Chapter Eight Class 9th

Transcription

Chapter Eight Class 9th
CHP # 8
Q.1
THERMAL PROPERTIES OF MATTER
Differentiate between heat and temperature?
(Ans) Heat
It can be defined as "the sum of kinetic energy of the molecules present in
a substance is called heat". Heat is a form of energy. It is denoted by "Q".
Its SI unit is joule "j".
Temperature
The degree of hotness or coldness of a body is called temperature. OR
It can be defined as "the average kinetic energy of the molecules present
in a substance is called temperature". It is denoted by "T". Its SI unit is
Kelvin "K".
Q.2
Explain the principle, calibration and various scales of temperature.
(Ans) Measurement of temperature
Temperature is measured by an instrument called thermometer.
Thermometer is based on the following principle.
Principle
Any property of a substance that increases or decreases uniformly with
temperature can be used for measurement of temperature.
Mercury and Alcohol thermometer are based on this principle.
Calibration
It involves two steps;
(i) Two standard reference temperatures are decided.
(a) Freezing point
(b) Boiling point
(ii) A scale of temperature is established by dividing the interval of
Temperature into a number of equal sub-intervals.
°
°
C
F
K
100
212
373
100
Equal parts
180
Equal
parts
100
Equal
parts
0
32
273
(1)




(2)




(3)




Scales of temperature
There are three scales of temperature.
Celsius or Centigrade scale
This scale was introduced by a Swedish Astronomer Celsius.
It is denoted by °C.
Its ice point is 0°C and steam point is 100°C.
The interval between ice point and steam point is divided into 100 equal
parts, each part is called a degree centigrade.
Fahrenheit scale
This scale was introduced by Fahrenheit.
It is denoted by °F.
Its ice point is 32°F and steam point is 212°F.
The interval between ice point and steam point is divided into 180 equal
parts, each part is called a degree Fahrenheit.
Kelvin or Absolute scale
This scale was introduced by Lord Kelvin.
It is denoted by K.
Its ice point is 273K and steam point is 373K.
The interval between ice point and steam point is divided into 100 equal
parts, each part is called one Kelvin.
Relationship between different scales of temperature.
5 
( F  32)
9
9

F  (   C )  32
5

K  C  273
Q.3

C

C  K  273
What is thermometer? On what principle does it work.
(Ans) Thermometer
An instrument used for measuring temperature is called thermometer.
Construction
A thermometer consists of a narrow glass tube having one end closed and
the other contains a metallic bulb. The tube is filled with a specific liquid.
Calibrations are placed on the glass tube on suitable scale which can be
read easily.
Working
It works on the principle that matter expands on heating and contract on
cooling. Thus the degree of expansion and contraction of matter can be
recorded by the calibration on the tube.
Q.4
What is meant by thermal expansion? Describe the thermal
expansion of solids.
(Ans) Thermal expansion
The process of increase in physical structure (length, area, volume) of a
body due to rise in temperature is called thermal expansion.
Thermal expansion of solids
The process of increase in physical structure (length, area, volume) of a
solid body due to rise in temperature is called thermal expansion.
Explanation
When heat is supplied to a body the K-E of molecules of the body
increases due to which amplitude of vibrations of molecules increases. As
a result the body expands.
Q.5
What do you meant by Linear thermal expansion? Derive an
expression for linear thermal expansion.
(Ans) Linear thermal expansion
The increase in length of a solid material due to rise in temperature is
called linear thermal expansion.
Mathematical derivation
Consider a rod of length “LO” having temperature “T1”. After heating the
temperature of the rod is raised to “T2” and its length becomes “L”.
LO
ΔL
L
Change in length = ΔL = L - Lo
Change in temperature = ΔT = T2 – T1
As from linear thermal expansion the change in length is directly
proportional to the original length and change in temperature.
Mathematically
L  L
L   T
L  L T
L  Const( L T )
L  L T
Where  is constant of proportionality and are called coefficient of linear
thermal expansion. Mathematically
L

L T
To find the final length of the rod we have;
L  L  L
L  L  L T
L  L  L T
L  L (1  T )
Q.6
What do you meant by Volumetric thermal expansion? Derive an
expression for volumetric thermal expansion.
(Ans) Volumetric thermal expansion
The increase in volume of a solid material due to rise in temperature is
called volumetric thermal expansion.
Mathematical derivation
Consider the original Volume of a solid is “VO” having temperature “T1”.
After heating the temperature of the rod is raised to “T 2” and its Volume
becomes “V”.
Change in Volume = ΔV = V - Vo
Change in temperature = ΔT = T2 – T1
As from Volumetric thermal expansion the change in Volume is directly
proportional to the original Volume and change in temperature.
Mathematically
V  V
V   T
V  V T
V  Const(V T )
V  V T
Where  is constant of proportionality and are called coefficient of
volumetric thermal expansion. Mathematically
V
 
V T
  3
To find the final volume of the rod we have;
V  V  V
V  V  V T
V  V  V T
V  V (1  T )
Q.7
Write some practical applications of thermal expansion of solids.
(Ans) Practical applications of solids are as under.
(i)
Railway lines
We know that solids expand when its temperature rises. So a gap is left
between the rails of a railway lines. In order to avoid bending of railway
line because in summer the temperature of the railway line rises. The gap
provides room for expansion, otherwise it will bend.
(ii)
To remove a tight glass stopper from a bottle
When a bottle top is too tight to open, we can loosen the top by placing it
in hot water for a while. The top expands and can be opened easily by
doing so.
(iii)
Fixing of an iron ring to a cart wheel
The diameter of the iron rim is slightly lesser at room temperature than the
diameter of the wooden wheel. The rim is heated, so it expands on
heating and can be placed around the wooden wheel. When the ring
comes to room temperature it contracts and produces a tight fit.
(iv)
Gaps between the roof girders and steel bridges
Gaps are left at the ends of metallic roof girders and bridges to give room
for expansion. Usually one end of the metallic structure is fixed and the
other end is allowed to expand in summer season.
(v)
Bimetallic strip
Two metallic strips whose co-efficient of expansion are different welded
together to form a bimetallic strip.
In a brass, iron strip brass expands more rapidly than iron. Thus the strip
bends as shown below.
Q.8 What is meant by a thermostat?
(Ans) thermostat
it is actually a bimetallic strip which may be used to open and close
electrical circuits. One end of the strip is open and the other end is fixed.
The moveable end of the strip changes its position with the change in
temperature. Hence the circuit is automatically opened or closed. Figure
Q.9 State and explain thermal expansion of liquids.
(Ans) Thermal expansion of liquids
The liquids also expand on heating. As the liquids have no definite shape,
they acquire the shape of the container. So we consider only volumetric
expansion in case of liquids.
The volumetric expansion of liquids is greater than the volumetric
expansion of solids, because in liquids the volumetric expansion occur in
two ways.
(i) Real expansion (ii) Apparent expansion
in case of real expansion we take the expansion of container as well as the
expansion of liquid.
While in case of apparent expansion we take only the expansion of liquid.
The real and apparent expansion of liquid can be demonstrated with the
help of the following experiment.
Experiment
Consider a flask which is filled with coloured water as shown in the
following figure.
Flask is fitted with a cork and a glass tube is passed through the tube. Let
the water rises to point “A” before heating. Now if we supply heat to the
flask, then the heat is first absorbed by the flask due to which the flask
expands and its volume increases. As a result the level of water falls from
“A” to “B”.
On further heating, the heat is transferred to the liquid. Now the liquid
expands and rises to point “A” and then “C”.
The expansion between point “A” and “C” is apparent expansion. While the
expansion between points “B” and “C” is known as real expansion.
Mathematically
Real expansion = apparent expansion + expansion of the flask
Apparent expansion = real expansion – expansion of the flask
Q.10 Define heat capacity and specific heat capacity?
(Ans) Heat capacity
The amount of heat required to raise the temperature of entire mass of the
substance by one Kelvin is called heat capacity. It is denoted by “Cm”.
Mathematically
Q
Cm 
T
The SI unit of heat capacity is joule/Kelvin (JK-1).
Specific heat capacity
The amount of heat required to raise the temperature of one kilogram
mass of the substance by one Kelvin is called specific heat capacity. It is
denoted by “C”. Mathematically
Q
C
mT
The SI unit of specific heat capacity is (JK-1Kg-1).
Q.11 What are the different effects of high specific heat of water?
(Ans) Water has particularly high specific heat. Its value is 4200Jk-1Kg-1. This
makes water a very useful material for storing and carrying energy. Water
is used for the following purposes due to high specific heat.
(i)
(ii)
(iii)
Hot water bottles
A hot water bottle remains warm for more than a hour and can be used for
therapeutic and other purposes.
Water as coolant in radiators
Because of high specific heat of water, it carries unwanted heat from the
engine of a car to the radiator.
Internal heating of buildings
In cold winter, water is used for internal heating. The hot water can carry a
large amount of water from the furnace to the room and keep them at a
moderate temperature.
Q.12 Define and explain Latent heat of Fusion of solids?
(Ans) Latent heat of fusion
The quantity of heat required to melt a unit mass (1Kg) of solid without
changing its temperature is called Latent heat of fusion of solid. It is
denoted by “Lf”. Mathematically
Q
Lf 
m
The SI unit of Latent heat of fusion is joule/kilogram (j/Kg).
Example
The amount of heat required to transfer one kilogram ice into water at 0°C
is known as latent heat of fusion of ice. Its value has been found to be
3  34  105 J / Kg .
Explanation
When heat is supplied to a solid body, then the K-E of the molecules of
the body increases and the molecules begin to separate from each other.
The solid begins to melt. This process continues until the whole solid is
converted into liquid. During melting process the temperature of the solid
remains constant. This heat is called latent heat of fusion of the solid.
Q.13 Define and explain Latent heat of vaporization of liquids?
(Ans) Latent heat of vaporization
The quantity of heat required to boil a unit mass (1Kg) of liquid without
changing its temperature is called Latent heat of vaporization of liquid. It is
denoted by “Lv”. Mathematically
Q
LV 
m
The SI unit of Latent heat of vaporization is joule/kilogram (j/Kg).
Example
The amount of heat required to transfer one kilogram water into steam at
100°C is known as latent heat of vaporization of water. Its value has been
found to be 2  26  105 J / Kg .
Explanation
When heat is supplied to a liquid, then the K-E of the molecules of the
body increases and the intermolecular attractive forces between the
molecules of the liquid becomes weaker. As a result the liquid begins to
boil. This process continues until the whole liquid is converted into steam.
During boiling process the temperature of the liquid remains constant. This
heat is called latent heat of vaporization of the liquid.
Q.14 Define the evaporation of liquids. On what factors does it depend?
(Ans) Evaporation
The process in which a liquid slowly change to vapours at any
temperature (below its boiling point) is called evaporation.
(i)
(ii)
(iii)
(iv)
(v)
Evaporation of liquids depends upon the following factors.
Nature of liquid
Liquids with low boiling points evaporate quickly than those of high boiling
points.
Example
Rate of evaporation of alcohol is higher than water.
Temperature of liquid and the surrounding
Liquids with high temperature evaporate quickly than liquids with low
temperature. Higher the temperature of the surrounding higher the rate of
evaporation and vice versa.
Example
Under a hot iron wet clothes dry out quickly as the water evaporates. Wet
clothes dry more quickly in hot summer season as compare to winter.
Presence of water vapours in air
Higher the amount of water vapours in air, smaller will be the rate of
evaporation and vice versa.
Example
Wet clothes dry slowly in rainy season as a lot of water vapours are
present in air.
Dryness of air
drier the air, the rapid will be the process of evaporation and vice versa.
Example
Wet clothes dry quickly on a dry day and slowly on a humid day.
Air pressure on the surface of liquid
Greater the air pressure on the surface of the liquid, lesser will be the rate
of evaporation.
Q.15 How evaporation produce cooling? Also write some applications of
cooling by evaporation.
(Ans) Cooling effect of evaporation
during evaporation more energetic molecules escape from the liquid
surface. Molecules that remain in the liquid have less K-E. as a result
cooling effect produced due to evaporation.
Applications of cooling by evaporation
(i)
Cooling by fans
We use fans in the hot summer season because the moving air increases
the rate of evaporation or perspiration from our bodies.
(ii)
Fever control
Usually a wet towel is placed on the forehead of a person having high
fever. It is because the water evaporates which take heat away from the
head. Thus the temperature of the head remains within the safe limits and
the patient does not suffer any brain damage.
(iii)
Refrigerators
the cooling effect in many refrigerators is produced by evaporation of a
volatile liquid called Freon.
The liquid Freon evaporates in the pipes of freezer compartment. As the
Freon evaporates, it draws the necessary latent heat from the food inside
the refrigerators.
CONCEPTUAL QUESTIONS
1.
A.
Why are small gaps left behind the girders mounted in walls?
See questions No.7 part (iv)
2.
A.
Why liquids have two co-efficient of expansion?
Liquids have no particular shape. So we can supply heat to a liquid in a
container. The liquid as well as container expands with heat. Therefore
liquids have two co-efficient of expansion.
3.
Water melts to form ice at 0°C. At what temperature does water
freezes to form ice?
The freezing or ice point of water is 0°C, same is the melting point of ice.
Because during melting the temperature remain constant and the heat is
absorbed by the ice to change the state of matter. Similarly liquid is
change into solid during freezing and the energy is losses by the water.
A.
4.
A.
Why is water used as coolant in radiators of automobile engines?
See question No.11 part (ii)
5.
A.
Why do we sweat in summer?
In summer the temperature of surrounding becomes more than the
temperature of our body. This high temperature affects our body. To keep
our body at suitable temperature the extra heat comes out in the form of
sweat from the pores of our body and evaporates from the skin.
6.
What is the effect of high specific heat of water on the climate of
coastal areas?
During the day the shines equally on the land and sea. The land heats up
more quickly than the sea, because of high specific heat of water. the hot
air over the land rises and the cold air from the sea blows to replace it.
Thus there is a sea breeze during the day.
At night the process is reversed, the land cools quickly than the sea. As a
result the cold air blows from the land and the hot air rises from the sea to
replace it.
Therefore the temperature of the coastal areas remains at moderate level.
A.
7.
A.
Why steam at 100°C produce more severe burns than boiling water at
100°C?
When the temperature of water reaches to 100°C, it is converted to steam.
At this time the temperature remains constant. Although heat is being
given to water which is absorbed as latent heat of vaporization
2  26  105 J / Kg . Therefore steam at 100°C produce more severe burns
than boiling water at 100°C.
8.
A.
Why does the temperature not change during a change of state?
The temperature does not change during change of state of matter,
because in this process the heat supplied is busy in breaking the bonds of
the molecules.
9.
An iron rim is fixed around a wooden wheel is heated before fixture.
Explain
See questions No.7 part (iii)
A.
10.
A.
11.
A.
12.
A.
13.
A.
14.
A.
If a hot piece of thick glass is dipped in cold water, it breaks. Give
reason
The K-E of the molecules of the hot thick glass is greater due to which the
molecules expands. When we dipped it in cold water it contracts quickly.
This sudden change in temperature causes cracks in the thick glass as a
result it breaks.
Why do soda water bottles often burst in summer? How can the
bursting be minimized.
Soda water bottles contain gases and it often burst in summer. Because in
summer the temperature rises which also affect the K-E of the gases
inside the bottled.
This bursting of the bottle can be minimized by leaving some space at the
top of soda water bottle.
Why is the water at the bottom of a water fall is warmer than water at
the top of water fall?
We know that temperature is directly proportional to the K-E. The K-E of
the molecule at the bottom is greater than the K-E of the molecules at the
top of the water fall. Therefore the water at the bottom of water fall is
warmer than the top water.
Does the land cools at slower rate than sea water? Give one reason
for your answer.
Land cools at faster rate than sea water.
Reason
It is because of high specific heat of water. Sea water takes a lot of time to
lose heat which is absorbed throughout the day.
Why we feel cool after perspiration?
See question No.15 part (i)
15.
A.
Why is ice at 0°C a better coolant of soft drink than water at 0°C?
Because for ice to transform into water it must absorb heat, called the
latent heat of melting. This heat is required to break down the crystalline
structure of the ice.
The heat is taken from the drinks/food and thus makes them colder than
they would be if kept in the same amount of cold water.
Numerical Problems
1.
2.
A person running fever has a temperature of 104°F. What is his
temperature in degree centigrade (°C)?
Given data
T = 104°F
T =? °C
Solution
5
C = ( F  32)
9
5
C  (104  32)
9
5
C  (72)
9
C = 40°C
A railway line 1200Km long is laid at 25°C. By how much will it
contract in winter when the temperature falls to 15°C? By how much
will it expand when the temperature rises to 40°C in summer?

(The coefficient of linear expansion is   12 106 C 1 )
Given data
L0  1200Km
L0  1200 1000m  1200000m
T0  25 C
TW  15 C
TS  40 C
LW  ?
L S  ?
Solution
To calculate the contraction in winter
T  T  TW  25  15  10 C
LW  LT
LW  1200000 12  106  10
LW  144000000 106  144m
To calculate the expansion in summer
T  T  TW  40  25  15 C
LS  LT
LS  1200000 12  106  15
3.
LS  216000000 106  216m
The volume of a brass ball is 800cm3 at 20°C. Find out the new
volume of the ball if the temperature is raised to 52°C? The
coefficient of volumetric expansion of brass is 57  106 ° C 1 .
Given data
V  800cm 3
T  20 C
T  52 C
  57  106 ° C 1 .
V  ?
Solution
T  T  T  52  20  32 C
V  V T
V  800  57  10 6  32
4.
V  1459200 10 6  1  46cm 3
New volume of ball = 800+1  46 =801  46cm 3
A brass disc at 293K has a diameter of 0  30m and a hole of diameter
010m is cut in the centre. Calculate the diameter of the hole when
the temperature of the disc is raised to 323K? Coefficient of linear
expansion of brass is 19  10 6 K 1 .
Given data
Diameter of disc = D = 0  30m
Diameter of hole = D = 010m
T  293K
T  323K
  19  10 6 K 1
Diameter of the hole =?
T  T  T  323  293  30 K
Solution
D  DT
D  0  10  19  10 6  30
5.
D  57  10 6
D  0  000057m
Diameter of the hole = 0 10  0  000057  0 100057m
0  5 Kg of copper needs 1950J of heat to raise its temperature
through 10°C. Calculate the heat capacity of sample?
Given data
m = 0  5 Kg
Q = 1950J
T  10 C
C ?
Solution
Q
C
T
1950
C
10
C  195JK-1