1. Thermodynamics & Phase Change

2. Heat can only do one thing at a time: either change the temperature OR change the state!

3. Changes of state – seen in phase change diagram temperature heat added

4. Diagonal regions: where thermal energy is added and particles increase in motion (temperature changes) Flat regions: where changes in state occur. All added thermal energy goes to overcome the forces holding the particles together.

5. Heat of fusion – The amount of energy required to melt one kilogram of a substance Heat of vaporization – the amount of energy required to vaporize one kilogram of a substance

6. Heat of Fusion & Vaporization for Water Heat of fusion for water: H f = 3.34x10 5 J/kg positive when melting, negative when freezing Heat of vaporization for water: H v = 2.26x10 6 J/kg positive when evaporating, negative when condensing

7. Heat required to melt: Q = mH f *use for melting & freezing! Heat required to vaporize: Q = mH v *use for evaporating & condensing!

8. Specific Heat Values C ice = 2060 J/kg·˚C C water = 4180 J/kg·˚C C steam = 2020 J/kg·˚C Use these values for temps in Celsius or Kelvin!

9. Example #1 You are asked to melt 0.100 kg of ice at its melting point and warm the resulting water to 20.0 ºC. How much heat is needed?

10. Calculate heat needed to melt the ice. Q 1 = mH f Calculate heat needed to raise water temperature. Q 2 = mCΔT Add all heats together to get the total heat needed. Q = Q 1 + Q 2 Example #1 Strategy:

11. Example #1 Solution: Q 1 = 33400 J Q 2 = 8360 J Q = 41,760 J

12. Example 2 How much energy will be used to convert a 3.5 kg block of ice at 240 K into steam at 390K?

13. Example 2 First, look at your phase change diagram and locate where you are starting in the process and where you are ending.

14. Temperature Heat H f = 3.34 x 10 5 H v = 2.26 x 10 6 Water Phase Change Diagram Solid Water/Ice Q=m(2060)ΔT Liquid Water Q=m(4180)Δt Water Vapor Steam Q=m(2020)Δt Q=m H f Q=m H v ⁰C = K - 273.15 Q=mCΔt Δt=t f -t i mCΔt a = -mCΔt b 0⁰ C 273K 100⁰ C 373K

15. Ex. 2 Cont’d Then write down all the formulas you will need Q=3.5(2060)(33) Q=3.5(3.34x10^5) Q=3.5(4180)(100) Q=3.5(2.26x10^6) Q=3.5(2020)(17)

16. Example #2 A 40.0-g sample of chloroform is condensed from a vapor at 61.6 ºC. It liberates 9,870 J of heat. What is the heat of vaporization of chloroform?

17. Example #2 Solution: Q = mH v -9870 = (.040)(H v ) H v = -246,750 J/kg (negative because it’s cooling)

19. 1 st Law of Thermodynamics The total increase in the thermal energy of a system is the sum of the heat added to it and the work done on it. Heat engines are devices which convert thermal energy to mechanical energy continuously along with some waste heat.

20. 2 nd Law of Thermodynamics Entropy of the universe is always maintained or increased. All things will become more and more disordered unless some action is taken.

21. Entropy is a measure of the disorder in a system. Entropy increases when heat is added and decreases when heat is removed. 2 nd Law of Thermodynamics

22. Thermal Engine efficiency Called Carnot efficiency, or the ideal efficiency of a heat engine (T hot -T cold )/T hot = Ideal efficiency Temperatures must be in Kelvin. Result is a percentage.