Why is water more stable as a solid below 0 o C but as a liquid above it?

What controls why only some things happen?

Energy is neither created or destroyed during chemical or physical changes, but it is transformed from one form to another.  E universe = 0

TYPES of ENERGY KineticPotential MechanicalGravitational ThermalElectrostatic ElectricalChemical Radiant

Dropping a rock? 1.Gravitational  Mechanical 2.Gravitational  Thermal 3.Gravitational  Gravitational

Using a flashlight? 1.Thermal  Radiant 2.Chemical  Thermal 3.Chemical  Radiant 4.Electrostatic  Radiant

Driving a Car?

One turn of a giant windmill produces 0.26 kWh of electricity. How many kJ is this?

Endo: heat added to system Exo: heat released by system SYSTEM and SURROUNDINGS System: The thing under study Surroundings: Everything else in the universe Energy transfer between system and surroundings:

Dissolution of NH 4 NO 3 HCl(aq) + NaOH(aq)  H 2 O(l) + NaCl(aq)

HEAT: What happens to thermal (heat) energy? Three possibilities: Warms another object Causes a change of state Is used in an endothermic reaction

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 Bill, go to the next slide. You know you want to.

Temperature Changes from Heat Exchange 5 g wood at 0 o C + 5 g wood at 100 o C 1.0 o C 2.33 o C 3.50 o C 4.67 o C o C 6.other

Temperature Changes from Heat Exchange 10 g wood at 0 o C + 5 g wood at 100 o C 1.0 o C 2.33 o C 3.50 o C 4.67 o C o C 6.other

Temperature Changes from Heat Exchange 5 g copper at 0 o C + 5 g copper at 100 o C 1.0 o C 2.33 o C 3.50 o C 4.67 o C o C 6.other

Temperature Changes from Heat Exchange 5 g wood at 0 o C + 5 g copper at 100 o C 1.0 o C 2.33 o C 3.50 o C 4.67 o C o C 6.other

Conclusion I: 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

Conclusion II: What controls the magnitude of an object’s temperature change?

Specific Heat Capacity The energy required to heat one gram of a substance by 1 o C. Usefulness: #J transferred = S.H. x #g x  T How much energy is used to heat 250 g water from 17 o C to 100 o C?

Quantitative: Calculating Heat Exchange: Specific Heat Capacity

Calculate the specific heat capacity of copper: Calculate the specific heat capacity of wood:

Experiment! What is the specific heat capacity of this g hunk of metal?

Which has the smallest specific heat capacity? 1.Wood 2.Copper 3.Silver 4.Water

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:

Conceptually Easy Example with Annoying Algebra: 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 (DH fus ) for Water: 333 J/g Boiling: Heat of Vaporization (DH vap ) for Water: 2256 J/g

Total 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 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 Reactions  H = energy needed to break bonds – energy released forming bonds Example: formation of water:  H = ?

Enthalpy 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.”

Experiment! Reaction of HCl + HNO 3

Calorimetry 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