A 50. 0 g ball is dropped from an altitude of 2. 0 km A 50.0 g ball is dropped from an altitude of 2.0 km. Calculate: Ui, Kmax, & W done through the fall

Chapter 12 Thermal Energy

Thermodynamics The movement of heat

Kinetic Theory All matter is made up of tiny particles All particles are in constant motion All collisions are elastic

A measure of average kinetic energy Temperature A measure of average kinetic energy

A measure of heat intensity Temperature A measure of heat intensity

Thermal Equilibrium When the average kinetic energy of two or more substances become equal; thus their particles have the same exchange rate

Because it is a measure of average kinetic energy, temperature is related to the motion of particles (atoms, molecules, ions, etc)

Thermometer A device, calibrated to some temp scale, that is allowed to come to thermal equilibrium with something else

Temperature Scales Celsius (oC) Kelvin (K) Based on MP & BP of water Based of absolute temperature

Temperature Scales K = oC + 273

Convert Temperatures 100 K = ___ oC 100 oC = ___ K

A form of energy that flows due to temperature differences Heat A form of energy that flows due to temperature differences

Heat (Q) Because particles at higher temp. move faster than particles at a lower temp., the net flow of heat is Hot to Cold

Heat (Q) Heat will continue to have net flow from H  C as long as there is a temperature difference

Heat (Q) When there is no temperature differences, the system has reached thermal equilibrium

The movement of energy by means other than temperature difference Work The movement of energy by means other than temperature difference

1st Law of Thermo. The increase in thermal energy = sum of heat added & work done to a system

1st Law of Thermo. DE = Q + W

In Most Engines Heat is added by some high energy source (gas) Work is done by the engine

In Most Engines DE = Q + W But W < 0

A measure of the disorder in a system Entropy A measure of the disorder in a system

In natural processes, entropy increases 2nd Law of Thermo. In natural processes, entropy increases

When fuel is burned, entropy is increased

The thermal energy required to raise 1 unit mass of matter 1 degree Specific Heat (C) The thermal energy required to raise 1 unit mass of matter 1 degree

The thermal energy required to raise 1 kg of matter 1 degree K Specific Heat (C) The thermal energy required to raise 1 kg of matter 1 degree K

Heat transfer = mass x specific heat x the temperature change Heat (Q or DH) Heat transfer = mass x specific heat x the temperature change Q = mCDT

Calculate the heat required to raise 50. 0 g of water from 25 Calculate the heat required to raise 50.0 g of water from 25.0oC to 65.0oC. Cwater = 4180 J/kgK

Calculate the heat required to raise 250.0 g of lead from -25.0oC to 175.0oC. Clead = 130 J/kgK

28 kJ of heat was required to raise the temperature of 100 28 kJ of heat was required to raise the temperature of 100.0 g of a substance from -125oC to 575oC. Calculate: C

3. 6 kJ of heat was required to raise the temperature of 10 3.6 kJ of heat was required to raise the temperature of 10.0 g of a substance from -22oC to 578oC. Calculate: C

The total energy of an isolated system is constant Conservation of Heat The total energy of an isolated system is constant

Because the total amount of heat is constant Conservation of Heat Because the total amount of heat is constant q or DHsystem = 0

Conservation of Heat q or DHsystem = 0 DHsys = DH1 + DH2 + .. qsys = q1 + Dq2 + ..= 0

qsys = q1 + Dq2 = 0 Conservation of Heat mCDT1 + mCDT2 = 0

qsys = qgained + qlost qgained = - qlost Conservation of Heat mCDTgain = - mCDTlost

A 50.0 g slug of metal at 77.0 oC is added to 500. g water at 25.0oC. Teq= 27.0oC. Calculate: Cmetal Cwater = 4180 J/kgK

A 200.0 g slug of metal at 77.5 oC is added to 400. g water at 25.0oC. Teq= 27.5oC. Calculate: Cmetal Cwater = 4180 J/kgK

Solving Mixture Temperatures qsystem = 0 qsystem = qhot + qcold mCDThot = -mCDTcold DT = Tf – Ti mC(Tf – Ti)hot = -mC(Tf – Ti)cold

Conservation of Heat mChTf - mChTh +mCcTf - mCcTc = 0

Conservation of Heat mChTf - mChTh = -mCcTf + mCcTc

20. 0 g of water at 25. 0oC is added to 30. 0 g water at 75. 0oC 20.0 g of water at 25.0oC is added to 30.0 g water at 75.0oC. Calculate: Teq Cwater = 4180 J/kgK

500. g of water at 75. 0oC is added to 300. g water in a 200 500. g of water at 75.0oC is added to 300. g water in a 200. g calorimeter all at 25.0oC. Calculate: Teq Cwater = 4180 J/kgK Ccal = 1000 J/kgK

A 500. 0 g slug of metal at 87. 5. oC is added to 4. 0 kg water in a 1 A 500.0 g slug of metal at 87.5.oC is added to 4.0 kg water in a 1.0 kg can at 25.0oC. Teq= 27.5oC. Calculate: Cmetal Cwater = 4180 J/kgK Ccan = 1.0 J/gK

States of Matter Solid Liquid Gas

Solid Has definite size & definite shape Particles vibrate at fixed positions

Liquid Has definite size but no definite shape Particles vibrate at moving positions

Gas Has neither size nor shape Particles move at random

When a substance changes from one state of matter to another Change of State When a substance changes from one state of matter to another

Change of state involves an energy change

Changes of State Melting-Freezing Boiling-Condensation Sublimation-Deposition

Melting Point The temperature at which a solid is at dynamic equilibrium with its liquid. Freezing Point (Same)

Boiling Point The temperature at which a liquid is at dynamic equilibrium with its gas. Condensation Point (Same)

Changes of State During changes of state, the temperature remains constant; all energy is used to change the state

Heat of Fusion (Hf) The heat required to melt one unit mass of a substance at its MP

Heat of Fusion (Hf) Hf water = 3.34 x 105 J/kg Hf water = 334 J/g

Heat of Vaporization (HV) The heat required to vaporize one unit mass of a substance at its BP

Heat of Vaporization (HV) Hv water = 2.26 x 106 J/kg Hv water = 2260 J/g

Change of State q = mH

Changes of State qf = mHf qv = mHv

Calculate the heat required to change 250 g ice to water at its MP: Hf = 3.34 x 105 J/kg

Calculate the heat required to boil 400 g of water at its BP: HV = 2 Calculate the heat required to boil 400 g of water at its BP: HV = 2.26 x 106 J/kg

Calculate the heat change when the temperature of 2 Calculate the heat change when the temperature of 2.0 kg H2O is changed from 50oC to 150oC:

Calculate the heat change when the temperature of 4 Calculate the heat change when the temperature of 4.0 kg H2O is changed from -25.0oC to 125.0oC:

Constants for Water Hf = 3.34 x 105 J/kg Hv = 2.26 x 106 J/kg Cice = 2060 J/kgK Cwater = 4180 J/kgK Csteam = 2020 J/kgK

Total DE equal work done plus heat added to it 1st Law of Thermo Total DE equal work done plus heat added to it DE = Q + W

Heat Engine Any engine that converts heat energy to mechanical energy (Steam, internal combustion, etc.)

Heat Pumps & Refrigerators Use pressure changes & the heat of vaporization to transfer heat from cold to hot

The total entropy of an isolated system always increases 2nd Law of Thermo The total entropy of an isolated system always increases

20. 0 g of lead at 75. 0oC is added to 100. 0 g water at 25. 0oC 20.0 g of lead at 75.0oC is added to 100.0 g water at 25.0oC. Calculate: Teq Cwater = 4180 J/kgK Clead = 130. J/kgK

50. 0 g of milk at 5. 00oC is added to 500. 0 g coffee in a 400 50.0 g of milk at 5.00oC is added to 500.0 g coffee in a 400.0 g cup at 75.0oC. Calculate: Teq Ccoffee = 4.00 J/gK Ccup = 1.50 J/gK Cmilk = 3.50 J/gK

Ti = 25.0oC Tf = 200.0oC BP = 100.0oC MP = 0.0oC Mass of H2O = 5.00 kg Calculate: Qtotal Cice= 2.06 J/gK, Hv = 2260 J/g Cwater= 4.18 J/gK, Hf = 334 J/g Csteam= 2.02 J/gK

Ti = -50.0oC Tf = 300.0oC BP = 100.0oC MP = 0.0oC Mass of H2O = 5.00 kg Calculate: Qtotal Cice= 2.06 J/gK, Hv = 2260 J/g Cwater= 4.18 J/gK, Hf = 334 J/g Csteam= 2.02 J/gK

20. 0 g of lead at 75. 0oC is added to 100. 0 g water at 25. 0oC 20.0 g of lead at 75.0oC is added to 100.0 g water at 25.0oC. Calculate: Teq Cwater = 4180 J/kgK Clead = 130. J/kgK

A 500. 0 g slug of metal at 86. 5. oC is added to 4. 0 kg water in a 2 A 500.0 g slug of metal at 86.5.oC is added to 4.0 kg water in a 2.0 kg can at 24.0oC. Teq= 26.5oC. Calculate: Cmetal Cwater = 4180 J/kgK Ccan = 1.0 J/gK

A 50. 0 g of ice at -20. 0 oC is added to 2. 0 kg water in a 1 A 50.0 g of ice at -20.0 oC is added to 2.0 kg water in a 1.0 kg can at 25.0oC. Calculate: Teq Cw = 4180 J/kgK Cc = 1.0 J/gK Cice = 2.06 J/gK Hf = 340 J/g

A 50. 0 g of steam at 120. 0 oC is added to 2. 0 kg water in a 1 A 50.0 g of steam at 120.0 oC is added to 2.0 kg water in a 1.0 kg can at 20.0oC. Calculate: Teq Cw = 4180 J/kgK Cc = 1.0 J/gK HV = 2260 J/g

Constants will be on the board A 400.0 g of steam at 125.0 oC is added to 2.0 kg ice in a 1.0 kg can at -20.0oC. Calculate: Teq Constants will be on the board