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Multi-step Problems

Multi-step Problems. Calculate the energy to heat 12.0 g of water from -100.0°C to 200.0°C. q=mc Δ T. q= ∆H x n. q=mc Δ T. q= ∆H x n. q=mc Δ T. 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:

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Multi-step Problems

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  1. Multi-step Problems

  2. Calculate the energy to heat 12.0 g of water from -100.0°C to 200.0°C. q=mcΔT q= ∆H x n 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 = 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!!!!

  4. 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 molx 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 molx 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!!

  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 = 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!!)

  8. 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 (q = Cm∆T) Total energy needed = 221 kJ

  9. 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

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

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