Chapter 10

Heat

Thermal Equilibrium

• Bring two objects into thermal contact.

– They can exchange energy.

• When the flow of energy stops, the objects are in thermal equilibrium.

Temperature• The quantity that tells how hot or

cold something is compared to the standard.

• Measure of average KE of particles

• Increase temp……matter expands

• decrease temp….matter shrinks

Temperature• A method of assigning numbers to objects

that tells us about their thermal equilibrium.

– If the two objects are in thermal equilibrium, they are assigned the same numbers.

– If not, energy flows from the one with the higher temperature to the one with the lower temperature.

Temperature

• Measure of the average KE of any substance.

• Thermal energy is proportional to the amount of matter, temp. is not.

Temperature

• Newton proposed a method in 1701– Used an alcohol-in-glass

thermometer– Mixture of ice and water ---- 0– Human body ----------------- 12– Mark off 12 equal divisions

0

12

Thermometer• Device used to measure

temp.

• Celsius, Kelvin and Fahrenheit

Temperature

• International scale…..Celsius

• 0oC…………..water freezes

• 100oC…………water boils

• United States……..Fahrenheit Scale

• Scientific research is the SI scale…..Kelvin

Temperature

• Gabriel Fahrenheit suggested zero correspond to mixture of ice and salt– Was lowest lab temperature– Avoided use of negative temperatures– Each of 12 degrees divided into 8– Human body -- 96 0

96

Fahrenheit Temperature• Neither of fixed points was

reproducible– fp of pure water at standard

atmospheric pressure = 32– bp of pure water at standard

atmospheric pressure = 212

• Best overall agreement

• Body temperature = 98.6oF

32

212

98.6

Celsius Temperature• Fixed points assigned new values

– 0oC -- freezing point

– 100oC -- boiling point

• Will leave conversions for you to learn.

0

100

37

Kelvin Temperature

• All gas thermometers give a lowest temperature of 273.15oC

• Choose this as the zero point.

• Use better fixed point triple point of water 0.01oC

• 0 K absolute zero273

373

310

Thermal Energy

• The total energy of particles in a material.

• KE and PE

Thermal Energy Units

• calorie (cal) - amount of heat required to raise the temperature of 1 g of water by 1 c.

• Calorie (food) = kilocalorie

• British thermal unit (Btu) - amount of heat required to raise the temperature of 1 lb of water by 1oF.

• joule - same as in mechanics

Mechanical Equivalent of Heat

• Can increase the internal energy of a body by – adding thermal energy– doing work on the body

• Therefore, there must be an equivalence– 1 cal = 4.186 J

Example

• A drop of water falls a distance of 100 m. How much will its temperature rise when it hits the bottom?

• Any assumptions?

mghW

ghm

W 222 /m 980m 100m/s 8.9 s J/kg

Heat• The energy transfer from one

object to another because of a temperature difference between them

• Matter doesn’t contain heat……..• Contains energy…..Heat is energy

in transit

Heat • Thermal Contact…..

• ……..When heat flows from one object to another that it contacts

• Energy that flows from areas that are higher in temp to areas that are lower in temp

• There is NO such thing as cold

Measurement of Heat• To quantify heat, we must specify

the mass and the kind of substance affected

• calorie….The amount of heat required to raise the temperature of 1 gram of water by 1oC

• Calorie………….1000 calories food

Specific Heat (c)• The amount of heat required

to raise the temp of 1mass unit of a substance by 1 temp. unit.

• Thermal inertia…… signifies the resistance of a substance to change in temp.

Computing thermal expansion

•Q = mCT• Q = heat gained or lost (J)

• m = mass (kg)T = change in temp (K)

• C = specific heat (J/kg • K)

Calorimeter

• Device used to measure changes in thermal energy

• it is a closed system to measure energy transfer

Heat of Fusion Hf

• Amount of energy needed to melt one kg of a substance

• ice= 3.34 x 105 J/kg

• ice= 80 cal

Melting Point

• Temp at which a substance changes from a solid to a liquid

• same temp. as freezing point

Heat of Vaporization Hv

• Amount of energy needed to vaporize a kg of liquid

• water = 2.26 x 106 J/kg

• water= 540 cal

Boiling Point

• Temp at which a substance changes from a liquid to a gas

• Same as temp for condensation to occur

• Q=mHf or Q=mHv

• Q….heat required to melt a solid or vaporize a liquid

• m….mass

Thermal Energy

• The total energy of particles in a material.

• KE and PE

Thermal Energy Transfer

• Occurs in 3 ways…

• CONDUCTION

• CONVECTION

• RADIATION

Conduction• Transfer of energy from molecule to

molecule

• Most common in solids

• KE is transferred when particles collide

• MUST have direct contact

Conductors

• Materials that conduct heat well

• Metals……….Silver

Insulators• A material that is a poor conductor

• Delays the transfer of heat

• Examples… wood, wool, straw, paper

• Poor conductor is a good insulator

Convection• Movement of all fluids (gas or liquid)

• Caused by substances having diff. Densities at diff. temps.

• Movement occurs in currents

• Example…….water, air

Radiation• Thermal energy transferred

through space in the form of electromagnetic waves.

• Examples ….Solar energy,

Example

• What does this mean?

• Convert to grams because we know what one cal does to one gram.

J 4.186

cal 1

kg

J980

kg

cal234

g 1000

kg 1

kg

cal234

g

cal234.0 C234.0 o T

Specific Heat• Heat Capacity = amount of heat required to

raise T by 1oC– Not very useful because it is not a property of the

material

• Specific Heat = Heat Capacity per unit mass– Intrinsic property of the material

Tm

Qc

TmcQ

Calorimetry

• Conservation of Energy

• Assume that no energy is lost to the surroundings

• Heat Gained = Heat Lost

Calorimetry

• Assume that we drop a lump of gold (m = 20 g) at 100oC into 50 g of water at 20oC. What is the final equilibrium temperature?

• Heat lost = mgcg(Tg Te)

• Heat gained = mwcw(Te Tw)

– note that both are positive

– equate and solve for Te

Calorimetry

• Do we need to use Kelvin temperatures?

wwgg

wwwggge cmcm

TcmTcmT

o o20 g 129 J/kg K 100 C 50 g 4186 J/kg K 20 C

20 g 129 J/kg K 50 g 4186 J/kg KeT

C21oeT

Calorimetry• Let’s check to see if this is correct.

• It checks! The small difference is due to roundoff.

o

Heat lost

0.02 kg 129 J/kg K 79 C 204 J

g g gm c T

o

Heat gained

0.05 kg 4186 J/kg K 1 C 209 J

w w wm c T

Latent Heat

• It requires heat to change the phase of a substance.

• Latent Heat is an intrinsic property– per unit mass

• Heat of fusion of water = 80 cal/g

• Heat of vaporization of water = 540 cal/g

• Q = mL

Latent Heat• How much heat released in converting 1 g

of steam at 120oC to ice at 30oC?

Cool steam, condense, cool water

TmcQ s

o1 g 0.5 cal/g K 20 C 10 cal

vmLQ cal 540cal/g 540g 1

TmcQ w

o1 g 1 cal/g K 100 C 100 cal

Freeze and cool ice

fmLQ

cal 80cal/g 80g 1

TmcQ i

o1 g 0.5 cal/g K 30 C 15 cal

cal 745total Q J 3120cal 1

J 186.4

Example

• A glass contains 120 g of ice at 0oC. What is the equilibrium temperature if you add 400 g of water at 20oC?

• Heat gained = heat lost

ewwwewifi TTcmTcmLm 0

Example

• What ?!?• Check for mistakes.

– There are no mistakes.

• We made the assumption that all of the ice melted. Actually only 100 g melted!

wwi

fiwwwe cmm

LmTcmT

C67.2 o