Chapter 12

Temperature and Heat

12.1 Common Temperature Scales

Temperatures are reported in degreesCelsius or degrees Fahrenheit.

Temperature changes, on theother hand, are reported in Celsiusdegrees or Fahrenheit degrees:

F 59C 1 =

100 oC or 212 oF

𝑇𝑇 ℃ − 0100 − 0

=𝑇𝑇 ℉ − 32℉(212 − 32)

Kelvin Scale

12.2 The Kelvin Temperature Scale

15.273+= cTT

Kelvin temperature

𝑇𝑇 𝐾𝐾 = 𝑇𝑇 ℃ + 273.15

0 K = -273.15 oCAbsolute Zero

12.1 Common Temperature Scales

Example 1 Converting from a Fahrenheit to a Celsius Temperature

A healthy person has an oral temperature of 98.6oF. What would thisreading be on the Celsius scale?

F 6.66F32F98.6 =−

degrees above ice point

( )

C 0.37

F C 1F 6.66

59

=

C 0.37

12.1 Common Temperature Scales

Example 2 Converting from a Celsius to a Fahrenheit Temperature

A time and temperature sign on a bank indicates that the outdoor temperature is -20.0oC. Find the corresponding temperature onthe Fahrenheit scale.

( )

F 0.36

C 1F C 0.20 5

9

=

degrees below ice point

F0.4F 0.36F 0.32 −=−

ice point

What is a healthy person’s oral temperature? (a) in oF(b) in 0C(c) In K

98.6 oF 37 0C

What is a Boiling water temperature ? (a) in oF(b) in 0C

What is water freezing temperature? (a) in oF(b) in 0C

212 oF

32 oF

100 0C

0 0C

310 K

373 K

273 K

12.4 Linear Thermal Expansion

NORMAL SOLIDS

12.4 Linear Thermal Expansion

oLL ∝∆

12.4 Linear Thermal Expansion

LINEAR THERMAL EXPANSION OF A SOLID

The length of an object changes when its temperature changes:

TLL o∆=∆ α

coefficient of linear expansion

Common Unit for the Coefficient of Linear Expansion: ( ) 1CC1 −=

12.4 Linear Thermal Expansion

12.4 Linear Thermal Expansion

THE EXPANSION OF HOLES

Conceptual Example 5 The Expansion of Holes

The figure shows eight square tiles that are arranged to form a square patternwith a hold in the center. If the tiled are heated, what happens to the size of the hole?

12.4 Linear Thermal Expansion

A hole in a piece of solid material expands when heated and contracts whencooled, just as if it were filled with the material that surrounds it.

12.5 Volume Thermal Expansion

VOLUME THERMAL EXPANSION

The volume of an object changes when its temperature changes:

TVV o∆=∆ β

coefficient of volume expansion

Common Unit for the Coefficient of Volume Expansion: ( ) 1CC1 −=

12.6 Heat and Internal Energy

DEFINITION OF HEAT

Heat is energy that flows from a higher-temperature object to a lower-temperature object because of a difference in temperatures.

SI Unit of Heat: joule (J)

The heat that flows from hot to cold originates in the internal energy ofthe hot substance.

It is not correct to say that a substancecontains heat.

12.7 Heat and Temperature Change: Specific Heat Capacity

SOLIDS AND LIQUIDS

HEAT SUPPLIED OR REMOVED IN CHANGING THE TEMPERATUREOF A SUBSTANCE

The heat that must be supplied or removed to change the temperature ofa substance is

TmcQ ∆=

specific heatcapacity

Common Unit for Specific Heat Capacity: J/(kg·Co)

12.7 Heat and Temperature Change: Specific Heat Capacity

12.7 Heat and Temperature Change: Specific Heat Capacity

Example 9 A Hot Jogger

In a half-hour, a 65-kg jogger can generate 8.0x105J of heat. This heatis removed from the body by a variety of means, including the body’s owntemperature-regulating mechanisms. If the heat were not removed, how much would the body temperature increase?

TmcQ ∆=

( ) ( )[ ]CkgJ3500kg 65J100.8 5

⋅×

==∆mcQT

C 5.3

12.7 Heat and Temperature Change: Specific Heat Capacity

GASES

The value of the specific heat of a gas depends on whether the pressure orvolume is held constant.

This distinction is not important for solids.

OTHER UNITS

1 kcal = 4186 joules

1 cal = 4.186 joules

12.8 Heat and Phase Change: Latent Heat

During a phase change, the temperature of the mixture does not change (provided the system is in thermal equilibrium).

12.8 Heat and Phase Change: Latent Heat

HEAT SUPPLIED OR REMOVED IN CHANGING THE PHASEOF A SUBSTANCE

The heat that must be supplied or removed to change the phaseof a mass m of a substance is

mLQ =

latent heat

SI Units of Latent Heat: J/kg

12.8 Heat and Phase Change: Latent Heat

12.8 Heat and Phase Change: Latent Heat

Example 14 Ice-cold Lemonade

Ice at 0oC is placed in a Styrofoam cup containing 0.32 kg of lemonadeat 27oC. The specific heat capacity of lemonade is virtually the same asthat of water. After the ice and lemonade reach an equilibriumtemperature, some ice still remains. Find the mass of the melted ice. Assume that mass of the cup is so small that it absorbs a negligible amount of heat.

( ) ( )

lemonadebylost Heat

lemonade

ice meltedby gainedHeat

iceTcmmLf ∆=

12.8 Heat and Phase Change: Latent Heat

( ) ( )

lemonadebylost Heat

lemonade

ice meltedby gainedHeat

iceTcmmLf ∆=

( )

( )[ ]( )( ) kg 11.0kgJ103.35

C0C27kg 32.0CkgJ4186

L

5

f

lemonade melted ice

=×

−⋅=

∆=

Tcmm

For Recitations Ch. 12FOC Q: 2,4, 7, 9 & 11.Problems: 11, 13, & following:

HWAssignment#4 will be active at 10 AM is due

by Thursday 11 PM

Exam# 4 on June 9: Ch. 12 to Ch. 17

Problem: How much heat is required to obtain 5.0 kg of steam of 120 oC from 5.0 kg of ice at -20 oC temperature. Assuming thatThe heat capacity of water = 4186 J/kgCO

The heat capacity of steam = 4186 J/kgCO

The heat capacity of ice = 2000 J/kgCO

Latent heat of fusion = 33.5 × 104𝐽𝐽𝑘𝑘𝑘𝑘

Latent heat of evaporation = 22.6 × 105𝐽𝐽/𝑘𝑘𝑘𝑘

Problem: How much heat is required to make 5 kg steam at 120 oC from 5 kg ice that was kept at -20 oC?Given that:

Specific heat capacity of water = 4186 J/kg/Co

Specific heat capacity of ice = 2000 J/kg/Co

Latent heat of evaporation is 2.26 *106 J/kgLatent heat of fusion of water is 3.35*105 J/Kg

Ice

-5 oC

Ice0 oC

Water Water Steam Steam

0 oC 100 oC 100 oC 110 oCPhase change Phase change

ΔT = 5 Co ΔT = 100 Co ΔT = 10 Co

Q = Q1 + Q2 + Q3 + Q4 + Q5

1.57 × 107𝐽𝐽