Physics - Ch6 Temperature and Heat

8/10/2019 Physics - Ch6 Temperature and Heat

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DAS 12603 TECHNICAL SCIENCE I

CHAPTER 6:

TEMPERATURE AND HEATNORAIHAN BINTI SALLEH HUDIN

DEPT OF SCIENCE AND MATHEMATICS

CENTRE FOR DIPLOMA STUDIES, UTHM


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LEARNING OBJECTIVES

The objectives of this chapter are:

i. To impart students with the basic knowledge in

science especially heat and thermal properties of

matter.

ii. To apply the concept of heat and thermal

properties in science and engineering courses.


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LEARNING OUTCOMES

After completing this chapter, students should be able to:

explain the definition of temperature and heat.

describe the thermal equilibrium and the zeroth law ofthermodynamics.

convert the temperature unit scales.

explain the specific heat (sensible heat) and latent heat.

calculate the problem of heat by using the calorimetricprinciple.

differentiate latent heat of fusion and latent heat ofvaporization.

solve problem of heat when a matter change from onephase to another phase.

List down three types of mechanisms of heat transfer.


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3.1 TEMPERATURE

Definition: the measure of the degree of hotness

and coldness of matter whether in solids, liquids or

gases.

When two or more objects are in contact with each

other and their temperature is equal, they are said

to be in thermal equilibrium.


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3.1.1 THERMAL EQUILIBRIUM

When two matters are in thermal contact to each other, the

energy (heat) from matter with higher temperature will flow tothe matter with lower temperature.

As energy continues to flow, temperature of the two matterswill approach each other.

At equal temperature, there are no longer net flow of energy

between them. The two matters are said to be in thermal equilibrium.

hotter cooler hotter cooler

thermal equilibrium


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3.1.2 THE ZEROTH LAW OF

THERMODYNAMICS

States that:

If two objects are in thermal equilibrium with a third

object, then the two objects are in thermal

equilibrium with each other.

AC C

B

A B

A in thermal equilibrium with C C in thermal equilibrium with B

Therefore, A in thermal equilibrium with B


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Thermometer Thermometry propertiesMercury thermometer The change in expansion of mercury in glass

column with temperature

Ideal gas thermometer The change in glass pressure with temperature

at constant volume OR the change in gas

volume with temperature at constant gaspressure

Thermocouple The change in electric motion force (emf)

through thin wires of different metals (welded

together at the ends to form two junctions) with

temperature

Electric resistance

thermometer (thermistor)

Electrical resistant changes with temperature

Thermogram The intensity of the radiation emitted by an

object (infrared radiation) increases

substantially as the temperature is raised

Types of thermometer with their thermometry properties

Thermometry propertyof a material: property of a material that is

temperature-dependent (change linearly with the changing of

temperature)


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3.1.3 TEMPERATURE SCALES

Temperature scales were built to determine thetemperature of an object quantitatively.

To built a temperature scale, several standard fixedpoints were used as calibration points, which are:

Ice point: the point where an ice (solid) and water

(liquid) are in equilibrium phase (atmospheric pressure,p= 1 atm)

Steam point: the point where the water (liquid) andsteam (gases) are in equilibrium phase (atmosphericpressure,p= 1 atm)

There are three temperature scales: Celsius (C)

Fahrenheit (F)

Kelvin (K)


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Temperature

scale

Symbol Ice Point Steam

Point

Difference

Celsius C 0 100 100 C

Fahrenheit F 32 212 180 F

Kelvin K 273.15 373.15 100 K

Ice point and steam point of temperature scales

The Kelvin temperature scale has been chosen as the

international standard temperature scale.

Thus, Kelvin (K) is the SI unit for temperature.

The temperature 0 K represents the absolute zero of

temperature (there are no temperature below 0 K)

Therefore, there are no negative temperature number in

Kelvin scale.


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The three temperature scales can be related by the

following equations:

Celsius to Fahrenheit:

Fahrenheit to Celsius:

Celsius to Kelvin:

1.8 32F CT T

32

1.8FC

TT

273.15K CT T


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Example:

A friend suffering from flu has a slight fever. His body temperature is

37.20C. What is his temperature in

i) Kelvinii) degree Fahrenheit, F.

Solution:

i)

Celsius to Kelvin conversion

i) Celsius to Fahrenheit conversion

273.15

37.2 273.15

310.35

K CT T

K

1.8 32

1.8(37.20) 32

98.96

F CT T

F


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3.1.4 DETERMINING TEMPERATURE BY

USING THERMOMETRY PROPERTIES

Thermometry property of a material can be exploited to

determine its temperature.

Firstly, the thermometer scale must be calibrated with

the melting point of 0C and boiling point of 100C of

water at atmospheric pressure of 1 atm.

The value of parameterX0andX100are assigned to the

melting and boiling point respectively.

Plot a graph of a straight line between these two points.


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Any unknown temperature Twith their correspondingparameterXTthat lies along the straight line can bedetermined by the graph or by using the equation:

0

100 0

100TX X

T C

X X

X0 XT X100

100C

TC

0C

Temperature, T

Thermometry

property


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Example:

A resistance of thermometer gives a reading of 30.00at

melting point, 41.58at boiling point and 34.59when

immersed into a beaker filled with oil. Calculate the oils

temperature.

Solution:

0

100

30.00

41.58

34.59T

X

X

X

0

100 0

100

34.59 30.00100

41.58 30.0039.64

TX XT CX X

C

C


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3.2 HEAT AND INTERNAL ENERGY

Heat: an energy that flows from a higher-temperature

object to a lower-temperature object because of the

difference in temperature.

SI unit for heat: Joule (J)

Internal energy (thermal energy): the sum of the

molecular kinetic energy, molecular potential energy and

other kinds of molecular energy in a substance.

Molecular kinetic energy: originates from the random motion (or

vibration) of the molecules.

Molecular potential energy: due to forces that acts between the

atom of a molecule and between molecules.


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3.3 HEAT & TEMPERATURE CHANGE.

3.3.1 HEAT CAPACITY

Heat capacity, C: the amount of heat required to raise the

temperature of a substance by 1 unit (JK-1)

The equation of heat capacity which relates heat, Qand

the temperature change,Tis given by:

Specific heat, c: heat capacity per unit mass (Jkg-1K-1)

QC

T

Qc

m T


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Substance Specific Heat, c

(Jkg-1 C-1)

Aluminium 910

Brass 380Copper 390

Lead 130

Mercury 140

Iron 470

Steel 450

Silver 270

Glass 700

Ice 2100

Water 4200

Steam 2000

Glycerine 2500

Alcohol 2430

Specific heat, cof some common substances


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Example:

How much heat is required to raise the temperature of

200g of mercury from 20C to 100C? Given cmercury

= 140

Jkg-1 C-1.

Solution:

(0.2)(140)(100 20)

2240

final initial

Q mc T

mc T T

J


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Example:

How much heat is required to raise the temperature of

200g of mercury from 250K to 330K? Given cmercury

= 140

Jkg-1 C-1.

Solution:


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Example:

A 1.5 kg material requires 20.25 kJ heat to raise its

temperature from 30C to 80C. State the name of the

material.

Solution:

11270

)3080)()(5.1(20250

CJkgc

CCkgJ

TmcQ


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Substance Specific Heat, c

(Jkg-1 C-1)

Aluminium 910

Brass 380Copper 390

Lead 130

Mercury 140

Iron 470

Steel 450

Silver 270

Glass 700

Ice 2100

Water 4200

Steam 2000

Glycerine 2500

Alcohol 2430

Specific heat, cof some common substances


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Example:

A 1.5 kg material requires 20.25 kJ heat to raise its

temperature from 30C to 80C. State the name of the

material.

Solution:

Therefore the material is silver.

11270

)3080)()(5.1(20250

CJkgC

CCkgJ

TmcQ


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Principle of Conservation of Energy: the heat lost by the

hotter object must equal to the heat gained by the coolerobject.

Heat lost = Heat gained

This principle is based on two assumptions:

No thermal energy lost from the system

No external energy comes into the system

.


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Example:

An aluminium cylinder is heated to 98C and then dropped

into a glass filled with 120g of water with initial temperature

20C. The equilibrium temperature of the mixture is 31C.Calculate the mass of the aluminium cylinder? Neglect the

glass heat capacity.

Solution:

heat lost by aluminium = heat gained by water

(910)(31 98) (0.12)(4200)(31 20)

60970 5544

0.091

Al Al Al w w w

Al

Al

Al

m c T m c T

m

m J

m kg

What if the glass is not negligible?


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To be continued.


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3.4 HEAT AND PHASE CHANGE

There forms of material: solid, liquid or gases.

Solid: atoms or molecules are strongly binding each other

in a rigid crystalline structure by their mutual attraction.

Liquid: molecules have more energy, thus more free to

move.

Gases: molecules have even more energy and are free of

one another.


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Heating a material: temperature of the material will

gradually increaseuntil it reaches a certain stage at which

the phase of the material changes:

Melting point: phase change from solid to liquid

Boiling point: phase change from liquid to gases

Cooling a material: temperature of the material will

gradually decrease until it reaches a certain stage at

which the phase of the material changes: Condensing point: phase change from gases to liquid

Freezing point: phase change from liquid to solid

At melting/freezing point and boiling/condensing point, the

temperature of the phase does not change even if theheat is supplied continuously.

This is because the energy absorbed by the material is

used to separate the molecules further apart than its

previous phase.


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Latent heat: the amount of heat, Qrequired to change

the phase of a unit mass of a substance

Latent heat of fusion, Lf: heat per unit mass for the solid-liquid phase transition (in either direction)

Latent heat of vaporization, Lv: heat per unit mass for the

liquid-gases phase transition (in either direction)

SI unit for latent heat: Jkg-1

QL

m


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Substance Melting

Point

(C)

Latent heat of

fusion, Lf

(kJkg-1)

Boiling

Point

(C)

Latent Heat of

Vaporization, Lv

(kJkg-1)

Aluminium 660 377 2467 11390

Copper 1083 134 2595 5075

Lead 327 25 1620 871

Mercury -39 11.7 357 293

Silver 961 88 2193 2340

Water 0 333 100 2264

Oxygen -219 13.8 -183 214

Nitrogen -210 25.5 -196 200

Table of melting and boiling points of few common substance

with their latent heat of fusion and vaporization


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Temperature, T

Quantity of

heat, Q

100 C

0 C

-20 C

Ice

only

Ice+

water

Water

only

Water

+

steam

Steam

only


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Example:

Calculate the energy required to change 100g of ice at -15C tosteam at 110C.

Explanation:

Calculate the energy according to this order:

To increase temperature from -15C to 0C.

To melt the ice at 0C.

To increase temperature from 0C to 100C.

To vaporize the water into steam at 100C.

To raise the temperature from 100C to 110C.

Hint

To calculate energy required to increase temperature, usespecific heat equation, Q= mcT.

To calculate energy require to change phase, use latent heatequation, Q

L m


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3.5 HEAT TRANSFER

Three principle methods by which heat transfer can

occur:

Conduction

Convection

Radiation


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3.5.1 CONDUCTION

Definition: Conduction is the process of heat flow

that occurs in all forms of material whether in solids,

liquids or gases.

Dominantly occurs in solids.

When cylindrical bar is heated at one end, heat will

flow from the hot end to the cold end by means of

the atomic/molecular collision.

Atoms/molecules that are vibrating at the higher

temperature end collide with atoms/molecules at the

lower temperature end resulting in a net transfer of

heat.


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In metal, conduction process is aided by the motion of

free electrons within the substance. Thus, metals are

good heat conductor. Liquid conduct less heat than solids because the forces

between the atoms are weaker.

Gases are even less efficient as heat conductor because

the atoms of gases are further apart. Hence, in order from the most conductive to the least

conductive:

Solids liquids gases


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A cylindrical bar with length dand uniform cross-sectionAismaintained at temperature Thotand Tcoldat its both end.

Evidences from experiments show that the quantity of heat, Q,transferred per unit time, t, is: Directly proportional to the temperature difference between the two ends;

(T=Thot- Tcold)

Directly proportional to the cross-section area,A.

Inversely proportional to the length, d.

The above statements can be expressed in form of equation:

k= thermal conductivity coefficients of material (Wm-1K-1)

= temperature gradient.

.

A

Direction of heat flow

d

Thot Tcold

hot cold T TQ kAt d

hot cold T T

d


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Substance Thermal Conductivity, k

(Wm-1K-1)

Silver 418

Copper 385

Aluminium 238

Brass 122

Iron 80

Steel 46

Lead 38Mercury 8

Concrete 1.7

Glass (pyrex) 1.1

Brick 1

Water 0.6

Asbestos 0.17

Dry soil 0.14

Wood 0.13

Air 0.023


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Example:

A block of concrete has a cross section of 5m2and

a thickness of 10cm. One side is at 40C and theother is at 20C.What is the rate of heat transfer?

Given kconcrete= 1.7 Wm-1K-1.

Solution:

40 20(1.7)(5)

0.1

1700

hot cold T TQ

kAt d

W


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3.5.2 CONVECTION

Convection of thermal energy occurs in a fluid

(liquids or gases) when warm fluids flow, carrying

energy with them as to replace cooler fluids.

Cannot happen in solids since atoms/molecules in

solids are tightly bound together in their position.

Convection occurs in two forms:

Natural convection, e.g. atmospheric convection and

daily weather variations.

Forced convection, e.g. by means of pumps, fans, etc.to propel fluids and create artificially induced convection

current.


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Heated water at the bottom of the beaker will expandand become more buoyant due to the increase ofvolume per unit mass.

This will cause the density of the hot water to decrease.Hot water will rise towards the upper side of the beaker.

The cooler and denser water from the top will go downto replace the hot water.

So the pattern of circulation is formed within the beakerof water.


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3.5.3 RADIATION Radiation is the heat transfer by the emission of

electromagnetic (EM) waves which carry energy away from theemitting object.

In contrary to the conduction and convection, radiation requiresno medium to transfer heat.

The radiation can travel through vacuum, e.g. from the sun to

the earth. The rate, P, at which an object emits energy via EM radiation is

given by:

(unit: Watt,W)

A= surface area (m2)

T = absolute temperature (K)

= Stefan-Boltzmann constant = 5.67 x 10-8 Wm-2K-4.

= emissivity of the emitting object which has the value

between 0 and 1.

4P AT


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An surface with the maximum emissivity of 1 is said to be

a blackbody radiator, but such surface is ideal and does

not occur in nature.

Any object with absolute temperature above 0K emits

thermal radiation, at the same time it absorbs thermal

energy from surroundings.

Thus, net energy radiated per second by such object isgiven by:

4 4EP A T T


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Example:

A spherical body with diameter of 6.0 cm is maintained at

200C. Assuming that the spherical body is an idealblackbody radiator, at what rate (in Watt) is energy

radiated from the sphere?

Solution:Temperature, T = 200C = 473.15 K

Surface area,A= 4r2= 4(0.03m)2 = 1.13 x 10-2m2

Emissivity for blackbody, = 1

4

8 2(5.67 10 )(1)(1.13 10 )(473)

32

P AT

W


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EXERCISE1. How much heat must be added to 3kg of water to raise its

temperature from 20C to 80C?

2. A total of 0.8kg of water at 20C is placed in a 1.2kW electrickettle. How long a time is needed to raise the temperature of thewater to 100C?

3. In preparing tea, 600g of water at 90C is poured into a 200g chinapot (cpot= 0.84 kJkg

-1C-1) at 20C.What is the final temperature ofthe water?

4. Five kilograms of water at 40C is poured on a large block of ice at0C. How many kilograms of ice melts?

5. 500 kcal of heat is added to 2kg of water at 80C. How muchsteam is produced?

6. Find the minimum amount of ice at -10C needed to bring thetemperature of 500g of water at 20C down to 0C.

7. A 30g ice cubes at 0C is dropped into 200g of water at 30C.What is the final temperature?

8. How much steam at 150C is needed to melt 1kg of ice at 0C?

9. A copper ball of 2cm radius is heated in a furnace at 400C. If itsemissivity is 0.3, at what rate does it radiate energy?

Physics - Ch6 Temperature and Heat

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