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CHAPTER 11 Heat

Units

• Heat As Energy Transfer • Internal Energy • Specific Heat • Calorimetry – Solving Problems • Latent Heat • Heat Transfer: Conduction • Heat Transfer: Convection • Heat Transfer: Radiation

Heat As Energy Transfer

We often speak of heat as though it were a material that flows from one object to another; it is not. Rather, it is a form of energy. Unit of heat: calorie (cal) 1 cal is the amount of heat necessary to raise the temperature of 1 g of water by 1 Celsius degree. Don’t be fooled – the calories on our food labels are really kilocalories (kcal or Calories), the heat necessary to raise 1 kg of water by 1 Celsius degree.

Units of Heat

• US Customary Unit – BTU • BTU stands for British Thermal Unit

– A BTU is the amount of energy necessary to raise the temperature of 1 lb of water from 63° F to 64° F

• 1 cal = 4.186 J – This is called the Mechanical Equivalent of Heat

James Prescott Joule

• 1818 – 1889 • British physicist • Conservation of Energy • Relationship between heat and other forms of energy

transfer

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If heat is a form of energy, it ought to be possible to equate it to other forms. The experiment below found the mechanical equivalent of heat by using the falling weight to heat the water:

Joule’s experiment

Heat as Energy Transfer

Definition of heat: Heat is energy transferred from one object to another because of a difference in temperature.

• Remember that the temperature of a gas is a measure of the kinetic energy of its molecules.

• In SI units, the unit for heat is the joule.

Example 1: Working off the Calories You decided to eat at the school cafeteria and had 500 Calories over the limit. You want to do an equivalent amount of work climbing stairs or a mountain. How much total height must you climb? Take your mass to be 60kg.

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2

2.1 103600 (11,000 )

(60 )(9.8 / )

W x Jh m ft

mg kg m s

Body only 20% efficient – energy wasted (0.2)(3600m)=700m

Example 2: Working off breakfast. A student eats a breakfast consisting of two bowls of cereal and milk, containing a total of 320 Calories of energy. He wishes to do an equivalent amount of work in the gym by doing curls with a 25.0-kg barbell. How many times must he raise the weight to expend that much energy? Assume that he raises it through a vertical displacement of 0.40-m each time, the distance from his lap to his upper chest. Convert his breakfast Calories, E, to joules

Use the work-energy theorem to find the work necessary to lift the barbell up to its maximum height.

Student must expend same energy up and down -2mgh times number of times n.

6(500 )(4186 / ) 2.1 10kcal J kcal x J

61000 4.186320 1.34 10

1.00

cal JE Cal x J

Cal cal

(0 0) ( 0)W KE PE mgh mgh

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If the student does one repetition every five seconds, it will take him 9.5 hours to work off his breakfast. In exercising, a large fraction of energy is lost through heat, however, due to the inefficiency of the body in doing work. So only a little over 2 hours needed.

Internal Energy

The sum total of all the energy of all the molecules in a substance is its internal (or thermal) energy. Temperature: measures molecules’ average kinetic energy Internal energy: total energy of all molecules Heat: transfer of energy due to difference in temperature Internal energy of an ideal (atomic) gas: But since we know the average kinetic energy in terms of the temperature, we can write:

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2U nRT

If the gas is molecular, rather than atomic, rotational and vibrational kinetic energy need to be taken into account as well.

a) Rotational KE b) Vibrational KE & PE

Specific Heat

• Every substance requires a unique amount of energy per unit mass to change the

temperature of that substance by 1° C • The specific heat, c, of a substance is a measure of this amount.

SI units J / kg °C

(2 )n mgh E

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2

1.34 106840

2 2(25.0 )(9.80 / )(0.400 )

E x Jn times

mgh kg m s m

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The amount of heat required to change the temperature of a material is proportional to the mass and to the temperature change: The specific heat, c, is characteristic of the material. Some values are listed at left.

• Q = m c ΔT • ΔT is always the final temperature minus the initial

temperature • When the temperature increases, ΔT and ΔQ are considered

to be positive and energy flows into the system • When the temperature decreases, ΔT and ΔQ are considered

to be negative and energy flows out of the system

A Consequence of Different Specific Heats

• Water has a high specific heat compared to land • On a hot day, the air above the land warms faster • The warmer air flows upward and cooler air moves toward

the beach

Example 3: How heat transfer depends on specific heat How much heat input is needed to raise the temperature of an empty 20-kg vat made of iron from

10 to 90o oC C ? The specific heat of iron = 450 / oJ kg C

What if the vat was filled with 20 kg of water?

Specific heats of gases are more complicated, and are generally measured at constant pressure (cP) or constant volume (cV).

(20 )(450 / )(80 ) 720o oQ mc T kg J kg C C kJ

(20 )(4186 / )(80 ) 6700o oQ mc T kg J kg C C kJ

720 6700 7400kJ kJ kJ

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Some sample values:

Calorimeter A calorimeter is a vessel that is a good insulator which allows a thermal equilibrium to be achieved between substances without any energy loss to the environment

Calorimetry – Solving Problems Closed system: no mass enters or leaves, but energy may be exchanged Open system: mass may transfer as well Isolated system: closed system where no energy in any form is transferred For an isolated system, Energy out of one part = energy into another part Or: heat lost = heat gained The instrument to the left is a calorimeter, which makes quantitative measurements of heat exchange. A sample is heated to a well-measured high temperature, plunged into the water, and the equilibrium temperature measured. This gives the specific heat of the sample.

Example 4: If 3200cm of tea at 95oC is poured into a

150-g glass cup initially at 25oC , what will be the common final temperature T of the tea and cup when equilibrium is reached? Since tea is mainly water, its specific heat is

4186 / oJ kg C and its mass m is its density times its volume

3 6 3( 200 200 10 )V cm x m 3 3 6 3(1.0 10 / )(200 10 ) 0.20m V x kg m x m kg

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Heat loss by tea = heat gained by cup

( 840J / for glass)o

cupc kg C

(0.20 )(4186J / )(95 ) (0.15 )(840J )( 25 )o o o okg kg C C T kg kg C T C

Example 5: An engineer wishes to determine the specific heat of a new metal alloy. A 0.150-kg

sample of the alloy is heated to540oC . It is then quickly placed in 400 g of water at10oC , which

is contained in a 200-g aluminum calorimeter cup. The final temperature of the system is30.5oC . Calculate the specific heat of the alloy.

(0.40 )(4186J / )(30.5 10.0 )o o okg kg C C C

(0.20 )(900J / )(30.5 10.0 )o o okg kg C C C

76.4 (34,300 3700)J/ )o

ac kg C

500J / o

ac kg C

Phase Changes

• A phase change occurs when the physical characteristics of the substance change from

one form to another • Common phase changes are

– Solid to liquid – melting – Liquid to gas – boiling

• Phase changes involve a change in the internal energy, but no change in temperature

Latent Heat

• During a phase change, the amount of heat is given as – Q = ±m L

• L is the latent heat of the substance – Latent means hidden – L depends on the substance and the nature of the phase change

• Choose a positive sign if you are adding energy to the system and a negative sign if energy is being removed from the system

• SI units of latent heat are J / kg • Latent heat of fusion, Lf, is used for melting or freezing

(95 ) ( 25 )o o

tea tea cup cupm c C T m c T C

79,500J (837J / ) (126J / ) 3150Jo oC T C T 86oT C

heat lost heat gained heat gained by

by alloy by water calorimeter cup

a a a w w w cal cal calm c T m c T m c T

(0.150 )( )(540 30.5 )o o

akg c C C

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• Latent heat of vaporization, Lv, is used for boiling or condensing

Sublimation

• Some substances will go directly from solid to gaseous phase

– Without passing through the liquid phase • This process is called sublimation

– There will be a latent heat of sublimation associated with this phase change Example 6: Boiling Liquid Helium Liquid helium has a very low boiling point, 4.2K, as well as a low latent heat of vaporization,

42.09 10 /x J kg , if energy is transferred to a container of liquid helium at the boiling point from an

immersed electric heater at a rate of 10.0W, how long does it take to boil away 2.00-kg of the liquid? Note: No change of temperature was involved. During such process, the transferred energy goes into changing the state of the substance involved.

Graph of Ice to Steam

Warming Ice

• Start with one gram of ice at –30.0º C • During A, the temperature of the ice changes from –30.0º C to 0º C • Use Q = m c ΔT • Will add 62.7 J of energy

4 4(2.00 )(2.09 10 / ) 4.18 10vQ mL kg x J kg x J 44.18 10

10.0

vmLQ x Jt

P P W

34.18 10 69.7mint x s

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Melting Ice

• Once at 0º C, the phase change (melting) starts • The temperature stays the same although energy is still being added • Use Q = m Lf

• Needs 333 J of energy

Warming Water

• Between 0º C and 100º C, the material is liquid and no phase changes

take place • Energy added increases the temperature • Use Q = m c ΔT • 419 J of energy are added

Boiling Water

• At 100º C, a phase change occurs (boiling) • Temperature does not change • Use Q = m Lv • 2 260 J of energy are needed

Heating Steam

• After all the water is converted to steam, the steam will heat up • No phase change occurs • The added energy goes to increasing the temperature • Use Q = m c ΔT • To raise the temperature of the steam to 120°, 40.2 J of energy are needed

Methods of Heat Transfer

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• Need to know the rate at which energy is transferred • Need to know the mechanisms responsible for the transfer • Methods include

– Conduction – Convection – Radiation

Conduction

• The transfer can be viewed on an atomic scale

– It is an exchange of energy between microscopic particles by collisions – Less energetic particles gain energy during collisions with more energetic particles

• Rate of conduction depends upon the characteristics of the substance

Conduction example

• The molecules vibrate about their equilibrium

positions • Particles near the stove coil vibrate with larger

amplitudes • These collide with adjacent molecules and transfer

some energy • Eventually, the energy travels entirely through the

pan and its handle • In general, metals are good conductors

– They contain large numbers of electrons that are relatively free to move through the metal

– They can transport energy from one region to another • Conduction can occur only if there is a difference in temperature between two parts of the

conducting medium

Conduction, equation

The slab allows energy to transfer from the region of higher temperature to the region of lower temperature.

h cT TQkA

t L

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• A is the cross-sectional area • L = Δx is the thickness of the slab or the length of a rod • P is in Watts when Q is in Joules and t is in seconds • k is the thermal conductivity of the material

– Good conductors have high k values and good insulators have low k values

Convection

• Energy transferred by the movement of a substance

– When the movement results from differences in density, it is called natural conduction

– When the movement is forced by a fan or a pump, it is called forced convection

• Air directly above the flame is warmed and expands • The density of the air decreases, and it rises • The mass of air warms the hand as it moves by

• The radiator warms the air in the lower region of the

room • The warm air is less dense, so it rises to the ceiling • The denser, cooler air sinks • A continuous air current pattern is set up as shown

Radiation

• Radiation does not require physical contact • All objects radiate energy continuously in the

form of electromagnetic waves due to thermal vibrations of the molecules

• Rate of radiation is given by Stefan’s Law • The electromagnetic waves carry the energy

from the fire to the hands • No physical contact is necessary • Cannot be accounted for by conduction or convection

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Energy Absorption and Emission by Radiation

• With its surroundings, the rate at which the object at temperature T with surroundings at To

radiates is – – When an object is in equilibrium with its surroundings, it radiates and absorbs at

the same rate • Its temperature will not change

Ideal Absorbers

• An ideal absorber is defined as an object that absorbs all of the energy incident on it – e = 1

• This type of object is called a black body • An ideal absorber is also an ideal radiator of energy

Ideal Reflector

• An ideal reflector absorbs none of the energy incident on it

– e = 0

Applications of Radiation

• Clothing

– Black fabric acts as a good absorber – White fabric is a better reflector

• Thermography – The amount of energy radiated by an object can be measured with a thermograph

• Body temperature – Radiation thermometer measures the intensity of the infrared radiation from the

eardrum

Resisting Energy Transfer

• Dewar flask/thermos bottle • Designed to minimize energy transfer to surroundings • Space between walls is evacuated to minimize conduction

and convection • Silvered surface minimizes radiation • Neck size is reduced

4 4= net oAe T T

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Global Warming

• Greenhouse example – Visible light is absorbed and re-emitted as infrared radiation – Convection currents are inhibited by the glass

• Earth’s atmosphere is also a good transmitter of visible light and a good absorber of infrared radiation