MODULE IN QUANTITY OF HEAT

Heat, Specific Heat and

Change of Phase

By:

Ma. Oliva Diana D. Castro

 INTRODUCTION

Heat is a world familiar to everyone. It plays a very important role in

our daily lives, from conversion of water to steam to freezing of water. The

precise definition of heat and its measurement as well as the quantities that

affect its transfer such as heat capacity needs to be examined in detail.

Likewise, its effect on the change of phase of a substance has to be

understood in order to eliminate the possibility of having misconceptions

among physics students.

This module presents a precise meaning of heat and related concepts.

It consists of two units. Unit I deals with the concept of Heat and Specific

Heat and Unit II deals with the concept of Phase Change. Self Assessment

Questions (SAQs) in the form of problems and other practical applications

of Heat were asked at some points of the module in order to enhance

learning and answers were provided at the end of each unit.

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 PRE- TEST

Directions: Each of the items below is followed by three or more choices.

Select the best answer and write the letter corresponding to corresponding to
your choices. Select the best answer and write the letter corresponding to
your choice on the space provided for in your answer.

MODULE ON QUANTITY OF HEAT

I. Instructions
This module was designed to help you develop your understanding on
Heat, Specific Heat and its Change of Phase. A series of activities has been
prepared for you to undertake as follows:
A. Pre-test on Quantity of Heat to determine your prior views and
misconceptions on the topics;
B. Reading activities on the topic presented step-by-step and as simple
and understandable as possible. Everything has been presented
conceptually. Problem solving are also present in this module.
C. Guide questions and self-assessment questions were also prepared for
you to deeply ponder. This will further enhance your knowledge
about the Quantity of Heat. The module will help you look for the
explanations of questions that seems difficult for you to answer; and
D. A post-test will be given at the end of this activity to determine the
extent the module helped you build your understanding of the topic
on Quantity of Heat.

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II. Purpose and Aim of the Module

The purpose of this module is to provide you with enough information

on Quantity of Heat and eliminate whatever misconceptions you previously

had. After this activity, you can over your prior knowledge on Heat, Specific

Heat and Change in Phase to determine the extent to which you have

changed conceptually. This module aims to:

A. determine your prior views on Heat, Specific Heat and Phase Changes
of Matter;
B. check your misconceptions on Heat, Specific Heat and Phase Changes
of Matter;
C. provide you with enough reading articles that will impart on you the
right concepts associated with Quantity of Heat;
D. provide you the proper attitude for the much needed conceptual
change.

III. Prerequisite Knowledge and Skills

Before this activity it is presumed that you had acquired enough
understanding and problem solving skills on:
A. mass conversion;
B. temperature conversion.

IV. Instructional Objectives

After reading the module, you should be able to:
• define heat as a form of energy

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• define specific heat

• solve problems on quantity heat and phase change and

• apply the concepts of heat, specially heat and latent heat to practical

situation

Unit I – The Concept of Heat and Specific Heat

Heat

Every matter of matter contains internal energy as well as any kinetic or

potential energy it may have because of its motion and position. The internal energy

belongs to the particles of which body is composed and which are in constant motion.

Heat is a term use only with the reference to transfer or flow. In other words, a

heat transfer is energy transfer brought about solely by temperature difference. The

amount of heat transferred maybe described as quantity of heat.

A unit of quantity of heat maybe defined with reference to temperature change of

any particular material.

The calorie (cal) is the amount of heat required to heat 1 g of water from 14.5 ° C

to 15.5 ° C (this leads to the “15 degree calorie”).

The British Thermal Unit (BTU) is the amount of heat needed to rise the

temperature of 1lb of water from 63° F to 64 ° F.

The relations among different heat units are as follows:

1 Btu = 252 cal = 0.252 kcal

The unit of heat is the fundamental unit of energy. Thus:

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 1 cal = 4.186 Joules = 4.186 J

1kcal = 4186 J

1Btu= 778 ff- lb = 1055 J

The international committee on Weight and Measures recommends the Joule to

be used for quantity of heat as well as all other forms of energy.

Specific Heat

Substances react when heat energy is added to or removed from them. The

temperature of a unit mass of a given substance increases by a unit temperature when it

absorbs a certain amount of heat. Conversely, the temperature of the same mass of

substance decreases when it gives off a certain amount of heat.

Various substances require different amounts of heat to raise the temperature of a

given mass of substance by the same amount. For example 1 kcal of energy is required to

raise the temperature of 1 kg of aluminum by 1° C.

Water has high specific heat. It has large capacity to hold a large amount of heat

energy. This is the reason why it is often used for cooling machines.

We often notice that at noon time, the sand has much higher temperature than

water, but in the late afternoon the sand (shore) has lower temperature than water. This

property of a substance is called its specific heat. The specific heat of a substance is the

amount of heat needed to change the temperature of a unit mass of a substance by one

degree. It is actually the capacity to store energy

We express the specific heat of an object in an equation form as:

Q
c= Eq II-1
mΔT

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 Equation II-1 may be written as:

Q = mcΔT Eq. II -2

Where:

Q – heat added or removed from the object

m- mass of the object

c – specific heat of the object

ΔT – temperature change of the object

If Tf and Ti are the objects final and initial temperature respectively, then

ΔT = Tf - Ti Eq. III – 3

Equation II-2 tells us that when the temperature of an object increases (i.e. T f >

Ti) both ΔT and Q are positive. A positive value of Q corresponds to heat energy being

absorbed by the object. Conversely when the temperature of the object decreases (i.e. T f

< Ti) both T and Q are negative. A negative value of Q corresponds to heat energy given

off by object.

Typical values of the specific heat of various substances are given in table I-1.

Table I-1. Specific heat of various substances.

Specific Heat

Substance kcal/kg C° kJ/kg C°

Alcohol 0.580 2.430
Aluminum 0.220 0.920
Concrete 0.700 2.900
Copper 0.093 0.390
Glass 0.200 0.840
Gold 0.030 0.130
Granite 0.190 0.800
Human body 0.830 3.470
Ice 0.500 2.090

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 Iron (steel) 0.113 0.460
Lead 0.031 0.130
Mercury 0.033 0.140
Silver 0.056 0.230
Steam 0.480 2.010
Water 1.000 4.190
Wood 0.420 1.760

Example:

A 325 g sample of an unknown substance cools from 85.5 °C to 25.5 °C when

4.00 kcal is removed from it. What is the specific heat of the unknown substance?

Given:

Mass of sample, m = 32.5 g = 0.325 kg

Initial temperature, Ti = 85.5 °C

Final temperature, Tf = 25.5 °C

Heat energy, Q = 4.0 kcal

The sign Q is negative because heat energy flows out of the substance (its

temperature decreases).

Find: Specific Heat: c = ?

Solution: substituting values of Ti and Tf from equation II -3, to I-1, we get

Q
c= but ΔT = Tf - Ti
mΔT

Working equation:
Q
c=
m (Tf - Ti )
Substitution:
-4.0 kcal
c=
(0.325 kg) (25.5 °C - 85.5 °C)

Answer: c = 0.205 kcal/kg C°

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 SAQ 1. When you take a bite of hot pizza, the sauce can burn your lips but the crust
does not. Why is it so?

SAQ 2. Why is water a better coolant than most liquids?

SAQ 3. Find (a) the energy required to raise the temperature of 100g of water from
25 °C to 55 °C, (b) the heat energy required to raise the temperature of 100 g
of iron through the same temperature interval.

Answers to Self Assessment Questions:

SAQ 1. The sauce which is mostly water has greater specific heat capacity than the crust.

SAQ 2. Because among all liquids water has the greatest value of specific heat. It take

more time for it to cool and to get hot than the most liquids.

SAQ 3. Given:

Mass of water, mw = 100 g = 0.100 kg

Mass of iron, mi = 100 g = 0.100 kg

Initial temperature, Ti = 25 °C

Final temperature, Tf = 55 °C

Specific heat of water, cw = 1.00 kcal/kg C°

Specific heat of iron, ci = 0.113 kcal/kg C°

Find: a) heat energy to raise temp. of water: Qw

b) heat energy to raise temp. of iron: Qi

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 Solution a: to find the heat energy needed to raise the temperature of water, we

use equation II-2

Working Equation: Qw = mwcw (Tf - Ti )

= (0.100 kg) (1.00 kcal/kg C°) (55 °C- 25 °C)
Qw = 3.00 kcal

Solution b: to determine the heat energy added to raise the temperature of water,

we use the same equation as in (a)

Working Equation: Qi = mi ci (Tf - Ti )

= (0.100 kg) (0.113 kcal/kg C°) (55 °C- 25 °C)
Qi = 0.339 kcal

Summary:

Heat is energy in transit. It is used only with reference to transfer or flow. It is the

energy that is transferred which is brought about solely by temperature difference.

The calorie is a unit of heat and it is the amount of heat needed to raise the

temperature of 1 g of water from 14.5 °C TO 15.5 °C. The Btu is the amount of heat

needed to raise the temperature of 1 lb of water from 63 °F to 64 °F.

1 Btu = 252 cal = 0.252 kcal

The Joule is the recommended unit of heat and other forms of energy by the

International Committee on Weights and Measures.

The specific heat capacity of an object is the amount of heat needed to change the

temperature of a unit mass of a substance by one degree. This is actually its capacity to

store energy. Specific heat capacity of a substance is determined by getting the ratio of

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the heat gained (or given off) to the product of the mass and the change in temperature in

formula:

Q
c=
mΔT

Different substances have different specific heat capacities. Among liquids, water

has the greatest specific heat. This is the reason why it is often used as a coolant.

UNIT II- Change of Phase

If we continually add heat to solid or liquid, the solid or liquid will eventually

change phase. A solid will liquefy and a liquid will vaporize. Energy is required for both

liquefaction of solid and vaporization of liquid. Conversely, energy must be extracted

from a substance to change its state in the direction from gas to liquid to solid (see Figure

below).

Energy is absorbed when change of phase is in this direction

SOLID LIQUID GAS

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 Energy is released when change of state is in this direction

The addition or removal of heat energy to or from an object does not always result

in temperature change. When solid (ice) absorbs heat, it changes to liquid (water) and

when liquid gives off heat, it changes to solid (ice). When liquid absorbs heat, it turns to

gas and when gas (vapor) gives off heat, it changes to liquid (water).

Sometimes, a solid can directly change to gas. When solid (dry ice) absorbs heat,

it changes directly to gas.

In this module, we will examine the energy required during the following phase

changes:

Melting – is the change of phase from solid to liquid.

Ex: the turning of ice to water

Freezing – is the change of phase from liquid to solid.

Ex: the freezing of water to ice

Condensation – is the change of phase from gas to liquid.

Ex: water vapor in the air turns to liquid, i.e. water droplets are
seen on surfaces of cold soft drinks bottles

Sublimation – is the change of phase directly from solid to vapor or gas

Ex: evaporation of naphthalene balls

It has been revealed in the concept map at the start of this module that phase

change is reversible process. The amount of heat needed to change it to solid (heat of

fusion). Similarly, the amount of heat needed to change liquid at its boiling point

completely to gas (heat of vaporization) is equal to the amount needed to change vapor to

solid.

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 FUSION

The molecules of a solid are relatively close together. These are rigidly held in

position by molecular force. As the temperature of the solid increases, the molecules

vibrate and move farther apart. Work is done against the binding molecular forces in

order to separate the molecules. This work is total to the heat energy that must be

supplied to the solid (at the melting point) in order to melt it.

The amount of heat that must be added in a substance at its melting point (without

changing its temperature) in order to melt a unit quantity of the substance is called its

latent heat of fusion Lf. If the substance changes from liquid to solid, each kilogram of a

substance gives off (liberates) an amount of energy equal to the latent heat of fusion.

SAQ 1. What really is freezing? Melting?

The S1 unit for latent heat of fusion, Lf is kcal/kg or cal/g. values of latent heat of

fusion for common substance are given in Table II-1.

Table II-1. Values of latent heat of fusion (Lf) and heat of vaporization (Lv) for common
substance at atmospheric pressure.

Melting Lf Boliling Lv
SUBSTANCE Point, °C J/g cal/g Point, °C J/g cal/g
Ethyl alcohol -114 105 25 78 854 204
Copper 1083 134 32 1187 5069 1211
Lead 330 25 509 1170 870 208
Mercury -39 12 208 358 297 71
Nitrogen -210 26 601 -196 201 48
Oxygen -219 14 303 -183 213 51
Silver 961 88 21 2193 2335 558
Tungsten 3410 184 44 5900 4813 1150
Water 0 335 80 100 2260 540

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 The total energy absorbed or liberated by an object undergoing a phase change

from solid to liquid (or liquid to solid) is given by the equation

Qf = m Lf (D1)

Where: Qf - the total heat energy absorbed (or given off) by the object undergoing the

phase change

Qf - is + for melting

Qf - is – for freezing

m – the mass of the object undergoing phase change

Lf – latent heat of fusion of the object undergoing phase change

Example: How much heat is required to completely melt 50 g of ice at 0 °C?

Given: mass of the ice, mi = 50 g = 0.050 kg

Temperature, T = 0 °C

Heat of Fusion, Lf = 80 cal/g

Find: Heat energy required to melt the ice, Qf

Working equation: Qf = m Lf

Substitution: = 50 g x 80 cal/g

Answer: Qf = 4,000 cal = 4 kcal

SAQ 2. What would be the amount of heat needed to change 50 g of water

at 0 °C completely to ice at 0 °C?

VAPORIZATION

As revealed by the concept map at the start of the module, a liquid absorbs latent

heat of vaporization when it is completely changed to gas (vapor) at its boiling point and

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a gas liberates (gives off) latent heat of vaporization when it is completely changed to

liquid.

The latent heat of vaporization Lv, is the amount of heat energy that must be

added to (or removed from) a substance at its boiling point (without temperature change)

in order to vaporize (or condense) it.

If the phase change is from liquid to gas (evaporation), heat energy is absorbed by

the substance. If the phase change is from gas to liquid (condensation), heat energy is

given off by the substance.

The total heat energy absorbed (or given off) by the substance that is undergoing a

phase change from liquid to gas (or from gas to liquid) is given by:

Qv = m L v

Where: Qv - total energy absorbed or liberated by the object undergoing the phase

change

Qv - is + for vaporization
Qv - is – for condensation

m – the mass of the object undergoing phase change

Lv – latent heat of vaporization of the object undergoing phase change

SAQ 3. How does vaporization differ from condensation?

Since the average distance between molecules in a gaseous phase of a substance is

much greater than the average distance between molecules of either liquids or solids, the

latent heat of vaporization for a given substance is greater than its latent heat of fusion.

For example, the latent heat of vaporization of matter is 540 cal/g but its latent heat of

fusion is 80 cal/g.

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Example:

How much heat energy is required to completely vaporize 125 g of ice at 100 °C?

Given: mass of water, mw = 125 g = 0.125 kg

Temperature, T = 100 °C

Heat of Vaporization, Lv = 540 cal/g

Find: Heat energy needed to completely vaporize the water, Qv = ?

Working equation: Qv = m L v

Substitution: = 125 g x 540 cal/g

Answer: Qv = 67,500 cal = 67.5 kcal

SAQ 4. How much amount of heat energy is needed to completely change

125 g of vapor to 100 °C to water?

In sample problem above, note that if only half of the required energy is added,

then only half of the water vaporized. The resulting mixtures would be half water and

half steam both at 100 °C. The temperature of the steam will not increase until all of the

water will vaporize.

Let us examine the processes that take place when a solid such as ice is changed

to steam. Figure D-1 below shows the relationship between heat energy absorbed and

temperature as ice (below 0° C) is heated from solid to gas. Notice that each section of

the graph has a definite slope.

Figure D-1

f
vapor (steam)
warming
100 °C _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ d e
water vaporizing

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 Ice melting water heating
0 °C _ _ _ _b c

ice warms

a
Heat Energy (kcal)

Points a – b = the ice below 0° C and reached 0° C

Qi = mi ci (Tf - Ti )

b-c = the ice at 0° C melts completely no temperature change

Qf = m Lf

c–d = the water at 0° C warms to 100° C

Qw = mw cw (Tf - Ti )

d–e = the water at 100° C completely turns to steam

Qv = m L v

e–f = the steam at 100° C warms when heat is added

Qs = ms cs (Tf - Ti )

In this type of problem, the best procedure is to find the heat energy needed for

each stage. The total required heat is then the sum of the heat required during each stage.

Example: How much heat energy is required to completely change 1 kg of ice at -10° C?

Solution:

T
E
M
P 17
E
R
A
 110 °C_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Qs

100 °C _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Qv

Qw

0 °C _ _ _ _ Qf

Qi

-10 °C
HEAT, kcal

The total required energy is the sum of energies during each stage

Hence: QT = Qi + Qf + Qw + Qv + Qs

But: Qi = mi ci (Tf - Ti )
= (1kg) (0.5 kcal/kg C°) [0° C- (-10° C)]
Qi = 5 kcal

And : Qf = m Lf
= 1 kg x 80 kcal/kg
Qf = 80 kcal

Qw = mw cw (Tf - Ti )
= (1 kg) (1 kcal/kg C°) (100° C - 0° C)
Qw = 100 kcal

And : Qv = m Lv
= (1 kg) (540 kcal/kg)
Qv = 540 kcal

Note: mass of steam is equal to mass of the water

Qs = ms cs (Tf - Ti )
= (1 kg) (0.98 kcal/ kg C°) (110° C - 100° C)
And: Qs = 4.8 kcal

Thus: QT = Qi + Qf + Qw + Qv + Qs

Substitution: QT = 5 kcal + 80 kcal + 100 kcal + 540 kcal + 4.8 kcal

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Answer: QT = 729.8 kcal

SAQ 4. How much heat energy is needed to completely change 10 g of ice

at 0 °C to steam at 100 °C?

SUBLIMATION

Under certain condition, most substances can pass directly from solid to gaseous

phase without passing through the liquid phase. The process is called sublimation. For

example, solid carbon dioxide (dry ice) sublimes at normal atmosphere pressure and

temperatures above -78.5 °C. Under very dry conditions, water (ice) will also sublime.

SUMMARY

The addition or removal of heat from a body does not always cause a change in

temperature. A certain amount of energy is needed to change the phase of a substance

without change in temperature. This amount of heat is called latent heat. The amount of

heat needed to melt a substance completely is the latent heat of fusion and the amount of

heat needed to completely vaporize substance is the latent heat of vaporization.

Phase change is reversible. The amount of heat needed to melt a substance is

equal to the amount of heat needed to completely change the same substance to solid.

The total energy absorbed (or given off) by the object undergoing phase change

from solid or liquid or vise versa is given by the equation: Q f = m Lf. Qf is + for melting

and – for freezing. Similarly, the total energy absorbed (or given off) by the object

undergoing phase change from liquid to vapor or vise versa is given by: Q v = m Lv. Qv is

+ for vaporization and – for condensation.

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 ANSWERS TO SELF ASSESSMENT QUESTIONS (SAQs)

1. Melting is the change in phase from solid to liquid. Conversely freezing is the
change of phase from liquid to solid.
2. – 4 kcal
3. Evaporation is the change of phase from liquid to gas. Conversely, condensation
is the change of phase from gas (steam) to liquid.
4. 67.5 kcal
5.

T
E
M 100 °C _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Qv
P
E
R
A Qw
T
U 0 °C _ _ _ _ Qf
R
E

HEAT, cal

QT = Qf + Qw + Qv
Working Equation: QT = m Lf + mw cw (Tf - Ti ) + m Lv
Substitution: QT = (10 g) (80 cal/g) + (10 g) (1 cal/g C°) (100°C – 0°C) + (10 g)
(540 cal/g)
= 800 cal + 1000 cal + 540 cal
Answer: QT = 7,200 cal
QT = 7.2 kcal

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References:
Beizer, A., 1991. Physics. Massachusets.
Addison-Wesley Publishing Co.

Young, H. and Freedman, R. 1996. University Physics.

Massachusets. Addison-Wesley Publishing Co.

Sears, F. et.al. 1988. University Physics.

Massachusets. Addison-Wesley Publishing Co.

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