2. Ability to do work Units– Joules (J), we will use “kJ” Can be converted to different types Energy change results from forming and breaking chemical bonds in reactions

3. 1) Kinetic Energy– energy of “motion” 2) Potential Energy– “stored” energy

5. 1) Open System 1) Closed System 1) Isolated System

7. Energy transfer between a system and the surroundings Transfer is instant from high----low temperature until equilibrium Temperature— Measure of heat, “hot/cold” the average kinetic energy of molecules

8. Kinetic theory of heat Heat increase resulting in temperature change causes an increase in the average motion of particles within the system. Increase in heat results in Energy transfer Increase in both potential and kinetic energies

9. First Law of Thermodynamics Energy is conserved in a reaction (it cannot be created or destroyed)---sound familiar??? Math representation: ΔE total = ΔE sys + ΔE surr = 0 Δ= “change in” ΔΕ= positive (+), energy gained by system ΔΕ= negative (-), energy lost by system Total energy = sum of the energy of each part in a chemical reaction

10. Heat = transfer of energy Temperature = measurement of heat System = the area or space we focus on Surroundings = everything else apart from the system Boundary = separates system and surroundings

11. How do we find the change in energy/heat transfer that occurs in chemical reactions???

12. Experimentally “measuring” heat transfer for a chemical reaction or chemical compound Calorimeter Instrument used to determine the heat transfer of a chemical reaction Determines how much energy is in food Observing temperature change within water around a reaction container ** assume a closed system, isolated container No matter, no heat/energy lost Constant volume

14. Amount of heat required to increase the temperature of 1g of a chemical substance by 1°C Units: cal/g-K or J/g-K 4.184 J = 1 cal, K = 273 + °C Allows us to calculate how much heat is released or absorbed by a substance ! ! ! Unique to each chemical substance Al (s) = 0.901J/g  °K H 2 O (l) = 4.18 J/g  °K

15. q = smΔΤ s/C p = specific heat (values found in reference table) m = mass in grams ΔΤ= change in temperature

16. Q = ? m = 420 g C (H 2 O (l)) = 4.18 J/g  C ΔT = 37-25 = 12  C Q = mc ΔT Q = (420 g)(4.18 J/g  C)(12  C) Q = 21067 J or 21 kJ

17. Q = ? m = 755 g c Fe = 0.45 J/g  C ΔT= 132  C - 12  C = 120  C Q = mc ΔT Q = (755g)(0.45 J/g  C)(120  C) Q = 40,700 J

18. Styrofoam cup with known water mass in calorimeter Assume no heat loss on walls Initial water temp and then chemical placed inside Final temperature recorded Any temperature increase has to be from the heat lost by the substance SOOO All the heat lost from the chemical reaction or substance is transferred to H 2 O in calorimeter

20. q chemical = -q water

21. The specific heat of gold is 0.128 J/g  ° C. How much heat would be needed to warm 250.0 g of gold from 25°C to 100°C?

22. Calorimetry Worksheet

23. Fusion means melting/freezing Vaporization means boiling/condensing H f and H v - amount of energy needed to melt/freeze or boil/condense 1g of a substance Different for every substance – look on reference tables Q = mH f Q = mH v

24. Calculate the mass of water that can be frozen by releasing 49370 J. Calculate the heat required to boil 8.65 g of alcohol (H v = 855 J/g). Calculate the heat needed to raise the temperature of 100. g of water from 25  C to 63  C.

25. The flat points represent a phase change – temperate does not change while a phase change is occurring even though heat is being added. Diagonal points represent the 3 phases