Chapter Chapter 1010

Ch 10-1: Temperature andThermal Equilibrium

• Determining an object’s temperature with precision requires

• a definition of temperature

• established measurements that determine how “hot” or “cold” objects are

Heat Energy

• Energy must be added to or removed from a substance to change its temperature.

• Temperature is PROPORTIONAL to the kinetic energy of atoms.

Thermal Energy…

• Internal EnergyInternal Energy – the energy of a substance due to the random motions of its component particles and equal to the total energy of those particles.

Internal Energy

• For an ideal gas, the internal energy depends only on the temperature of the gas.

• For gases with 2 or more atoms per molecule, as well as for liquids and solids, other properties besides temperature contribute to the internal energy.

INTERNAL ENERGY

• The symbol, U, stands for internal energy

• Thus, delta U, stands for the change in internal energy

Heating and Cooling

• If an object has become hotterhotter, it means that it has gainedgained heat energy.

• If an object cools downcools down, it means it has

lostlost energy

Thermal Equilibrium

• The state in which two bodies in physical contact with each other have identical temperatures

Thermal Equilibrium

• Ex: When you put a warm coke can into a large beaker of cold water, there is a noticeable temperature difference between the two.

• After about 15 minutes, the can of soda will be cooler and the water surrounding it will be slightly warmer. Eventually, both the can and the water will be at the same temperature.

• This temperature will not change as long

as conditions remain unchanged in

the beaker. This is Thermal EquilibriumThermal Equilibrium.

• Thermal Equilibrium is the basis for measuring temperature with thermometers.

• This is because the thermometer is at the same temperature, or is in thermal equilibrium with, the object.

HEAT ENERGY

What is HEAT?

• Form of energy and measured in JOULES• Particles move about more and take up more room if heated – this is why things expand if heated• It is also why substances change from:

solids liquids gases when heated

Matter and Temperature• Matter Expands as its temperature

increases

• This is known as Thermal Expansion

Heating and Cooling cont…

• Heat energy always moves from:

HOTHOT object COOLER object

e.g. Cup of water at 20 °C in a room at 30°C - gains heat energy and heats up – its temperature rises

Cup of water at 20 °C in a room at 10°C loses heat energy and cools down – its temperature will fall.

Measuring Temperature

• Units depend on scale used:– Fahrenheit, Celsius, and Kelvin (or absolute)

scales.– Fahrenheit commonly used in US– Fahrenheit and Celsius temps can be

converted…

The FahrenheitFahrenheit temperature scale is an English measurement scale. FREEZING POINT:FREEZING POINT: Water Water freezes at 32 freezes at 32 ooFF and and BOILING BOILING POINT:POINT: boils at 2l2 boils at 2l2 ooFF..

The Fahrenheit degree is smaller than the Celsius degree or Kelvin.

On the Celsius temperature scale, (the METRIC SCALEMETRIC SCALE based on based on

the the freezing and boiling point freezing and boiling point of waterof water)

water’s freezing point is 0 oC

& water’s boiling point is l00 oC

The Kelvin (K) scale is based on the boiling and freezing point of water and absolute zero.

The Kelvin scale does not use the degree symbol. The units are Kelvins (K), not degrees Kelvin.

1 Kelvin unit = 1 1 Kelvin unit = 1 ooC scaleC scale..

On this scale the lowest possible On this scale the lowest possible temperature is temperature is absolute zeroabsolute zero, written ”0 K."

At absolute zero, the average the average kinetic energy of particles is kinetic energy of particles is zerozero.

Absolute zero on the Kelvin scale is equal to -273 degrees Celsius.

Temperature Conversion Formulas:

ooCC == K – 273K – 273

K = K = ooC + 273C + 273

ooC = C = ((ooF-32)F-32)1.81.8

ooF = (1.8F = (1.8ooC) + 32C) + 32

HOMEWORK:HOMEWORK:Page 363: #1-3 allPage 363: #1-3 all

THIS WILL BE FOR A GRADE WHEN YOU WALK INTO CLASS TOMORROW!!!

Must show all work

QOTD

• What is the boiling point of water in degrees Fahrenheit?

• What is the freezing point of water in degrees Celsius?

• What is the value of absolute zero?

QOTD

• What is the boiling point of water in degrees Fahrenheit?

• 212 degrees

• What is the freezing point of water in degrees Celsius?

• 0 degrees

• What is the value of absolute zero?• 0 Kelvin

10-1 Review Questions

• Which of the following is true for the water molecules inside popcorn kernels during popping?

A: Their Temperature Increases.

B: They are destroyed.

C: Their Kinetic Energy Increases.

D: Their mass changes.

HEAT and ENERGYHEAT and ENERGY

• HEAT: the energy transfer between two objects because of a difference in their temperatures.

• Energy transferred as heat ALWAYS transfers from an object of a higher temp to that of a lower temp.

Internal Energy

• If the can and the water in the glass reach thermal equilibrium, which has greater internal energy?

Internal Energy

• If the can and the water in the glass reach thermal equilibrium, which has greater internal energy?

• A: The glass of water because of its greater mass

Thermal energy is unique in that it does NOT exist by itself. Thermal energy is present Thermal energy is present ONLY when another form of ONLY when another form of

energy is present.energy is present.

Thermal energy is unique in that it does NOT exist by itself. Thermal energy is present ONLY when another

form of energy is present.

Ex. Thermal energy like body heat only exist from breaking chemical bonds in food. Ex. Heat from the motion of frictionfriction Ex. Electrical items get hot from something resisting the flow of electricity. …….Etc.

Key Points to remember…

• TemperatureTemperature – measures the average KE of molecules in an object

• HEATHEAT – energy transferred from one object to another

• Internal EnergyInternal Energy – sum of the energies of the molecules.

The ccalorie – the quantity of heat needed to raise the temperature of one gram of water 10C.

1 calorie = 4.18 joules

The ccalorie –

The CCalorie (also called a kilocalorie) – often measures the energy content in food 1 CCalorie = 1000 ccalories

There is a differencedifference!

1 CCalorie = 1000 ccalories

There is a differencedifference!

Example item: That means a 190 Calorie bowl of cereal REALLY has 190,000 has 190,000 caloriescalories of energy!

Heat - units of Energy

• Heat Energy (heat flow): Q

• Internal energy: U

calorie (cal) = 4.186 J

kilocalorie (kcal) = 4186 J

Calorie (dietary Calorie) = 1 kcal = 4186 JCalorie (dietary Calorie) = 1 kcal = 4186 J

Thus, for every Calorie you consume you Thus, for every Calorie you consume you actually obtain 4186 J of energyactually obtain 4186 J of energy

Heat and Work

• Friction can increase a substances internal energy

• For solids, internal energy can be increased by deforming their structure.

• EX: Rubber band stretched or a piece of metal bent.

Rubber Band Activity…• Purpose: To demonstrate how work increases an

object’s internal energy.• Procedure: Hold the rubber band between your

thumbs.Touch the middle section of the rubber band to your lip and note how it feels.Rapidly stretch the rubber band and keep it stretched.Touch the middle section of the rubber band to your lip again.Notice is the rubber band’s temperature has changed.

(you may have to repeat this procedure several times before you can clearly distinguish the temperature difference.)

Conservation of Energy

• If changes in internal energy are taken into account along with changes in mechanical energy, the total energy is a universally conserved property

PE + KE + U = 0

Ex: Conservation of Energy

• A 0.10 kg ball falls 10 m onto a hard floor and then bounces back up 9.0 m. How much of its mechanical energy is transformed to the internal energy of the ball and the floor?

ball

Ex: Conservation of EnergyA 0.10 kg ball falls 10 m onto a hard floor and then bounces back up 9.0 m. How much of its mechanical energy is transformed to the internal energy of the ball and the floor?

Givens: m = 0.10 kg g = 9.81 U = ?h = 10.0 m PE = ? KE=?

Formula: PE + KE + U = 0Or

PEi + KEi + Ui = PEf + KEf + Uf

Ex: Conservation of EnergyA 0.10 kg ball falls 10 m onto a hard floor and then bounces back up 9.0 m. How much of its mechanical energy is transformed to the internal energy of the ball and the floor?

Givens: m = 0.10 kg g = 9.81 U = ?h = 10.0 m PE = ? KE= ?

Formula: Delta PE + Delta KE + Delta U = 0or

PEi + KEi + Ui = PEf + KEf + Uf

mg( h) + 0 + Uf-Ui = 0 mg ( h) = U 0.10 (9.81) (1) = U

0.98 J0.98 J

HOMEWORK

• Page 370 Practice: # 1

• Page 370 Review: # 1 and 3

QOTD

1: A substance’s temperature increases as a direct result of:

a. Energy being removed from the particles of the substance

b. Kinetic Energy being added to the particles of the substance

c: A change in the number of atoms and molecules in a substance

QOTD CONT

2: What happens to the internal energy of an ideal gas when it is heated from 0 degrees Celsius to 4 degrees Celsius?

a: increases

b: decreases

c: stays the same

d: impossible to determine

QOTD cont

• Which of the following is proportional to the KE of atoms and molecules?

a: Elastic Energy

b: Thermal Equilibrium

c: Potential Energy

d: Temperature

QOTD

• Heat Flow occurs between two bodies in thermal contact when they differ in which of the following properties?

A: mass

B: density

C: temperature

D: specific heat

QOTD

• Which of the following is equivalent to 88 degrees Fahrenheit?

A: 49 degrees C

B: 31 degrees C

C: 16 degrees C

D: 58 degrees C

10-3: Changes in temperature and phase

• Specific Heat Capacity:the quantity of energy needed to raise the temperature of 1 kg of substance by 1 degree Celsius at constant pressure.

Specific Heat Capacity (c)

• Relates mass, temperature change, and energy transferred as heat.

• Specific heat capacity equation:

cp = Q/m(T)

Where the subscript p represents heat capacity measured at a constant pressure

Specific Heat

• Though heat and temperature are not the same thing, there is a correlation between the two, captured in a quantity called specific heat, c.

• Specific heat measures how much heat is required to raise the temperature of a certain mass of a given substance.

See page 372

• Table 10-4 Specific Heat capacities

• **You must know cp for water = 4.186 x 103 J/(kg * C)

Calorimetry

• An experimental procedure used to measure the energy transferred from one substance to another as heat.

The higher the specific heat, the more energy the material contains when it is at the same temperature as a material with a lower specific heat.Ex. 2 kg of Copper ( 385 J/kg*oC)

2 kg of Glass [664 J/kg*oC ]

If both are 20 oC, Glass has more energy!

Determining Specific Heat Capacity• For simplicity, a subscript w will always stand for

“water” in problems involving specific heat capacities.

Energy absorbedabsorbed by water = energy releasedreleased by the substance

Or

Qw = Qx

Where Q = mc ΔTQ = mc ΔT

Specific Heat Capacity

• The equation applies to both substances that absorb energy AND those that transfer energy to their surroundings.

• When temp increases; change in T and Q are taken to be positive.

• When temp decreases; they are negative and energy is transferred from the substance.

Specific Heat

• Specific heat units: J/kg ·ºC or cal/g ·ºC.

• Every substance has a different specific heat, but specific heat is a constant for that substance.

• For instance, the specific heat of water, cwater , is 4.19 x 103 J/kg · ºC or 1 cal/g · ºC. That means it takes 4.19 x 103 joules of heat to raise one kilogram of water by one degree Celsius.

• Substances that are easily heated, like copper, have a low specific heat, while substances that are difficult to heat, like rubber, have a high specific heat.

SPECIFIC HEAT

• Specific heat allows us to express the relationship between heat and temperature in a mathematical formula:

Q = mc TQ is the heat transferred to a material

m is the mass of the material,

c is the specific heat of the material

T is the change in temperature.

Calculating Thermal EnergyCalculating Thermal EnergyQ = mc ΔTQ = change in thermal energy

Joules ( J )

m = mass (kg )

c = specific heat ( J/kgoC ) ΔT = change in temperature oC

Thermal Energy Problem A.Ex. How much thermal energy is needed to raise the temp. of 20 kg of iron from 35oC up to 50oC if the specific heat for iron is 450 J/kgoC ?given: formula: sub: answer & unit

Q = Q = mc Q = mc ΔΔTT

m=

c =

ΔT =ΔT =

Thermal Energy Problem A.Ex. How much thermal energy is needed to raise the temp. of 20 kg of iron from 35oC up to 50oC if the specific heat for iron is 450 J/kgoC ?

given: formula: sub: answer & unit

Q = Q Q = mc Q = mc ΔΔTT

m= 20 kg

c = 450 J/kgoC

ΔT = ΔT = 50 - 35 = 15 oC

Thermal Energy Problem A.Ex. How much thermal energy is needed to raise the temp. of 20 kg of iron from 35oC up to 50oC if the specific heat for iron is 450 J/kgoC ?

given: formula: sub: answer & unit

Q = Q Q = m c Q = m c ΔΔTT

m= 20 kg

C = 450 J/kgoC

ΔT = 50 - 35 = 15ΔT = 50 - 35 = 15 oC

BE SURE TO BE SURE TO WRITE THE WRITE THE UNITS IN UNITS IN YOUR NOTES !YOUR NOTES !

Thermal Energy Problem A.Ex. How much thermal energy is needed to raise the temp. of 20 kg of iron from 35oC up to 50oC if the specific heat for iron is 450 J/kgoC ?

given: formula: sub: answer & unit

Q = Q Q = mc Q = mc ΔΔTT

m= 20 kg Q = 20 x 450 x 15Q = 20 x 450 x 15

C = 450 J/kgoC

ΔT = 50 - 35 = 15ΔT = 50 - 35 = 15 oC

Thermal Energy Problem A.Ex. How much thermal energy is needed to raise the temp. of 20 kg of iron from 35oC up to 50oC if the specific heat for iron is 450 J/kgoC ?

given: formula: sub: answer & unit

Q = Q Q = mc Q = mc ΔΔTT

m= 20 kg Q = 20 x 450 x 15 Q = 20 x 450 x 15 = 135,000 J

C = 450 J/kgoC

ΔT = 50 - 35 = 15ΔT = 50 - 35 = 15 oC

A 0.05 kg metal bolt is heated to an unknown initial temperature. It is then dropped into a beaker containing 0.15 kg of water with

an initial temperature of 21.0 C. The bolt and the water then reach a final temperature of 25.0 C. If the metal has a specific heat capacity of 899 J/kg *C, find the initial temperature of the

metal.

Givens: m metal = 0.05 kg cp, m = 899 J/kg *C

m water = 0.15 kg cp, w = 4186 J/kg *C

T water = 21.0 C T final = 25.00 C

Unknown T metal = ?

A 0.05 kg metal bolt is heated to an unknown initial temperature. It is then dropped into a beaker containing 0.15 kg of water with an initial temperature of 21.0 C. The bolt and the water then reach a final temperature of 25.0 C. If the metal has a specific heat capacity of 899 J/kg *C, find the initial temperature of the metal.

Givens: m metal = 0.05 kg cp, m = 899 J/kg *C

m water = 0.15 kg cp, w = 4186 J/kg *C

T water = 21.0 C T final = 25.00 C

Unknown T metal = ?Unknown T metal = ?

energy removed from the metal = energy absorbed by water

Qm = Qw

mmcp, m ΔΔTTm = = mw cp, w ΔΔTTw

ΔΔTTm = = mw cp, w ΔΔTTw

mmcp, m

A 0.05 kg metal bolt is heated to an unknown initial temperature. It is then dropped into a beaker containing 0.15 kg of water with an initial temperature of 21.0 C. The bolt and the water then reach a final temperature of 25.0 C. If the metal has a specific heat capacity of 899 J/kg *C, find the initial temperature of the metal.

Unknown T metal = ?Unknown T metal = ?

energy removed from the metal = energy absorbed by water

Qm = Qw

mmcp, m ΔΔTTm = = mw cp, w ΔΔTTw

ΔΔTTm = = mw cp, w ΔΔTTw

mmcp, m

Next step…Plug and Chug!

A 0.05 kg metal bolt is heated to an unknown initial temperature. It is then dropped into a beaker containing 0.15 kg of water with an initial temperature of 21.0 C. The bolt and the water then reach a final temperature of 25.0 C. If the metal has a specific heat capacity of 899 J/kg *C, find the initial temperature of the metal.

Unknown T metal = ?Unknown T metal = ?

energy removed from the metal = energy absorbed by water

Qm = Qw

mmcp, m ΔΔTTm = = mw cp, w ΔΔTTw

ΔΔTTm = = mw cp, w ΔΔTTw

mmcp, m

Not done yet….looking for INITIAL TEMP OF THE METAL!!!

ΔΔTTm = (0.15) (4186) (4) = (0.15) (4186) (4) = 56 = 56 o oCC

(.05) (899)(.05) (899)

A 0.05 kg metal bolt is heated to an unknown initial temperature. It is then dropped into a beaker containing 0.15 kg of water with an initial temperature of 21.0 C. The bolt and the water then reach a final temperature of 25.0 C. If the metal has a specific heat capacity of 899 J/kg *C, find the initial temperature of the metal.

ΔΔTTm = (0.15) (4186) (4) = (0.15) (4186) (4) = 56 = 56 o oCC

(.05) (899)(.05) (899)

Δ Δ TTm = T = Tf + + Δ Δ TTm

Δ Δ TTm = = 25.0 C + 56.0 CΔ Δ TTm ==81.0 81.0 C

EXAMPLE

• 4190 J of heat are added to 0.5 kg of water with an initial temperature of 12ºC. What is the temperature of the water after it has been heated?

• By rearranging the equation above, we can solve for :T = Q/(mc)T = Q/(mc)

• The temperature goes up by 2 Cº, so if the initial temperature was 12ºC, then the final temperature is 14ºC. Note that when we talk about an absolute temperature, we write ºC, but when we talk about a change in temperature, we write Cº.

Homework Part 1

• Page 374 #1,2