A 50.0 g ball is dropped from an altitude of 2.0 km.

Calculate: Ui, Kmax, & W done through the fall

Chapter 12Thermal Energy

Thermodynamics

•The movement of heat

Kinetic Theory1)All matter is made up of

tiny particles2)All particles are in

constant motion3)All collisions are elastic

Temperature•A measure of average kinetic

energy

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• Celcius (oC)

–Based on MP & BP of water

• Kelvin (K)–Based of absolute temperature

Temperature Scales

•K = oC + 273

Convert Temperatures

100 K = ___ oC

100 oC = ___ K

Heat•A form of energy

that flows due to temperature differences

Heat (Q)•Because particle at higher temp. move

faster than particles at a lower temp., the net

flow of heat is H C

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

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.

E = Q + W

In Most Engines•Heat is added by some

high energy source (gas)

•Work is done by the engine

In Most Engines

E = Q + WBut W < 0

Entropy

•A measure of the disorder in a

system

2nd Law of Thermo.

•In natural processes,

entropy increases

Entropy

•When fuel is burned, entropy is

increased

Specific Heat (C)•The thermal energy

required to raise 1 unit mass of matter

1 degree

Specific Heat (C)•The thermal energy

required to raise 1 kg of matter 1

degree K

Heat (Q or H)•Heat transfer = mass

x specific heat x the temperature change

Q = mCT

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 100.0 g of a substance from

-125oC to 575oC. Calculate: C

3.6 kJ of heat was required to raise the

temperature of 10.0 g of a substance from -22oC to 578oC.

Calculate: C

Conservation of Heat

•The total energy of an isolated system

is constant

Conservation of HeatBecause the total amount of heat is

constant

q orHsystem = 0

Conservation of Heatq orHsystem = 0

Hsys = H1 + H2 + ..

qsys = q1 + q2 + ..= 0

Conservation of Heatqsys = q1 + q2 = 0

mCT1 + mCT2 = 0

mCT1 = - mCT2

Conservation of Heatqsys = qgained + qlost

qgained = - qlost

mCTgain = - mCTlost

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

mCThot = -mCTcold

T = 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. Calculate: Teq

Cwater = 4180 J/kgK

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/kgKCcal = 1000 J/kgK

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/kgKCcan = 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

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

another

Change of State•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.

•Condensationing 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

Temperature vs Heat Plot

160

210

260

310

360

410

460

510

0 50 100 150 200 250 300 350 400

Heat (kJ)

Tem

pera

ture

(K)

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 = 2.26 x 106 J/kg

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 4.0 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

1st Law of ThermoTotal E equal work done

plus heat added to it

E = 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

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. Calculate: Teq

Cwater = 4180 J/kgKClead = 130. J/kgK

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/gKCcup = 1.50 J/gKCmilk = 3.50 J/gK

Ti = 25.0oC Tf = 200.0oC BP = 100.0oC MP = 0.0oCMass of H2O = 5.00 kgCalculate: Qtotal

Cice= 2.06 J/gK, Hv = 2260 J/gCwater= 4.18 J/gK, Hf = 334 J/gCsteam= 2.02 J/gK

Ti = -50.0oC Tf = 300.0oC BP = 100.0oC MP = 0.0oCMass of H2O = 5.00 kgCalculate: Qtotal

Cice= 2.06 J/gK, Hv = 2260 J/gCwater= 4.18 J/gK, Hf = 334 J/gCsteam= 2.02 J/gK

20.0 g of lead at 75.0oC is added to 100.0 g

water at 25.0oC. Calculate: Teq

Cwater = 4180 J/kgKClead = 130. J/kgK

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/kgKCcan = 1.0 J/gK

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 1.0 kg can at 20.0oC.

Calculate: Teq

Cw = 4180 J/kgKCc = 1.0 J/gKHV = 2260 J/g

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