 2530 J
Step 2: ∆Hfus x n
12.0 g x 1 mol x 6.02 kJ = 4.01 kJ
18.02g 1 mol
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 = J/g°C x 12.0 g x 100.0°C
q = 2.40 x 103 J
Now convert all values to kJ and add together!!

5. How much heat is required to heat 12.0 g of water from -100°C to 200°C?
TOTAL energy required =
41.1 kJ

6. 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!

7. 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 = J/g°C
Specific Heat of water = J/g°C
Specific Heat of steam = J/ g °C
Heat of fusion for ice = 6.02 kJ/mol
Heat of Vaporization for water = 40.7 kJ/mol
Draw a heating curve
Determine the number of steps needed
Determine the equation needed for each step
Find the energy needed for each step
Add all energy values together. (make sure they are all in the same units!!)

8. Calculate the energy required to heat 45. 0 g of water from 37
Calculate the energy required to heat 45.0 g of water from 37.0°C to 103.0°C.
Given:
Specific Heat of ice = J/g°C
Specific Heat of water = 4.18 J/g°C
Specific Heat of steam = J/ g °C
Heat of fusion for ice = 6.02 kJ/mol
Heat of Vaporization for water = 40.7 kJ/mol
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
Total energy needed = 221 kJ

9. Phase Changes
How many joules are required to raise the temperature of g of ice at -10.0˚C to 50.0˚C?
Cice = 20.9 J/g ˚C Cwater = J/g ˚C
ΔHfusion = 6.01 kJ/mol
Answer: kJ

10. (hint: draw a heating curve to help)
How much energy is required to bring 45 g of benzene from - 10˚C to 70˚C?
(hint: draw a heating curve to help)