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

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 12Thermal Energy

Temperature Scales•Kelvin•Fahrenheit•Celsius

Absolute Zero - The temperature at which all the thermal energy has been removed from the gas.

Absolute Zero -273.15º C or 0 K

Tk = Tc + 273

Temperatures Fahrenheit ºF Celsius ºC KelvinH20 Boils 212 100 373H20 Freezes 32 0 273CO2 Freezes -189 -123 150Nitrogen Boils -320 -196 77Absolute Zero -459 -273 0


•Rumford’s Experiment•Disproved Caloric Theory•Showed a relationship between heat and work

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

•Joules’ Experiment

•Related heat to energy

•N(mgh) converts to heat


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.



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)


How much heat is needed to raise 20 grams of water from 40ºCto 70 ºC?

m = 20 gC = 4.18 J/g C ºTi = 40 ºCTf = 70 ºC

ΔQ = mCΔT

ΔQ = 2500 J

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


Find the specific heat of tungsten if it takes 100 joules of energyto raise 20 grams of the material from 20 ºC to 57 ºC.

m = 20 gC =Tf = 57 ºCTi = 20 ºCQ = 100 J

ΔQ = mCΔT

C = .135 J/g ºC

C = 100 J/(20 g*37 ºC)

C = Q/(mΔT)


149,400 J of heat are added to a 5 kg mass of a substance that raisesthe temperature from -25ºC to 20º C. What is the material?

m = 5000 gC =Tf = 20 ºCTi = -25 ºCQ = 149,400 J

ΔQ = mCΔT

C = .644 J/g ºC

C = 149,400 J/(5000 g*45 ºC)

C = Q/(mΔT)

The material is glass


Method of mixtures Heat gained plus heat loss in a closed systemis zero.

ΔQgained+ΔQlost = 0


What is the final temperature of a mixture where 100 grams of ironat 80ºC is added to 53.5 g of water at 20ºC?


(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


What is the final temperature of a mixture where 400 grams of alcohol at 16ºC is added to 400 g of water at 85ºC?


(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

Tf= 59.5 ºC

980 Tf – 15680 + 1672 Tf -142120 = 0 2652 Tf = 157800

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?

Qmct

c Q

mt

c 2500J

(10000g)(70C 10C)

c .0042J

gC


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 a 50.0 g sample of water at 20.0°C. What is the final temperature of the metal and the water?

Qgained Qlost 0

mctgained mctlost 0

(50g)(4.18J

gC)(TF 20C) (1000g)(.5

J

gC)(TF 100C) 0


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 a 50.0 g sample of water at 20.0°C. What is the final temperature of the metal and the water?

(50g)(4.18J

gC)(TF 20C) (1000g)(.5

J

gC)(TF 100C) 0

209TF 4180 500TF 50000 0

709TF 54180

TF 76C


Chapter 12Change of State

Solid Liquid Gas

Melting

CondensationFreezing

Sublimation

Supercooled

Vaporization


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 vaporizea 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


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 waterat 100 ºC to steam?

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


How much energy is needed to melt 40 grams of ice at -60ºCto steam at 150ºC?

1. Warm the ice Q = mCΔT = (40 g)(2.06 J/gºC)(60ºC) = 4944 J2. Melt the ice Q = mHf = (40 g)(334 J/ g) = 13360 J 3. Warm the water Q = mCΔT = (40g)(4.18 J/g ºC)(100 ºC) = 16720 J4. Vaporize the water Q = mHv = (40g)(2260 J/g) = 90400 J5. 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


How much energy is needed to melt 20 grams of ice at -10ºCto steam at 130ºC?

1. Warm the ice Q = mCΔT = (20 g)(2.06 J/gºC)(10ºC) = 412 J2. Melt the ice Q = mHf = (20 g)(334 J/ g) = 6680 J 3. Warm the water Q = mCΔT = (20g)(4.18 J/g ºC)(100 ºC) = 8360 J4. Vaporize the water Q = mHv = (20g)(2260 J/g) = 45200 J5. 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

chapter 12 thermal energy

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