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 ∆Hfus = 6.02 kJ/mol Cwater = 4.18 J/g°C ∆Hvap = 40.7 kJ/mol Csteam = 1.996 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!!!!

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 -100.0°C to 200.0°C. Given: C ice = 2.108 J/g°C ∆Hfus = 6.02 kJ/mol Cwater = 4.18 J/g°C ∆Hvap = 40.7 kJ/mol Csteam = 1.996 J/g°C 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 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 = 1.996 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!!

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 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!!)

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 = 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 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

Phase Changes How many joules are required to raise the temperature of 200.0 g of ice at -10.0˚C to 50.0˚C? Cice = 20.9 J/g ˚C Cwater = 4.184 J/g ˚C ΔHfusion = 6.01 kJ/mol Answer: 150.4 kJ

(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)