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Chapter 12 Thermal Energy. Thermodynamics - The study of heat. Kinetic Theory - All matter consists of minute particles which are in constant motion. Thermal Energy - The overall energy of motion of the particles that make up an object. Temperature Scale. Chapter 12 Thermal Energy.

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chapter 12 thermal energy
Chapter 12 Thermal Energy

Thermodynamics - The study of heat

Kinetic Theory - All matter consists of minute

particles which are in constant motion

Thermal Energy - The overall energy of motion

of the particles that make up an object.

Temperature Scale

chapter 12 thermal energy2
Chapter 12 Thermal Energy

Temperature - The average kinetic energy

of the particles.

Heat - The transfer of thermal energy because of a difference in temperature.

  • Thermal Energy Transfer
  • Conduction - The transfer of energy when particles collide.
  • Convection - The transfer of heat by means of motion in a fluid.
  • Radiation - The transfer of energy by electromagnetic waves.
chapter 12 thermal energy3
Chapter 12Thermal Energy
  • Temperature Scales
  • Kelvin
  • Fahrenheit
  • Celsius

Absolute Zero

-273.15º C or 0 K

Tk = Tc + 273

Absolute Zero - The temperature at which all the

thermal energy has been removed

from the gas.

chapter 12 thermal energy4
Chapter 12 Thermal Energy

Temperatures

Fahrenheit ºF Celsius ºC Kelvin

H20 Boils 212 100 373

H20 Freezes 32 0 273

CO2 Freezes -189 -123 150

Nitrogen Boils -320 -196 77

Absolute Zero -459 -273 0

chapter 12 thermal energy5
Chapter 12 Thermal Energy

1st Law of Thermodynamics – When energy is converted to heat, all energy is conserved.

  • Rumford’s Experiment
    • Disproved Caloric Theory
    • Showed a relationship between heat and work
  • Joules’ Experiment
    • Related heat to energy
    • N(mgh) converts to heat
chapter 12 thermal energy6
Chapter 12 Thermal Energy

2nd Law of Thermodynamics – Heat flows from

hot to cold.

Law of Entropy The universe is continuously

going from a state of order to

disorder.

chapter 12 thermal energy7
Chapter 12 Thermal Energy

ΔQ The transfer of thermal energy, measured in joules

1 calorie = 4.18 joules

  • Specific Heat ( C ) The amount of energy needed to to raise the temperature of a unit mass one temperature unit.
  • Specific Heat is measured in J/kg·K or J/kg·°C
  • Table Pg 279

Heat Transfer Q = mCΔT = mC(Tfinal-Tinitial)

chapter 12 thermal energy8
Chapter 12 Thermal Energy

How much heat is needed to raise 20 grams of water from 40ºC

to 70 ºC?

m = 20 g

C = 4.18 J/g C º

Ti = 40 ºC

Tf = 70 ºC

ΔQ = mCΔT

ΔQ = (20g)(4.18 J/g C)(70 ºC - 40 ºC )

ΔQ = 2500 J

chapter 12 thermal energy9
Chapter 12 Thermal Energy

Find the specific heat of tungsten if it takes 100 joules of energy

to raise 20 grams of the material from 20 ºC to 57 ºC.

m = 20 g

C =

Tf = 57 ºC

Ti = 20 ºC

Q = 100 J

ΔQ = mCΔT

C = Q/(mΔT)

C = 100 J/(20 g*37 ºC)

C = .135 J/g ºC

chapter 12 thermal energy10
Chapter 12 Thermal Energy

149,400 J of heat are added to a 5 kg mass of a substance that raises

the temperature from -25ºC to 20º C. What is the material?

m = 5000 g

C =

Tf = 20 ºC

Ti = -25 ºC

Q = 149,400 J

ΔQ = mCΔT

C = Q/(mΔT)

C = 149,400 J/(5000 g*45 ºC)

C = .644 J/g ºC

The material is glass

chapter 12 thermal energy11
Chapter 12 Thermal Energy

Method of mixtures Heat gained plus heat loss in a closed system

is zero.

ΔQgained+ΔQlost = 0

chapter 12 thermal energy12
Chapter 12 Thermal Energy

What is the final temperature of a mixture where 100 grams of iron

at 80ºC is added to 53.5 g of water at 20ºC?

ΔQgained+ΔQlost = 0

(53.5 g)(4.18 J/gºC)(Tf - 20ºC ) + (100g)(.45 J/g ºC)(Tf - 80ºC) = 0

(223.63 J/ºC)(Tf – 20ºC )+(45 J/ ºC)(Tf – 80 ºC) = 0

Tf = 30ºC

chapter 12 thermal energy13
Chapter 12 Thermal Energy

What is the final temperature of a mixture where 400 grams of

alcohol at 16ºC is added to 400 g of water at 85ºC?

ΔQgained+ΔQlost = 0

(400 g)(2.45 J/g ºC)(Tf - 16 ºC ) + (400g)(4.18 J/g º C)(Tf - 85 ºC) = 0

(980 J/ ºC)(Tf – 16 ºC )+(1672 J/ ºC)(Tf – 85 ºC) = 0

980 Tf – 15680 + 1672 Tf -142120 = 0

2652 Tf = 157800

Tf= 59.5 ºC

slide14

Chapter 12 Thermal Energy

What is the specific heat of a substance that absorbs 2.5 x 103 joules of heat when a sample of 1.0 x 104 g of the substance increases in temperature from 10.0º C to 70.0° C?

chapter 12 thermal energy15
Chapter 12 Thermal Energy

A 1.0 kg sample of metal with a specific heat of 0.50 J/g°C is heated to 100.0º C and then placed in a50.0 g sample of water at 20.0°C.What is the final temperature of the metal and the water?

chapter 12 thermal energy16
Chapter 12 Thermal Energy

A 1.0 kg sample of metal with a specific heat of 0.50 J/g°C is heated to 100.0º C and then placed in a50.0 g sample of water at 20.0°C.What is the final temperature of the metal and the water?

chapter 12 change of state
Chapter 12Change of State

Sublimation

Melting

Vaporization

Solid Liquid Gas

Freezing

Condensation

Supercooled

chapter 12 thermal energy18
Chapter 12Thermal Energy

Heat of Fusion (Hf) The amount of energy needed to melt a unit

mass of a substance. Melting Point

Heat of Vaporization (Hv) The amount of heat needed to vaporize

a unit mass of a liquid. Boiling Point

There is no temperature change in changing between states.

Q = m Hf

Q = m Hv

Chart Pg 287

chapter 12 thermal energy19
Chapter 12Thermal Energy

How much energy is needed to melt 20 grams of ice at 0º?

Q = mHf = (20g)(334 J/g) = 6680 J

How much energy is needed to change 30 grams of water

at 100 ºC to steam?

Q = mHv = (30g)(2260 J/g) = 67800 J

chapter 12 thermal energy21
Chapter 12Thermal Energy

How much energy is needed to melt 40 grams of ice at -60ºC

to steam at 150ºC?

  • Warm the ice Q = mCΔT = (40 g)(2.06 J/gºC)(60ºC) = 4944 J
  • Melt the ice Q = mHf = (40 g)(334 J/ g) = 13360 J
  • Warm the water Q = mCΔT = (40g)(4.18 J/g ºC)(100 ºC) = 16720 J
  • Vaporize the water Q = mHv = (40g)(2260 J/g) = 90400 J
  • Warm the steam Q = mCΔT = (40g)(2.02 J/ gºC)(50ºC) = 4040 J

Qtotal = 4944 J + 13360 J + 16720 J + 90400 J + 4040 J

= 129464 J

chapter 12 thermal energy22
Chapter 12Thermal Energy

How much energy is needed to melt 20 grams of ice at -10ºC

to steam at 130ºC?

  • Warm the ice Q = mCΔT = (20 g)(2.06 J/gºC)(10ºC) = 412 J
  • Melt the ice Q = mHf = (20 g)(334 J/ g) = 6680 J
  • Warm the water Q = mCΔT = (20g)(4.18 J/g ºC)(100 ºC) = 8360 J
  • Vaporize the water Q = mHv = (20g)(2260 J/g) = 45200 J
  • Warm the steam Q = mCΔT = (20g)(2.02 J/ gºC)(30ºC) = 1212 J

Qtotal = 412 J + 6680 J + 8360 J + 45200 J + 1212 J

= 61864 J