# A 50.0 g ball is dropped from an altitude of 2.0 km. Calculate: U i, K max, & W done through the fall.

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A 50.0 g ball is dropped from an altitude of 2.0 km. Calculate: U i, K max, & W done through the fall

Chapter 12 Thermal Energy

Thermodynamics The movement of heat

Kinetic Theory 1)All matter is made up of tiny particles 2)All particles are in constant motion 3)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 ( o C) –Based on MP & BP of water Kelvin (K) –Based of absolute temperature

Temperature Scales K = o C + 273

Convert Temperatures 100 K = ___ o C 100 o C = ___ 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

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

1 st 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 + W But W < 0

Entropy A measure of the disorder in a system

2 nd 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 = mC  T

Calculate the heat required to raise 50.0 g of water from 25.0 o C to 65.0 o C. C water = 4180 J/kgK

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

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

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

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

Conservation of Heat Because the total amount of heat is constant q or  H system = 0

Conservation of Heat q or  H system = 0  H sys =  H 1 +  H 2 +.. q sys = q 1 +  q 2 +..= 0

Conservation of Heat q sys = q 1 +  q 2 = 0 mC  T 1 + mC  T 2 = 0 mC  T 1 = - mC  T 2

Conservation of Heat q sys = q gained + q lost q gained = - q lost mC  T gain = - mC  T lost

A 50.0 g slug of metal at 77.0 o C is added to 500. g water at 25.0 o C. T eq = 27.0 o C. Calculate: C metal C water = 4180 J/kgK

A 200.0 g slug of metal at 77.5 o C is added to 400. g water at 25.0 o C. T eq = 27.5 o C. Calculate: C metal C water = 4180 J/kgK

Solving Mixture Temperatures q system = 0 q system = q hot + q cold mC  T hot = -mC  T cold  T = T f – T i mC(T f – T i ) hot = -mC(T f – T i ) cold

Conservation of Heat mC h T f - mC h T h +mC c T f - mC c T c = 0

Conservation of Heat mC h T f - mC h T h = -mC c T f + mC c T c

20.0 g of water at 25.0 o C is added to 30.0 g water at 75.0 o C. Calculate: T eq C water = 4180 J/kgK

500. g of water at 75.0 o C is added to 300. g water in a 200. g calorimeter all at 25.0 o C. Calculate: T eq C water = 4180 J/kgK C cal = 1000 J/kgK

A 500.0 g slug of metal at 87.5. o C is added to 4.0 kg water in a 1.0 kg can at 25.0 o C. T eq = 27.5 o C. Calculate: C metal C water = 4180 J/kgK C can = 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 (H f ) The heat required to melt one unit mass of a substance at its MP

Heat of Fusion (H f ) H f water = 3.34 x 10 5 J/kg H f water = 334 J/g

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

Heat of Vaporization (H V ) H v water = 2.26 x 10 6 J/kg H v water = 2260 J/g

Change of State q = mH

Changes of State q f = mH f q v = mH v

Calculate the heat required to change 250 g ice to water at its MP: H f = 3.34 x 10 5 J/kg

Calculate the heat required to boil 400 g of water at its BP: H V = 2.26 x 10 6 J/kg

Calculate the heat change when the temperature of 2.0 kg H 2 O is changed from 50 o C to 150 o C:

Calculate the heat change when the temperature of 4.0 kg H 2 O is changed from -25.0 o C to 125.0 o C:

Constants for Water H f = 3.34 x 10 5 J/kg H v = 2.26 x 10 6 J/kg C ice = 2060 J/kgK C water = 4180 J/kgK C steam = 2020 J/kgK

1 st Law of Thermo Total  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

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

20.0 g of lead at 75.0 o C is added to 100.0 g water at 25.0 o C. Calculate: T eq C water = 4180 J/kgK C lead = 130. J/kgK

50.0 g of milk at 5.00 o C is added to 500.0 g coffee in a 400.0 g cup at 75.0 o C. Calculate: T eq C coffee = 4.00 J/gK C cup = 1.50 J/gK C milk = 3.50 J/gK

T i = 25.0 o C T f = 200.0 o C BP = 100.0 o C MP = 0.0 o C Mass of H 2 O = 5.00 kg Calculate: Q total C ice = 2.06 J/gK, H v = 2260 J/g C water = 4.18 J/gK, H f = 334 J/g C steam = 2.02 J/gK

T i = -50.0 o C T f = 300.0 o C BP = 100.0 o C MP = 0.0 o C Mass of H 2 O = 5.00 kg Calculate: Q total C ice = 2.06 J/gK, H v = 2260 J/g C water = 4.18 J/gK, H f = 334 J/g C steam = 2.02 J/gK

20.0 g of lead at 75.0 o C is added to 100.0 g water at 25.0 o C. Calculate: T eq C water = 4180 J/kgK C lead = 130. J/kgK

A 500.0 g slug of metal at 86.5. o C is added to 4.0 kg water in a 2.0 kg can at 24.0 o C. T eq = 26.5 o C. Calculate: C metal C water = 4180 J/kgK C can = 1.0 J/gK

A 50.0 g of ice at -20.0 o C is added to 2.0 kg water in a 1.0 kg can at 25.0 o C. Calculate: T eq C w = 4180 J/kgK C c = 1.0 J/gK C ice = 2.06 J/gK H f = 340 J/g

A 50.0 g of steam at 120.0 o C is added to 2.0 kg water in a 1.0 kg can at 20.0 o C. Calculate: T eq C w = 4180 J/kgK C c = 1.0 J/gK H V = 2260 J/g

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

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