2. Section 15-1 The Nature of Energy Energy is the ability to do work or produce heat.Energy weightless, odorless, tasteless Two forms of energy exist, potential and kinetic. Potential energy is due to composition or position. Chemical potential energy is energy stored in a substance because of its composition. Kinetic energy is energy of motion.

3. Section 15-1 The Nature of Energy (cont.) The law of conservation of energy states that in any chemical reaction or physical process, energy can be converted from one form to another, but it is neither created nor destroyed.law of conservation of energy Heat is energy that is in the process of flowing (transferring) from a warmer object to a cooler object. q is used to symbolize heat.

4. Essentially all chemical reactions and changes in physical state involve either: release of heat, or absorption of heat Endothermic and Exothermic Processes

5. In studying heat changes, think of defining these two parts: the system - the part of the universe on which you focus your attention the surroundings - includes everything else in the universe Together, the system and it’s surroundings constitute the universe Endothermic and Exothermic Processes

6. Heat flowing into a system from it’s surroundings: defined as positive q has a positive value called endothermic system gains heat (gets warmer) as the surroundings cool down Endothermic and Exothermic Processes

7. Heat flowing out of a system into it’s surroundings: defined as negative q has a negative value called exothermic system loses heat (gets cooler) as the surroundings heat up

8. Section 15-1 Measuring Heat A calorie is defined as the amount of energy required to raise the temperature of one gram of water one degree Celsius.calorie Food is measured in Calories, or 1000 calories (kilocalorie). A joule is the SI unit of heat and energy, equivalent to 0.2390 calories.joule 1 calorie = 4.184 J or 1 J = 0.2390 calories

9. Section 15-1 Measuring Heat (cont.)

10. Example: A candy bar has 245 Calories. Convert this to calories and then to Joules of energy.

11. Section 15-1 Specific Heat The specific heat of any substance is the amount of heat required to raise one gram of that substance one degree Celsius.specific heat Some objects require more heat than others to raise their temperature.

12. Section 15-1 Specific Heat (cont.) Calculating heat absorbed and released –q = c × m × ΔT –q = heat absorbed or released (in Joules) –c = specific heat of substance –m = mass of substance in grams –ΔT = change in temperature in Celsius

13. Examples: How much heat does a 20.0 g ice cube absorb as its temperature increases from (-27.0 o C) to 0.0 o C? Give your answer in both joules and calories. q = c × m × ΔT Specific Heat of Ice = 2.03 J/g o C 1 calorie = 4.184 J Specific Heat (cont.)

14. Example Cont.  q = ?  c = 2.03 J/g o C  m = 20.0 grams  ΔT = Final Temp (0.0 o C) – Initial Temp (-27.0 o C) = Change (27.0 o C)  q = c × m × ΔT q = (2.03 J/g o C)(20.0g)(27.0 o C)= Specific Heat (cont.)

15. Example 2: A 5.00 gram sample of a metal is initially at 55.0 ºC. When the metal is allowed to cool for a certain time, 98.8 Joules of energy are lost and the temperature decreases to 11.0º C. What is the specific heat of the metal? What metal is it? q = c × m × ΔT To make the problem easier, solve for the unknown BEFORE you plug in the numbers.

16. For Water during a phase change: –The Heat of Fusion (melting) is 334 j/g –The Heat of Solidification (freezing) is 334 j/g They are the same value (energy in or out) –The Heat of Vaporization is 2260 j/g –The Heat of Condensation is 2260 j/g They are the same value (energy in or out) Measuring Heat

17. Example – Phase change Calculate the amount of energy needed to convert 55.0 grams of ice to all liquid water at its normal melting point. Using the same amount of water calculate the energy needed to completely vaporize the water at its normal boiling point. Why is there such a large difference in energy needed?

18. The solid temperature is rising from -20 to 0 o C (use q = mass x ΔT x C) The solid is melting at 0 o C; no temperature change (use q = mass x ΔH fus. ) The liquid temperature is rising from 0 to 100 o C (use q = mass x ΔT x C) The liquid is boiling at 100 o C; no temperature change (use q = mass x ΔH vap. ) The gas temperature is rising from 100 to 120 o C (use q = mass x ΔT x C) The Heat Curve for Water, going from -20 to 120 o C, 120

19. Section 15-2 Calorimetry Calorimetry - the measurement of the heat into or out of a system for chemical and physical processes. A calorimeter is an insulated device used for measuring the amount of heat absorbed or released in a chemical reaction or physical process.calorimeter Based on the fact that the heat released = the heat absorbed

20. Section 15-2 Chemical Energy and the Universe (cont.) Chemists are interested in changes in energy during reactions. Enthalpy is the heat content of a system at constant pressure.Enthalpy Enthalpy (heat) of reaction is the change in enthalpy during a reaction symbolized as ΔH rxn.Enthalpy (heat) of reaction ΔH rxn = H final – H initial ΔH rxn = H products – H reactants

21. Changes in enthalpy =  H q =  H These terms will be used interchangeably in this textbook Thus, q =  H = m x C x  T  H is negative for an exothermic reaction  H is positive for an endothermic reaction

22. Section 15-2 Chemical Energy and the Universe (cont.)

23. Section 15-2 Chemical Energy and the Universe (cont.)

24. The products are lower in energy than the reactants Thus, energy is released. ΔH = -395 kJ –The negative sign does not mean negative energy, but instead that energy is lost. Exothermic

25. The products are higher in energy than the reactants Thus, energy is absorbed. ΔH = +176 kJ –The positive sign means energy is absorbed Endothermic

26. Chemistry Happens in MOLES  An equation that includes energy is called a thermochemical equation  CH 4 + 2O 2  CO 2 + 2H 2 O + 802.2 kJ 1 mole of CH 4 releases 802.2 kJ of energy. When you make 802.2 kJ you also make 2 moles of water and 1 mole of CO 2

27. Thermochemical Equations  The heat of reaction is the heat change for the equation, exactly as written The physical state of reactants and products must also be given. Standard conditions (SC) for the reaction is (1 atm.) and 25 o C (different from STP)

28. CH 4(g) + 2 O 2(g)  CO 2(g) + 2 H 2 O (l) + 802.2 kJ  If 10.3 grams of CH 4 are burned completely, how much heat will be produced? 10. 3 g CH 4 16.05 g CH 4 1 mol CH 4 802.2 kJ = 514 kJ ΔH = -514 kJ, which means the heat is released for the reaction of 10.3 grams CH 4 Ratio from balanced equation 1 Start with known value Convert to moles Convert moles to desired unit