Last time…  Course policies  Thermodynamics  Types of Energy  First Law of Thermodynamics (Conservation of Energy)

JkJcalkcalBTUkWh 1 J = x x kJ = x cal = x x kcal = x kWh =3.6 x x x x Energy Units  Read as 1J = cal

Example-Energy Unit Conversion The amount of energy required to pump one sodium ion out of a cell is 114 “milli electron Volts” 114meV=114 x eV 1eV = x J The Chocolate Chip Challenge: How many grams of salt would you need to give you the number of sodium ions that could be transported by the energy in one chocolate chip? (1 chocolate chip has 3.5 Calories)

Temperature and Heat  Heat (q) and Temperature (T) are not the same!  The more thermal energy something has, the greater the motion of its atoms  The total thermal energy in an object is the sum of the individual energies of all the atoms, molecules, or ions

Which statement below best describes the process of placing a thermometer initially at 22ºC into ice water? 1.Some of the thermal energy of the ice water is transferred to the thermometer. 2.Some of the thermal energy of the thermometer is transferred to the ice water, melting some of the ice. 3.The atoms of mercury begin to move faster as a result of the thermal energy transfer between the thermometer and the ice water. 4.The mercury in the thermometer begins to expand as a result of the thermal energy transferred.

What happens to thermal (heat) energy?  Warms another object (transfer)  Causes a change of state  Is used in an endothermic reaction

Heat Transfer  If heat (q) is transferred, in which direction does it go?  From hotter to cooler (related to the 2 nd Law, but we’ll get to that later)  Heat lost = Heat gained (1 st Law)  Thermal equilibrium= when two objects in contact reach the same temperature

System and Surroundings  System = Thing or things being studied  Isolated: Neither energy or matter can be transferred to surroundings  Closed: Energy, but not matter can be transferred to surroundings  Surroundings = Everything else in the universe

Energy transfer between system and surroundings  Internal energy=Energy of a closed system  We can measure changes  Change occurs if heat is transferred to/from system or if work is done on/by system

How to Determine the Sign of q and w  Endothermic- system takes in heat Ammonium thiocyanate and barium hydroxide hydrate  Exothermic- system gives off heat + - System  E=q+w Heat transferred in q>0 Heat transferred in q>0 Work done on w>0 Work done on w>0 Heat transferred out q<0 Heat transferred out q<0 Work done by w<0 Work done by w<0 EndothermicExothermic

Internal Energy Change Example: A gas is compressed and during this process the surroundings do 143 J of work on the gas. At the same time, the gas absorbs 212 J of heat from the surroundings. What is the change in the internal energy of the gas?

Heat Transfer and Specific Heat Capacity  When I heat an object, what happens and how much energy does it require? It depends... 1.Quantity (How much stuff do I have?) 2.Amount of heat energy added 3.Identity of the material  Heat capacity is the energy needed to raise 1g by 1ºC.

How much energy does it take to boil water for tea?

The instructions for baking brownies say to heat the oven to 350ºF if using an aluminum pan, but to heat the oven to 325ºF if using a glass pan. Why is this?  Glass heats up faster because it has a lower heat capacity.  Glass heats more slowly because it has a lower heat capacity.  Aluminum heats more slowly because it has a lower heat capacity.  Aluminum heats up faster because it has a higher heat capacity.

The instructions for baking brownies say to heat the oven to 350ºF if using an aluminum pan, but to heat the oven to 325ºF if using a glass pan. Why is this? 1.Glass heats up faster because it has a lower heat capacity. 2.Glass heats more slowly because it has a lower heat capacity. 3.Aluminum heats more slowly because it has a lower heat capacity. 4.Aluminum heats up faster because it has a higher heat capacity.

Enthalpy

Quantitative: Calculating Heat Exchange: Specific Heat Capacity

Example 1: 5 g wood at 0 o C + 5 g wood at 100 o C Example 2: 10 g wood at 0 o C + 5 g wood at 100 o C Example 3: 5 g copper at 0 o C + 5 g copper at 100 o C Example 4: 5 g wood at 0 o C + 5 g copper at 100 o C Clicker Choices: 1: 0 o C 2: 33 o C 3: 50 o C o C 5: 100 o C 6: other Temperature Changes from Heat Exchange

What happens to thermal (heat) energy?What happens to thermal (heat) energy? When objects of different temperature meet:  Warmer object cools  Cooler object warms  Thermal energy is transferred  q warmer = -q cooler specific heat x mass x  T = specific heat x mass x  T warmer object cooler object

Heat transfer between substances:Heat transfer between substances:

Example  If we mix 250 g H 2 O at 95 o C with 50 g H 2 O at 5 o C, what will the final temperature be?

Thermal Energy and Phase Changes First: What happens?

Thermal Energy and Phase Changes First: What happens?

Thermal Energy and Phase Changes First: What happens?

Warming : Molecules move more rapidly Kinetic Energy increases Temperature increases Melting/Boiling : Molecules do NOT move more rapidly Temperature remains constant Intermolecular bonds are broken Chemical potential energy (enthalpy) increases But what’s really happening?

Energy and Phase Changes: Quantitative Treatment Melting: Heat of Fusion (  H fus ) for Water: 333 J/g Boiling: Heat of Vaporization (  H vap ) for Water: 2256 J/g

Total Quantitative AnalysisTotal Quantitative Analysis Convert 40.0 g of ice at –30 o C to steam at 125 o C Warm ice: (Specific heat = 2.06 J/g- o C) Melt ice: Warm water (s.h. = 4.18 J/g- o C)

Total Quantitative AnalysisTotal Quantitative Analysis Convert 40.0 g of ice at –30 o C to steam at 125 o C Boil water: Warm steam (s.h. = 1.92 J/g- o C)

Enthalpy Change and Chemical ReactionsEnthalpy Change and Chemical Reactions  H = energy needed to break bonds – energy released forming bonds Example: formation of water:  H = ?

Enthalpy Change and Chemical ReactionsEnthalpy Change and Chemical Reactions  H is usually more complicated, due to solvent and solid interactions. So, we measure  H experimentally. Calorimetry Run reaction in a way that the heat exchanged can be measured. Use a “calorimeter.”

Calorimetry ExperimentCalorimetry Experiment N 2 H O 2  2 NO H 2 O Energy released = E absorbed by water + E absorbed by calorimeter E water = E calorimeter = Total E =  H = energy/moles = g N 2 H g water 420 J/ o C

Hess’s Law If reactions can be “added” so can their  H values.

Fig. 5-4, p. 183

Table 5-2, p. 195