Heat Content (internal thermal energy)  
Heat content is the energy possessed by a body due to its temperature.  
o
Factors that determine the heat Content  
•
You might have felt that the land is cool in the morning and hot during day time. But, water in a  
lake will be almost at a particular temperature both in the morning as well as in the afternoon.  
Both are subjected to same amount of heat energy from the Sun, but they react differently. It is  
because both of them have different properties. In general, the amount of heat energy absorbed  
or lost by a body is determined by three factors.  
•
(i)  
Mass of the body – The amount of heat energy required is directly proportional to the mass  
of the substance.  
(ii)  
(iii)  
Nature of material of the body – The amount of heat energy required depends on the  
nature on the substance and it is expressed in terms of its specific heat capacity C.  
Rise in temperature of the body – The amount of heat energy required is directly  
proportional to the rise in temperature.  
Heat Capacity of a Substance  
✓
Heat Capacity – Is the amount of heat required to raise the temperature of a given mass of a  
substance by 1Kelvin (K). It is denoted by letter  
C,  
✓
The SI unit for the heat capacity of a substance is Joule per Kelvin (J/K or JK−1). When the  
temperature is increased by 1°C is the same as an increase of 1 K, which is the same unit as  
J/°C  
.
✓
Mathematically  
:
풄풉풂풏품풆 풊풏 풕풆풎풑풆풓풂풕풖풓풆(∆휽)  
Worked Examples,  
1. In an experiment to determine the heat capacity of steel, 100KJ of heat energy was supplied to  
a block of steel initially at 22°C. If the final temperature of the block was 2190 C, determine the  
heat capacity of steel.  
Soln:  
Given: Q = 100 KJ = 100000 J, T1 = 220C, T2 = 2220 C, C =?  
From: Heat capacity, C = 풒풖풂풏풕풊풕풚 풐풇 풉풆풂풕 풂풃풔풐풓풃풆풅(푸)  
풄풉풂풏품풆 풊풏 풕풆풎풑풆풓풂풕풖풓풆(∆휽)  
100000  
∴
Heat capacity, C =  
= 500 J/K  
222−22  
Class Activity – 8:1  
1. Define the term heat capacity. 3000 J of heat is lost when the temperature of an iron rod  
reduces from 500 C to 30.50 C. Determine its heat capacity. (ANS: C = 153.85 J/K)  
A solid with a heat capacity of 320 J/K requires 2000 J of heat to raise its temperature to 800 C.  
2.  
Find its original temperature (ANS: T1 = 73.750 C)  
3. An iron ball requires 5000 J heat energy to raise its temperature by 20ºC. Calculate the heat  
capacity of the iron ball. [ANS; 250J/K]  
Specific heat Capacity  
•
Consider the following two observations below  
(i) Quantity of heat required to raise the temperature of 1 litre of water will be more than the  
heat required to raise the temperature of 500 ml of water. If Q is the quantity of heat  
absorbed and  
(ii) Quantity of heat energy (  
풎
is the mass of the body, then; 푸 ∝ 풎  
푸
) required to raise the temperature of 250 ml of water to 100 0C  
0
is more than the heat energy required to raise the temperature to 500 C. Here, 푸 ∝ ∆푻  
,
where ΔT is the change in temperature of the body.  
•
Hence, from the above two observations, heat lost or gained by a substance when its  
temperature changes by ΔT is; 푸 ∝ 풎푻  
•
•
When removing the proportionality sign, then; Q = mCΔT  
From the above equations, the absolute temperature and energy of a system are proportional  
to each other. The proportionality constant is the specific heat capacity (푪) of the substance.  
•
In order to understand the specific heat capacity of the substance, think of heating 500 ml of  
water and 500 ml of oil. Which will be heated first? Why? It is because heat gained by a body  
depends upon the nature of the substance (specific heat capacity), i.e,.. A substance with low  
specific heat capacity heats up and cools down rapidly while a substance with high  
specific heat capacity heats up and cools down slowly  
•
•
The capacity of a substance to gain heat energy is denoted by the term specific heat capacity.  
Mathematically it is derived from the equation above as,  
풉풆풂풕 풆풏풆풓품풚 풔풖풑풑풍풊풆풅(푸)  
Specific heat capacity, C  
=
풎풂풔풔 × 풕풆풎풑풆풓풂풕풖풓풆 풄풉풂풏품풆(∆푻)  
•
•
•
Therefore; Specific heat capacity of a substance is defined as the amount of heat required to  
raise the temperature of 1 kg of the substance by 1˚C or 1K  
.
The SI unit of specific heat capacity is Jkg-1 K-1. The most commonly used units of specific  
heat capacity are J/kg˚C and J/g0C  
.
Among all the substances, water has the highest specific heat capacity and its value is 4200  
J/kg 0C. So, water absorbs a large amount of heat for unit rise in temperature. Thus, water is  
used as a coolant in car radiators and factories to keep engines and other machinery parts  
cool. It is because of the same reason the temperature of water in the lake does not change  
much during day time.  
•
Specific heat capacities of some common substances are given in Table below.  
Materials  
Specific heat capacity  
(J/kgK)  
Materials  
Specific heat capacity  
(J/kgK)  
Water  
Sea water  
4200  
3900  
Glass  
Steel  
700  
500  
Paraffin  
Methylated spirit 2500  
Ice  
Mercury  
2200  
Copper  
Brass  
Iron  
Lead  
Zinc  
390  
320  
480  
130  
380  
2100  
1395  
900  
Aluminium  
Worked Examples:  
1. An energy of 84000 J is required to raise the temperature of 2 kg of water from 60° C to  
70° C. Calculate the specific heat capacity of water.  
Solution;  
풉풆풂풕 풆풏풆풓품풚 풔풖풑풑풍풊풆풅(푸)  
From: Specific heat capacity, C  
=
풎풂풔풔 × 풕풆풎풑풆풓풂풕풖풓풆 풄풉풂풏품풆(∆푻)  
ퟖퟒퟎퟎퟎ  
= ퟒퟐퟎퟎJkg-1K-1  
푪 =  
ퟐ×(ퟕퟎ−ퟔퟎ)  
2. The specific heat capacity of a metal is 160J kg-1K-1. Calculate the amount of heat energy  
required to raise the temperature of 500 gram of the metal from 125° C to 325° C.  
Solution:  
From; 푄 = 풎∆푻푪 = ퟎ. ퟓ × (ퟑퟐퟓ − ퟏퟐퟓ) × ퟏퟔퟎ = 16,000J  
3. Calculate the specific heat capacity of a body of mass 3 kg if it requires 6000 J of heat to  
raise its temperature from 300 C to 340 C  
Soln:  
Given: m = 3 kg, H = 6000 J,  
T1 = 300 C, T2 = 340 C  
From: Specific heat capacity, C  
풉풆풂풕 풆풏풆풓품풚 풔풖풑풑풍풊풆풅(푸)  
=
풎풂풔풔 × 풕풆풎풑풆풓풂풕풖풓풆 풄풉풂풏품풆(∆푻)  
ퟔퟎퟎퟎ  
푄
∴
Specific heat capacity, C  
=
=
ퟑ 풙 (ퟑퟒ−ퟑퟎ) = ퟓퟎퟎ푱/풌품푲  
푚∆푻  
Class Activity – 8:2  
1. Calculate the heat energy required to raise the temperature of 2kg of water from 10ºC to  
50ºC. Specific heat capacity of water is 4200 JKg-1 K-1. [ANS; 336,000J]  
2. How much heat is required to raise the temperature of a 25kg sample of mercury from 20°C  
to 30°C?  
(ANS: H = 348750J)  
3. The temperature of a 6kg block of copper rises from 15°C to 30°C on being heated.  
Determine the amount of heat energy supplied to the block. (Specific heat capacity of block is  
390Jkg°C) (ANS: H = 35 100J)  
4. How much heat energy is given out by an iron block of 20g mass when it cools from 920°C to  
20°C. (ANS: H = 8 640J)  
5. How much heat energy is required to raise the temperature of a 3 kg sheet of glass from  
24°C to 36°C?  
[Specific heat capacity of glass = 840 J kg-1 °C-1] [ANS; 30240J]  
6. Water in an ice maker of a refrigerator has a mass of 0.4 kg and a temperature of 22°C.  
What is the temperature of the water after 33 600 J of heat has been removed from it?  
[Specific heat capacity of water = 4200 J kg-1 °C-1] [ANS; 20C]  
7. A piece of copper of mass 40 g at 2000 C is immersed into a copper calorimeter of mass 60 g  
containing 50 g of water at 250 C .Neglecting heat losses ,what will the final temperature of  
the mixture be ? (ANS: 36.00 C)  
8. A brass cylinder of mass X was heated to 1000 C and then transferred into a thin aluminium  
can of negligible heat capacity containing 150 g of paraffin at 110 C .If the final steady  
temperature of the paraffin attained was 200 C, Determine the value of X (A: X = 0 . 116 Kg)  
9. Some heat energy is given to 120g of water and its temperature rises by 10K. When the  
same amount of heat energy is given to 60g of oil, its temperature rises by 40K. The specific  
heat capacity of water is 4200JKg-1 K-1. Calculate:  
(i) The amount of heat energy in joule given to water.[ANS; 5040J]  
(ii) The specific heat capacity of oil. [ANS; 2100J/kgK]  
Determination of Specific Heat Capacity of Substance  
•
In determination of specific heat capacity of substance two methods are used  
a) Method of mixtures  
b) Electrical method  
(a) Method of mixture  
•
If the heat loss controlled when mixing the water, the heat energy gained by the cold water is  
equal to the heat energy lost by hot water due to the principle of conservation of energy.  
Calorimetry  
•
Calorimetry means the measurement of the amount of heat released or absorbed by  
thermodynamic system during the heating process.  
•
When a body at higher temperature is brought in contact with another body at lower  
temperature, the heat lost by the hot body is equal to the heat gained by the cold body. No  
heat is allowed to escape to the surroundings. It can be mathematically expressed as  
푸푮풂풊풏 = −푸푳풐풔풕  
→ 푸푮풂풊풏 + 푸푳풐풔풕 = ퟎ  
•
Heat gained or lost is measured with a calorimeter. Usually the calorimeter is an insulated  
container of water as shown in Figure below.  
Calorimeter  
Is a device used to control the losses of heat energy when determining specific heat  
capacities of substances  
•
How Specific heat Capacity is determined  
•
If a liquid of known mass (mL) and initial temperature  
say a calorimeter of mass mC at i and a hot substance of known mass (mS) at its initial  
temperature, ( S) is added to the liquid.  
N.B: the initial temperature of a liquid i  
If is the final temperature of the mixture and assume some heat is being absorbed by the  
(i) is put in the inner container (Let’s  
)
•
•
(
)
=Initial temperature of the calorimeter (i)  
calorimeter ,then to determine the specific heat capacity of A SUBSTANCE, CS can be  
calculated as follows:  
Heat lost by substance = heat gained by liquid + heat gained by calorimeter  
(
)
(
)
(
)
풎푺푪푺 휽푺 − 휽 = 풎푳푪푳 휽 − 휽풊 + 풎푪푪푪 휽 − 휽풊  
(
)
(
)
풎푳푪푳 휽 − 휽풊 + 풎푪푪푪 휽 − 휽풊  
푪푺 =  
(
)
풎푺 휽푺 − 휽  
•
Therefore the specific heat capacity of the substance is given by:-  
(
)
(
)
풎푳푪푳 휽 − 휽풊 + 풎푪푪푪 휽 − 휽풊  
푪푺 =  
(
)
풎푺 휽푺 − 휽  
N.B  
•
Under the assumption that, if heat is not absorbed by any of the apparatus used to carry out  
this experiment. Then  
Heat lost by substance = heat gained by liquid  
(
)
(
)
풎푺푪푺 휽푺 − 휽 = 풎푳푪푳 휽 − 휽풊  
(
)
풎푳푪푳 휽 − 휽풊  
… … . . (풄풂풔풆 푰푰)  
푪푺 =  
(
)
풎푺 휽푺 − 휽  
Assumption made when using this method:  
•
•
No heat is lost to the surrounding  
Heat is not absorbed by any of the apparatus used to carry out this experiment.(case II  
)
Precautions to be taken when carrying out such experiments  
•
•
•
Use a highly polished calorimeter so as to minimize heat loss by radiation  
The calorimeter should be heavily lagged so as to minimize heat loss by conduction  
The calorimeter should be covered with a lid of poor conductor so as to prevent heat loss by  
evaporation and convection.  
Worked Examples  
1. A block of metal of mass 5 kg is heated to 1100 C and then dropped into 1.5 kg of water. The  
final temperature is found to be 500 C. What was the initial temperature of the water? (Specific  
heat capacity of metal = 460 Jkg-1 K-1)  
Soln: Let initial temperature of the water be  
Heat lost by block = mCΔ  
= 5 x 460 (110 – 50) = 5 x 460 x 60 =138 000J  
θ
θ
Heat gained by the water = mCΔ  
θ
= 1.5 x 4200 x (50 – ) = 6300 x (50 –  
θ
θ) = 315000 –6300  
θ
From: Heat gained by water = Heat lost by the metal block  
315 000 - 6300 = 138 000  
6 300 = 315 000 – 138 000  
θ
θ
ퟏퟕퟕퟎퟎퟎ  
ퟔퟑퟎퟎ  
∴ 휽 =  
= ퟐퟖ. ퟏퟎ푪  
2. Determine the final temperature when a 25.0 g piece of iron at 85.0°C is placed into 75.0  
grams of water at 20.0 °C. Given specific heat capacity of iron is 480J/kg/K and that of water is  
4200J/kg/K [ANS; 22.250C]  
3. If 5 L of water at 50 0C is mixed with 4L of water at 30 0C, what will be the final temperature of  
water? Take the specific heat capacity of water as 4184 J kg-1K-1. [ANS; 410C]  
NB; Suppose if we mix equal amount of water (m1 = m2) with 50°C and 30°C, then the final  
temperature is average of two temperatures.  
(b) By electrical method  
∴ 푰풕푽 = 풎푪∆휽  
Worked Examples  
:
1. A block of metal of mass 1.5 kg which is suitably insulated is heated from 300 C to 500 C in 8  
minutes and 20 seconds by an electric heater coil rated 54 W. calculate:  
(a) The quantity of heat supplied by the heater  
(b) The specific heat capacity of the block  
Solution:  
(a)  
Heat energy supplied = 풑풐풘풆풓 풙 풕풊풎풆  
→ 푸 = 푷풕 = 푰푽 풙 풕  
[(  
)
]
(
)
푸 = ퟓퟒ 풙 ퟖ 풙 ퟔퟎ + ퟐퟎ = ퟓퟒ 풙 ퟒퟖퟎ + ퟐퟎ = ퟓퟒ 풙 ퟓퟎퟎ = ퟐퟕ ퟎퟎퟎ 푱  
∴ 풉풆풂풕 풆풏풆풓품풚 풔풖풑풑풍풊풆풅 = ퟐퟕ ퟎퟎퟎ푱  
(b)  
From: Q = mcΔθ  
27 000J = 1.5 x C x (50 – 30)  
→
27 000 = 1.5 x C x 20  
ퟐퟕퟎퟎퟎ  
ퟑퟎ  
∴ 푪 =  
= ퟗퟎퟎ 푱/풌품푲  
Class Activity – 8:3  
1. 2.5 kg of a metal at 2000 C is immersed into 1.2 kg of cold water of temperature 200 C. The  
mixture, after thoroughly stirring attained a final temperature of 34.50 C. Given that the  
specific heat capacity of water is 4200 J/kgK, determine the specific heat capacity of the  
metal. (ANS: Cmetal = 176.63 J/kgK)  
2. A block of metal of mass 0.20kg at a temperature of 100°C is placed in 0.40kg of water at  
20°C.if the final steady temperature of the water is 24°C, determine the specific heat capacity  
of the metal. (Neglect heat absorber by the container) (ANS: c = 442.1 J/kg K)  
3. A block of aluminum of mass 0.5kg at a temperature of 100°C is dipped in 1.0kg of water at  
20°C. Assuming that no thermal energy is lost to the environment, what will the final  
temperature of the water be at thermal equilibrium? (A:θF = 27.7°C)  
4. A mechanic dropped a steel nut of mass 0.02 kg and temperature 90°C into 0.25 kg of water  
at 24°C in a polystyrene cup. What is the temperature when the steel nut and water have  
come to thermal equilibrium?  
[Specific heat capacity of water = 4200 J kg-1 °C-1; Specific heat capacity of steel = 450 J kg-  
1 °C-1] Assume that the exchange of heat is between the steel nut and water only.  
[ANS; 24.560C]  
5. An electric kettle with a power rating, P can heat up 4.0 kg of water from 30°C to 100°C in 10  
minutes.  
(a)  
(b)  
Calculate the power, P of the kettle.  
What assumption must you make to arrive at the answer?  
[Specific heat capacity of water = 4200 J kg-1 °C-1]  
ANS;  
풎푪∆휽  
풕
ퟒ×ퟒퟐퟎퟎ×ퟕퟎ  
(c)  
(d)  
푷풕 = 풎푪∆휽 → 푷 =  
=
= ퟏퟗퟔퟎ푾  
ퟏퟎ×ퟔퟎ  
All the heat supplied by the heater of the cattle is absorbed by the water. No heat is  
lost to the surroundings or absorbed by the kettle  
6. At a certain section of the Victoria Falls in Africa, water drops vertically through a  
height of 480 m.  
(a) Explain why the water at the base of the waterfall has a temperature slightly higher than  
the water at the top.  
(b) Estimate the maximum possible difference in the temperature between the water at the  
base and at the top of the waterfall. (Take g = 10 m s-2)  
ANS;  
(a) As the water falls, it loses potential energy. Some of its potential energy is converted  
heat energy to increase the temperature of the water a  
(b) Assume all the potential energy is lost by the falling water is converted to heat energy;  
품풉  
ퟒퟖퟎ×ퟏퟎ  
∴ 풎품풉 = 풎푪∆휽 → ∆휽 =  
=
= 1.140C  
푪
ퟒퟐퟎퟎ  
7. In a ballistics test, a bullet travelling at a velocity of 360 m s-1 is stopped by a stationary sand  
bag. If 20% of the energy lost by the bullet is converted to heat energy that is absorbed by  
the bullet. What is the increase in temperature of the bullet? [Specific heat capacity of the  
bullet = 150 J kg-1 °C-1  
]
ퟐ
ퟐ
ퟎ.ퟐ×ퟑퟔퟎ  
ퟐퟎ%×풗  
ퟐퟎ  
= 86.40C  
ANS;  
×
ퟏ 풎풗ퟐ  
=
= 풎푪∆휽 → ∆휽 =  
ퟐ푪  
ퟐ×ퟏퟓퟎ  
ퟏퟎퟎ  
ퟐ
8. A 10kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0kg. How  
much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used  
up in heating the machine itself or lost to the surroundings? Specific heat of aluminium =  
910Jkg-1 K-1. [ANS; 1030C]  
Change of State  
•
•
As you know, matter can exist in different states. These states include solids, liquids and  
gases  
When matter changes its state, its internal energy changes, so the kinetic energy of its  
constituent particles changes. As it is changing from one state to another, the change in  
energy is reflected in the bonds between the particles, and therefore the temperature of the  
object doesn’t change.  
•
Once the state change is complete, however, changes in energy are again observed in the  
form of changes in temperature. (see the figure below)  
Explanations from the figure above  
✓
From point A the temperature is rising steadily as more heat is added until point B where the  
solid starts to melt with steady temperature from point B to C  
✓
Once the melting is over, addition of heat leads to a steady rise in temperature in region CD.  
D is the boiling point of the liquid. Then the liquid is transforming into gas (after vaporization  
occurs in DE)  
✓
✓
As the point E is reached ,all the liquid has been converted into gas  
Reversing the process is also possible by removing the heat from gas reduces its  
temperature from point E to D,D to C,C to B and finally from B to A (Cooling takes place)  
Melting and freezing Point of a substance  
•
Melting and freezing refer to the changes in state which occur when the solid and liquid states  
interchange. Melting occurs when a solid is heated and turns to a liquid and freezing occurs  
when a liquid is cooled and turns to a solid.  
Melting Point  
•
•
Melting point is the temperature at which a substance changes from a solid to a liquid.  
At melting point the substance absorbs heat but the temperature does not change until the  
substance has completely melted.  
•
Pressure affects melting. Increase in pressure lowers the melting point of a substance while  
decreasing in pressure raises the melting point  
Freezing Point  
•
Freezing point is the temperature at which a liquid changes into a solid without a change in  
temperature.  
•
•
During solidification a substance loses heat to its surrounding but its temperature does not fall  
Freezing occurs at the same temperature as melting, hence, the melting point and freezing  
point of a substance are the same temperature. Example, water freezes and melts at 0°C  
Factors affecting freezing point  
•
The presence of impurities in a substance leads to a lower freezing point, this process is  
called freezing point depression. Freezing point depression is the reason why adding salt to  
frozen substances help to melt the ice. Freezing point depression occurs to the character of a  
material’s solid state.  
•
In terms of physical changes, you can also alter the freezing point by changing the pressure.  
Normally, if you reduce the pressure, you can reduce the temperature at which a liquid freezes.  
However, in the case of water, the more you increase the pressure, the more its freezing point  
decreases.  
Regelation  
Regelation is basically the phenomenon where the ice melts to the water below 0°C when  
pressure is applied and it refreezes back to ice when the pressure is removed (reduced).  
•
Consider an illustrative diagram below;  
In this experiment, an ice block is set on the table, a copper wire connected to weights is placed over  
the ice block.  
•
Because of the hanging weight, the wire exerts pressure on the ice block, the high pressure  
lowers the melting point of ice at the point of contact  
•
The ice block absorbs the heat from the wire so the region in contact with the wire melts and  
lowers itself to water below 0°C and as the wire passes through, the water freezed back again  
resulting in the successful passing through of the wire without cutting the ice block in half.  
•
At the end of the experiment, the ice block will still be in one piece, even though the wire  
would have completely passed through it.  
Effects of Regelation:  
•
Glacier acts as a source of a river due to Regelation. The mass of the glacier exerts pressure on  
the lower surface lowering the melting point of the ice at its base. This results in the melting of ice  
and propels the glacier to slide over the liquid. Under appropriate conditions, liquid water flows  
from the base of the glacier to lower altitudes when the temperature of the air is above the freezing  
point of water.  
•
•
The ice slab is shredded into pieces, and the shredded pieces are pressurized around the tip of a  
stick to prepare the ice ball. If two small pieces of ice are taken and pressed against each other,  
they stick to each other.  
Skating is possible on snow due to the formation of water only for the Regelation. Water is formed  
due to the increase of pressure and it serves as a lubricant.  
Boiling and Evaporation  
Boiling (Ebullition)  
•
Is the process by which a liquid turns into a vapor when it is heated to its boiling point  
Boiling Point  
•
Is the temperature at which all of a liquid changes into a gas  
OR  
•
Is the temperature at which its saturated vapour pressure becomes equal to the external  
atmospheric pressure.  
Mechanism of Boiling  
The molecules at the surface of the liquid gain more kinetic energy move faster and are able  
to overcome intermolecular forces holding them together and hence escape.  
•
What happens when a liquid boils?  
•
If a liquid is heated its temperature begins to rise, and therefore the saturated vapour  
pressure will increase. Ultimately, the saturated vapour pressure becomes equal o the  
external atmospheric pressure  
•
At this stage the further addition of heat will cause bubbles of vapour to form inside the body  
of the liquid and rise to the surface.  
Factors affecting boiling point  
The factors that affect the boiling point of a liquid are its pressure and Impurities.  
•
(i) Boiling Points Change with Changes in Pressure  
•
Since liquids boil when their vapor pressure becomes equal to surrounding pressure, then  
if the surrounding pressure is lower, liquids will boil at lower temperatures.  
•
At sea level water boils at 1000 C (The atmospheric pressure is at its maximum). As one  
goes to higher altitude, the atmospheric pressure keeps decreasing, thus decreasing the  
boiling point of water. For example, in cities whose altitude is around 5,000 feet, water  
boils at 950 C instead of at 1000 C and at 10,000 feet, water boils around 900 C  
.
•
Water boils faster at the top of mountain than at the bottom, this is because the higher  
the altitude, the lower the pressure. Thus at the top water will boil at lower temperature  
than it does at sea level (bottom).This means that it requires less energy and therefore a  
shorter heating time to reach its boiling point.  
•
Water in a pressure cooker boils at a very high temperature nearly 1200 C due to the high  
pressure created in the cooker. That is why food cooked in a pressure cooker or rice  
cooker takes less time to get ready than food cooked in common cooking pots  
(ii) Boiling point change with presence of impurities  
•
The boiling point of a liquid can be increased by adding impurities in the liquid. This occurs  
because the presence of impurities decreases the number of molecules available to become  
vaporized during boiling. Impurities include salt, sugar, and other dissolving molecules.  
Generally, impurities increase the boiling point of a liquid  
.
•
This explains why sea water boils at a higher temperature than pure / distilled water  
Applications of boiling at increased and reduced Pressure  
•
•
Boiling under increased pressure is important for fast cooking  
Boiling under reduced pressure is used in preparation of drug tablets  
Evaporation  
•
•
•
•
Evaporation – Is the process through which a liquid changes to vapour (gas) at a temperature  
below its boiling point  
Evaporation occurs when heat energy forces the bonds which hold the water molecules to  
break, causing the water to shift from its liquid state to a gaseous state  
If the water is instead kept in a closed container, the water vapor molecules do not have a  
chance to escape into the surroundings and so the water level does not change.  
Evaporation is more rapidly when there is windy sunny and less humidity  
,
Difference between Boiling and Evaporation  
Boiling  
Evaporation  
Occurs at a definite temperature which is  
boiling point  
Occurs at any temperature  
Accompanied by formation of bubbles  
No bubbles  
Occurs throughout the liquid  
Has no cooling effect  
Takes place rapidly  
Occurs at the surface of the liquid  
Has cooling effect  
Takes place slowly  
Class Activity – 8:4  
1. The figure below shows a block of ice with two heavy weights hanging such that the copper  
wire/string connecting them passes over the block of ice. It is observed that the wire gradually  
cuts its way through the ice block, but leaves it as one piece. Explain  
2. Explain why does food cook faster in a pressure cooker rather than in common cooking pot?  
3. Why it is advisable to use pressure cooker at high altitude?  
ANS; At higher altitude atmospheric pressure is lower than near the sea. This causes the  
water to boil at a lower temperature (below 100). Due to this it takes longer time for food to be  
cooked. A pressure cooker prevents boiling of water at lower temperature and leads to  
quicker cooking of the food. That’s why it is advisable to use pressure cooker at high altitude  
4. Explain the variation of boiling point with altitude and why are we using a pressure cooker at high  
altitude for cooking?  
Latent heat (Hidden heat)  
•
The energy required to change a specific material’s state is known as the material’s latent  
heat of transformation ). When an object changes from the solid to liquid phase, you use  
(
푳
the latent heat of fusion (푳푭).  
•
•
When an object changes from the liquid to the gaseous phase, you use the latent heat of  
vaporization  
You can calculate the energy required for a material to change phases using the following  
formula, where is the heat added, is the mass of an object, and is the specific latent  
heat of transformation of a material.  
(푳푽).  
푸
풎
푳
푸 = 풎푳  
NB;  
•
Latent heat is the heat absorbed or released when matter changes its state of matter without  
change in temperature.  
•
The latent heat associated with melting a solid or freezing a liquid is called The Latent Heat  
of Fusion WHILE. The latent heat associated with vaporizing a liquid or a solid or  
Condensing a vapor is called The Latent Heat of Vaporization  
Latent heat of fusion  
•
Latent heat of fusion – Is the quantity of heat energy required to change a solid to liquid at  
melting point without any change in temperature  
Specific Latent Heat of Fusion of a substance  
•
Specific latent heat of fusion – Is the quantity of heat energy required to change completely a  
unit mass (1kg) of the solid to liquid at its melting point.  
•
•
Its SI Unit is J/kg  
푯
=
Mathematically  
:
푳푭  
→ 푯 = 풎푳푭  
풎
•
Whereby, 푳푭 = Specific latent heat of fusion  
Melting /Freezing Point of some Substances at STP  
Substance  
melting/freezing  
point (°C)  
659  
1086  
1535  
latent heat of fusion  
(J/kg)  
396000  
134000  
293000  
Aluminum  
Copper  
Iron  
Water  
Mercury  
Ethyl alcohol  
0
335000  
11000  
105000  
-
-
39  
117  
Latent heat of Vaporization  
•
Latent heat of Vaporization – Is the quantity of heat energy required to change a liquid to  
vapour at boiling point without any change in temperature  
Specific Latent heat of Vaporization  
•
Specific latent heat of vaporization – Is the quantity of heat energy required to change  
completely a unit mass (1kg) of the liquid to vapour at its boiling point.  
•
•
Its SI Unit is J/kg  
Mathematically:  
푯
푯 = 풎푳풗 → 푳풗 =  
풎
•
Where by 푳풗 = Specific latent heat of vaporization  
Worked Examples  
1. Calculate the quantity of heat required to covert 50 grams of ice at 00 C to water at 00 C.  
(Specific latent heat of fusion of ice = 3.4 x 105 J/kg)  
Solution:  
From: 푸 = 풎푳푭  
∴ 푯 = 풎푳푭 = 0.05 × 3.4 × 105 = 17 000 퐽 = 17푘퐽  
→ 풎 = ퟓퟎ품 = ퟎ. ퟎퟓ 풌품, 푳푭 = ퟑ. ퟒ 풙 ퟏퟎퟓ 푱/풌품  
Why steam is hotter (dangerous) than boiling water (liquid)?  
•
Steam is hotter than boiling water because it has enough energy (Latent heat of vaporization) to  
escape the boiling water. The water remains at the boiling temperature of 1000 C and cannot get  
hotter till all the water has changed to steam.  
•
Since Steam has much more thermal energy than liquid that is why steam is used in  
engines to convert thermal energy to mechanical energy  
Class Activity – 8:5  
1. How much heat is required to change a 500 g ice cube at 00 to water at 00? (푳푭 = 3.36 x 105  
J/kg) [ANS: 186 kJ]  
2.  
H
ow much heat is required to change 100 g of ice from ice at 00C to vapour at 1000C (Lv =  
2.26 x 106 J/kg, 푳푭 = 3.36 x 105 J/kg, c = 4200 J/kgK). (ANS:301600 J)  
3. How much heat would be required to change 1.5kg of ice at -10°C to stream at 120°C.? The  
specific heat capacities of ice, water and stream are 2144J/kg° C, 4186 J/kg° C and 2010  
J/kg° C respectively (ANS: Ht = 4, 627, 860J)  
4. How much heat energy is required to melt 5 kg of ice? (Specific latent heat of ice = 336 Jg-1)  
ANS; H =mL= 5000x336 = 1680000J = 1.68 x106J  
5. How much boiling water at 100ºC is needed to melt 2kg of ice so that the mixture which is all  
water is at 0ºC? [Specific heat capacity of water 4.2 JKg-1 and specific latent heat of ice 336 Jg-1].  
′
풎 푳  
A; 풉풆풂풕 풍풐풔풕 = 풉풆풂풕 품풂풊풏풆풅 → 풎푪∆휽 = 풎′푳 → 풎 =  
= ퟏ600g (1.6kg)  
푪∆휽  
6. A copper block of mass 2.5kg is heated in a furnace to a temperature of 500°C and then  
placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat  
capacity of copper = 0.39Jg-1 K-1; latent heat of fusion of water = 335 Jg-1). [ANS; 1.5KG]  
7. Two accidents have happened. The first one with water at 100°C and the second one with  
steam at 100°C  
(a) Which is dangerous burn due to water at 100°C, and bum due to steam at 100°C?Why?  
(b) Latent heat of vapourisation of water is 2.26 x106J/kg. Explain the idea of latent heat of  
vopourisation.  
(c) Find the heat required to convert 1g of ice at 0°C to steam at 100°C  
[ANS; 푯 = 풎푳풇 + 풎푪∆푻 + 풎푳풗  
The Cooling effect of evaporation and mechanism of the Refrigeration  
•
Evaporation has a cooling effect because it takes thermal energy away from the surface. This  
is because the faster particles escape from the liquids surface leaving behind the slower ones,  
and that the faster ones have more energy, causing the temperature of the liquid to drop.  
Sweating (Perspiration) uses the cooling effect that comes from evaporation. The sweat,  
which is secreted from pores on the skin, can evaporate very easily. As the sweat evaporates, it  
takes away thermal energy from the skin to the surrounding air. This process causes the body to  
cool.  
•
•
•
For example, Dogs remove excess heat from their bodies by panting.  
Also, a volatile liquid like alcohol or ether poured on the skin feels much colder than water at the  
same temperature .This is because the alcohol evaporates quickly taking latent heat of  
vaporization from the skin ,thus cooling the skin  
•
Another example of cooling effect by evaporation is that; During the summer we wear  
cotton clothes. Cotton, being a good absorber of water allows more sweat to be in contact with  
the atmosphere, consequently helping in more evaporation. It is for this reason that we feel  
cooler when we wear cotton clothes. Similar example is when water is stored in earthen pots so  
as to make it cool. The pores of the earthen pot, just like the pores of cotton cloth provide a larger  
surface area for evaporation  
Mechanism of Refrigerator  
•
The basic principle used in refrigeration is cooling by absorption of latent heat (cooling effect  
by evaporation)  
•
•
Refrigerator – Is a machine which can enable heat to flow from a cold region to a hot region  
Refrigeration is the process in which work is done to move heat from one location to another  
In the refrigeration cycle ,there are five basic components:  
(i) Fluid refrigerant  
(ii) A compressor (controls the flow of refrigerant)  
(iii) The condenser coils (on the outside of the fridge)  
(iv) The evaporator coils (on the inside of the fridge)  
(v) Expansion valve (coolant)  
•
•
The figure above represents a typical refrigeration system. The electric compressor motor forces  
a gas at high pressure through a heat exchanger (condenser) on the rear outside wall of the  
refrigerator, where thermal energy (QH) is given off, and the gas cools to become liquid. The  
liquid passes from a high-pressure region, via a valve, to low-pressure tubes on the inside walls  
of the refrigerator; the liquid evaporates at this lower pressure and thus absorbs heat (QL) from  
the inside of the refrigerator cooling down the air.  
Lastly, the refrigerant evaporates to a gas, then flows back to the compressor, where the cycle  
begins again.  
Class Activity – 8:6  
1. Why does spirit poured on the skin feel much colder than water at the same temperature?  
2. Ice cream appears colder to the mouth than water at 00C. Give reason  
3. Why does a bottle of soft drink cool faster when surrounded by ice cubes than by ice cold water,  
both at 00C?  
Class Activity – 8  
Use the following constants where necessary when solving the questions below  
0
❖
❖
❖
❖
❖
Specific heat capacity of water = 4200J/ (kg C)  
0
Specific heat capacity of ice = 2100J/ (kg C)  
Specific heat capacity of steam = 2000 J/(kg C)  
0
Specific latent heat of fusion of ice 3.3 x 105 J/kg  
Specific latent heat of steam = 2.26 x 106 J/kg  
Specific latent heat of vaporization of water = 2.3 x 105 J/kg  
Acceleration due to gravity ,g = 10 m/s2  
❖
❖
0
1. Which contains the greater amount of heat – a lake of water at 20 C or a bowl of water at  
900C? Explain  
2. An iron bar of mass 80 g is heated from a temperature of 150 C to a temperature of 650 C .How  
0
much heat is absorbed by the bar ,given that iron has a specific heat capacity of 460 J/(kg C)  
3. Water of mass 20g at a temperature of 42˚C is poured into a well lagged copper vessel of mass  
27 g at a temperature of 20˚C.find the final temperature of the water.(specific heat capacity of  
copper = 400J/(kg˚C)  
4. Calculate the heat content of a piece of brass of mass 120g at a temperature of 20˚C .Find the  
final temperature of water. (specific heat capacity of copper = 400J/(kg ˚C)  
5. In an experiment to determine the specific heat capacity of a piece of metal, the following results  
were obtained:  
Mass of piece of metal =200g  
Initial temperature =25˚C  
Final temperature=80˚C  
Heat absorbed by the piece of metal = 1430 J  
Calculate the specific heat capacity of the piece of metal.  
6. Distinguish between latent heat of fusion and the specific latent heat of fusion of a substance.  
Find the amount of heat required to change 1kg of ice at 0˚C to water at the same temperature.  
7. Define the term latent heat of vaporization and specific latent heat of vaporization. Find the  
quantity of heat required to change 5kg of water at 60˚C into steam at 100˚C  
8. Determine the final temperature obtained when 500 g of water at 1000 C was mixed with 500 g  
of water at 100 C and well stirred. (ANS: Tf = 550 C)  
9. (a) Differentiate between heat and temperature  
(b) The specific heat capacity of a certain substance is 800J/kg0 C; what does this  
statement mean?  
(c) Calculate the specific heat capacity of mercury, if 980 J of heat is required to raise the  
temperature of 7 g of mercury from 00 C to 10000 C  
10. A piece of metal of specific heat capacity 840 J/(kg ˚C) and mass 30 g is heated to a  
temperature of 99˚C and then dropped into a cavity in a block of ice at 0˚C. Find the amount of  
ice that will melt  
11. Discuss how high specific heat capacity of water helps in formation of land and sea breeze.  
ANS;  
•
Specific heat capacity of land is 5 times less as compared to water. Thus, air above land  
becomes hot and light and rises up resulting in drop in pressure of land mass during day  
time. Thus, cool air from sea starts blowing towards and forming sea breeze.  
•
During night, land as well as sea radiates heat energy. However, temperature of land falls  
faster than sea water, because of high specific heat capacity of sea water. So, at night the  
temperature of sea water is more than land. Warm air above the sea rises and cold air  
from land starts blowing towards sea resulting in land breeze.  
12. A refrigerator can convert 0.4 kg of water at 20˚C to ice at -10˚C in 4 hours .Find the average  
rate of heat extraction from the water in joules per second.  
13. (a) what is meant by the following terms  
(b) Describe how to find the melting point of a substance by means of cooling curve  
14. Explain why:  
(i) melting point  
(ii) freezing point  
(a) Heat energy has to be supplied to a solid in order to change it into liquid.  
(b)Heat energy has to be supplied to a liquid in order to change it to vapor  
15. A 0.2kg block of ice at 0˚C is placed into a Styrofoam calorimeter cup with unknown mass of  
water at 20˚C .When thermal equilibrium is reached, the final temperature is measured to be  
5˚C.What was the mass of the water initially in the cup?  
16. A pressure cooker is a pot with a tight fitting lid that does not allow steam to escape until a  
preset pressure is reached. Explain how the pressure cooker can cook food faster than a sufuria  
with a loose  
–
fitting  
17. A container holds 1.5kg of ice initially at 40˚C.Heat is supplied to the container at the rate of  
12.6kJ per minute for 120 minutes.  
(a) Plot a graph of temperature versus time for the 120 minutes during which heat is  
supplied.  
(b) What will the temperature of the contents of the container be at the end of the 120  
minutes?  
(c) What will the mass of the steam in the container be at the end of the 120 minutes  
18. A 0.15kg aluminium cup holds 0.2kg water at 180 C. A 0.12 kg iron block at 85˚C is placed into  
the water and the entire system surrounded by an insulating jacket. What will be the final  
temperature of the system when thermal equilibrium is reached?  
19. The temperature of 500 g of a certain metal is raised to 1000 C and it is then placed in 200 g of  
water at 150 C. If the final steady temperature rises to 210 C, Calculate the specific heat capacity  
of the metal. (ANS: C = 128 Jkg-1K-1  
)
20. How much thermal energy is required to raise the temperature of 3kg of aluminium from 15˚C to  
25˚C?  
21. Explain the following:  
(b) When the brakes of a moving car are applied for an applicable time, they get hot  
(c) When the tyre of a car is pumped up, the pump gets warm  
22. A car of mass 1000 kg travelling at 72 km/h is brought to rest by applying the brakes. Assuming  
that the kinetic energy of the car becomes transferred to internal energy in four steel brake  
drums of equal mass, find the rise in temperature of the drums if their total mass is 20 kg, the  
specific heat capacity of steel is 450 J/kgK, and the work done is equal on all four drums  
.
(ANS: ∆휽 = ퟐퟐ. ퟐ 푲)  
23. A bath contains 100 kg of water at 600 C. Hot and cold taps are then turned on to deliver 20 kg  
per minute each at temperatures of 700 C and 100 C respectively. How long will it be before the  
temperature in the bath has dropped to 450? Assume complex mixing of the water and ignore  
heat losses.(ANS: t = 7.5 mins  
)
24. Some hot water was added to three times its mass of water at 100 C and the resulting  
temperature was 200 C. What was the temperature of the hot water. (ANS:T = 500 C)  
25. A piece of lead of mass 500 g and at air temperature falls from a height of 25 m. What is (a)  
Initial potential energy  
(b) Its kinetic energy on reaching the ground. Assume that all the  
energy becomes transferred to internal energy in the lead when it strikes the ground, calculate  
the rise in temperature of the lead if its specific heat capacity is 130 J/kgK. State the energy  
changes which occur from the moment the lead strikes the ground until it has cooled to air  
temperature again.(P.E =123 J ,K.E =123 J,∆휽 = ퟏ. ퟖퟗ 푲  
)
26. A waterfall is 100 m high and the difference in temperature between the water at the top and  
that at the bottom is 0.24 K. Obtain a value for the specific heat capacity of water in J/kgK  
explaining the steps in your calculations. Mention any assumptions you make.(C = 4100 J/kgK)  
27. A 0.5 kg block of aluminium at a temperature of 100˚C is placed in 1.0 kg of water at 20˚C.  
Assuming that no thermal energy is lost to the surroundings, what will the final temperature of  
the aluminium and the water be when they attain the same temperature?  
28. When a certain quantity of heat was supplied to a substance, its temperature rose from 5˚C to  
20˚C.What will the final temperature of the substance be if twice the amount of heat is removed  
from the sample?  
29. Why is water used as a coolant in car engines?  
30. State what changes, if any, take place in the following:  
(d) Melting point of ice when salt is added to the ice  
(e) The volume of water if it changes into ice  
(f) The boiling point of a liquid when the pressure on the liquid is reduced  
31. Two substances A and B have the same mass and are at the same temperature. Substance A  
has a higher specific heat capacity than substance B. Which substance will have a higher final  
temperature if the same amount of heat is supplied to each substance?  
32. An electric heater is rated at 250 W. Calculate the quantity of heat generated in 10 minutes  
(
ANS: H = 150 kJ)  
33. A tin contains water at 290 k and is heated at constant rate. It is observed that the water  
reaches boiling point after 2 minutes and after further 12 minutes it is completely boiled away  
.Calculate the specific latent heat of steam.(ANS: 2092kJ/kg)  
34. An insulated cup holds 0.1kg of water at 0˚C. 0.1 kg of boiling water at a temperature of 100˚C  
is poured into the cup. What will be the final temperature of the mixture at thermal equilibrium?  
35. A 50 watt heater is immersed in a 2 kg block of alluminium which also holds a thermometer .The  
temperature of the block rises by 8 k in 5 minutes. Neglect heat losses, Calculate the specific  
heat capacity of alluminium. (AN: 937.5 J)  
36. A metal sphere of unknown composition has a mss of 0.4kg.The sphere is heated in a furnace  
to a temperature of 150˚C and then dropped into an insulated cup holding 0.35 kg of water at  
20˚C.upon reaching thermal equilibrium ,the temperature of the system is measured to be  
32.4˚C.  
(a) Calculate the specific heat capacity of the metal.  
(b)Use the values of specific heat capacity in table 8.1 to identify the metal.  
37. Which contains the great amount of heat –a lake of water at 20˚C or a bowl of water at 90˚C?  
Explain.  
38. An iron bar of mass 80 g is heated from a temperature of 15˚C to a temperature of 65˚C.How  
much heat is absorbed by the bar, given that iron has a specific heat capacity of 460J/(kg ˚C)  
39. Water of mass 20 g at a temperature of 42˚C is poured into a well lagged copper vessel of mass  
27 g at a temperature of 20˚C. Find the final temperature of the water.(specific heat capacity of  
copper = 400J/ (kg˚C)  
40. Calculate the heat content of a piece of brass of mass 200 g at a temperature of 20˚C .Find the  
final temperature of water.(specific heat capacity of copper =400J/(kg˚C)  
41. Differentiate between  
(a) Melting point and boiling point  
(b) Freezing and vaporization  
(c) Evaporation and boiling  
(d) Melting and cooling  
42. Explain the following  
(a) The boiling point of water in Dar es salaam is higher than at the top of Mt. Kilimanjaro  
(b) Why does water boil faster at the top of a mountain than at the bottom?  
(c) Water being heated while covered boils faster than uncovered water.  
(d) When one wipes spirit on the skin he feels cold  
(e) When snow is pressed by the hands, it melts to water .The water then immediately  
freezes.  
(f) The use of ammonia as a household refrigerant is discouraged  
43. How much heat is required to change 40 g ice cube from ice at –100 C to steam at 1100 C?  
44. Describe how a household refrigerator preserves food.  
45. If 200 g of water is contained in a 500 g aluminium CAN at 100 C then an additional 100 g of  
water at 1000 C is added into the CAN , what is the final equilibrium temperature of the mixture?  
46. An unknown liquid of mass 400 g at a temperature of 800 C is poured into 400 g of water at 400  
C. The final temperature of the mixture is 490 C .What is the specific heat capacity of the  
unknown liquid  
47. 20 g of steam at 1000 C is added to 50 g of ice at 00 C .Find the amount of ice that is melted and  
the final temperature  
48. Explain how the following factors affect the melting and boiling points:  
(a) Pressure  
(b) Impurities  
49. Explain how a refrigerator works.  
50. An electric heater rated 1500 W is used to heat water in an insulated container of negligible heat  
capacity for 10 minutes .The temperature of water rises from 200 C to 400 C. Calculate the mass  
of water heated  
51. An electric kettle rated 2 kW is filled with 2.0kg of water and heated from 200 C to 980 C.  
Calculate the time taken to heat the water assuming that all the electrical energy is used to heat  
the water in the plastic kettle and the kettle has negligible heat capacity  
52. The following data was obtained from an experiment .Mass of copper metal block = 200 g, initial  
temperature of the block = 220 C, ammeter reading = 0.5 A, voltmeter reading = 3.0 v, final  
temperature of the block = 300 C, time of heating = 7 minutes .Use the data to calculate the  
specific heat capacity of copper .What does this value mean? (ANS: CC = 394 J kg-1K-1)  
53. In an experiment , the following data was obtained .Use the data to calculate the time taken by  
the heater to raise the temperature of water container and the stirrer from 200C to 230 C . What  
assumption have made in your calculations? Power of electric heater = 30 W ,mass of the  
container and the stirrer = 200 g ,specific heat capacity of the container and the stirrer = 400  
J/kgK ,mass of water in the container = 100g, specific heat capacity of water = 200 J/ kgK  
(
ANS: t = 50 s)  
54. A class of Physics students decided to determine the specific heat capacity of water in a  
waterfall. They used a sensitive thermometer to find the difference in temperature of water at the  
top and the bottom of the waterfalls and obtained the following results; height of the waterfalls =  
52 m, temperature of the water at the top = 21.540 C and that at the bottom = 21.670 C. Stating  
any assumptions made, calculate a value for specific heat capacity of water  
55. A 200 g of liquid at 210 C is heated to 510 C by a current of 5 A at 6 v for 5 minutes. What is the  
specific heat capacity of the liquid? (A:C = 1500 J/kg 0C  
)
56. An electric kettle rated 1 500 W is used to boil 500 g of water into steam at 1000 C .Calculate  
the time required to boil off water.  
57. Why steam is hotter than boiling water?  
58. Explain as fully you can what happens when a liquid boils. Why would you expect the boiling  
point of a liquid to be lowered when the pressure above the free surface is vapour