Short
Questions of Chapter No. 08
Q No. 1 Why does
heat flow from hot body to cold body?
Answer: Heat flows from hot body to cold body because
the energy state is higher. The bigger the difference in temperature between
two objects, the faster heat flows from them.
Q No. 2 Define the
terms heat and temperature.
Answer: Heat is the energy that is transferred from one body to the other
in thermal contact with each other as a result of the difference of the
temperature between them.
Temperature of a body is the coldness or hotness
of the body.
Q No. 3 What is
meant by internal energy of the body?
Answer: The sum of kinetic and potential
energy associated with the atoms, molecules and particles of a body is called
its internal energy.
Q No. 4 how does
heating affect the motion of molecules of a gas?
Answer: The larger the temperature of a gas
the faster the molecules will move and the larger the force they will exert,
and the higher the pressure.
Q No.5 What is thermometer? Why mercury is
preferred as a thermometric substance?
Answer: A device which is used to measure the
temperature of the body is called thermometer.
Mercury has uniform
thermal expansion, easily visible, has low freezing point, has high boiling
point and less specific heat. Due to these properties mercury is preferred as
thermometric substance.
Long Questions of chapter no. 8
Thermal properties of matter
Q No.1 Define thermometer. Write a note on glass thermometer.
Answer:
A device that is used to measure the temperature of a body is called thermometer.
Glass
thermometer:
A glass thermometer has a bulb with a long capillary tube of uniform and
fine bore. A suitable liquid is filled in the bulb. When a bulb contacts a hot
object, the liquid in it expands and rises in tube. The glass stem of a
thermometer is thick and acts as a cylindrical lens. This makes it easy to see
the liquid level in glass tube.
Mercury freezes at -39°C and boils at 357°C.
It has all the thermometric properties. Thus, mercury is the most suitable
thermometric materials. Glass thermometers are widely used in laboratories,
clinics, and houses to measure temperatures in the range from -10°C to 150°C.
Q No. 2 Write a note on scales of temperature.
Answer: A scale is marked on the thermometer. The
temperature of the body in contract with the thermometer can be read on that
scale. Three scales of temperature are in common use. These are:
i.
Celsius scale or centigrade scale
ii.
Fahrenheit
scale
iii.
Kelvin scale
Celsius Scale:
On Celsius
scale, the interval between lower and upper fixed points is divided into 100
equal parts. The lower fixed point is marked as 0°C and the upper fixed point
is marked as 100°C.
Fahrenheit Scale:
On Fahrenheit scale, the interval between
lower and upper fixed point is divided into 180 equal parts. Its lower fixed
point is marked as 32°F and upper fixed point is marked as 212°F.
Kelvin scale:
In SI units, the unit of temperature is
Kelvin (K) and its scale is called Kelvin scale of temperature. The interval
between the upper fixed point and lower fixed point is divided into 100 equal parts.
Thus, a change 1°C is equal to a change of 1K. The lower fixed point on this
scale corresponds to 273K and the upper fixed point is referred 373K.The zero
on this scale is called absolute zero and is equal to -273°C.
Q No. 3 Explain Specific Heat Capacity.
Answer: Specific heat of a substance is the
amount of heat required to raise the temperature
of 1 kg mass of that substance through 1 Kelvin.
Generally, when a body is heated, its temperature increases. Increase in
temperature of a body is found to be proportional to the amount of heat
absorbed by it. It has also been observed that the quantity of heat Δ Q
required to raise the temperature Δ T of the body is proportional to the mass m
of the body. Thus
ΔQ
m Δ T or ΔQ=c m Δ T
Mathematically: C=
Here Q
is the amount of heat absorbed by the body and c is the constant of
proportionality called the specific heat capacity or simple specific heat.
SI unit of specific heat is Jkg-1K-1.
Q No.4 Define heat capacity and derive its equation.
Answer: Heat Capacity of a body is the
quantity of thermal energy absorbed by it for one Kelvin (1K) increase in its
temperature.
Equation: Thus, if the
temperature of a body increases through Δ T on adding Δ Q amount of heat, then
its heat capacity will be:
Heat Capacity=
=m c ΔT
Heat capacity=m c
Equation shows that heat capacity of a body is equal to the product of its
mass of the body and its specific heat capacity. For Example, heat capacity of
5 kg of water is (5 kg × 4200Jkg-1K-1)
21000JK-1.
Thus larger is the quantity
of a substance, larger will be its heat capacity.
Q No.5 Explain Latent heat of Fusion and derive its equation.
Answer: Heat energy
required to change unit mass of a substance from solid to liquid state at its
melting point without change in its state temperature is called its Latent
heat of Fusion. It is denoted by Hf.
Equation: Hf=
OR ΔQf=mHf
Ice changes at 0°C into water. Latent heat
of fusion of ice is 3.36×105 Jkg-1.
When a substance is changed from solid to
liquid state by adding heat, the process is called Melting or fusion.
The temperature at which a solid starts melting is called its Fusion point
or melting point. When a process is reversed i.e. when a liquid is cooled,
its changes into solid state. The temperature at which a substance changes from
liquid to solid state is called its freezing point. Different substances
have different melting points. However, the freezing point of a substance is
the same as its melting point.
Q No.6 Explain latent heat of vaporization
and derive its equation.
Answer: The quantity of heat that changes unit
mass of a liquid completely into gas at its boiling point without any change in
its temperature is called its Latent heat of vaporization. It is denoted
by Hv.
Equation: Hv=
OR ΔQv=mHv
When water is heated, it boils at 100°C under
standard pressure. Its temperature remains 100°C until it is changed completely
into steam. Its latent heat of vaporization is 2.26×106 JKg-1.That
is; one kg of water requires 2.26×106 joule heat to change it
completely into gas (steam) at its boiling point.
Q No.7 What is meant by evaporation and on what factors the evaporation of
liquid depends?
Answer: Evaporation is the changing of liquid
into vapours (Gaseous state) from the surface of the liquid without heating it.
Factors: The rate of
evaporation is affected by various factors.
Temperature: Wet clothes
dry up more quickly in summer than in winter. At high temperature, more
molecules of a liquid are moving with high velocities. Thus, evaporation is
faster at high temperature than at low temperature.
Surface area: Water
evaporates faster when spread over large area. Larger is the surface area of a
liquid, greater number of molecules has the chance to escape from its surface.
Wind: Wind blowing over the
surface of a liquid sweeps away the liquid molecules that have just escaped
out. This increases the chance for more liquid molecules to escape out.
Nature of the liquid: Liquids differ in the rate at which they evaporate. Spread a few drops of
ether or spirit on your palm. You feel cold due to the evaporation.
Q No.8 What is Thermal Expansion and also write its types.
Answer: On heating substances i.e. solids, liquids and
gases expand. Their expansion is called thermal expansion. The Thermal
Expansion is of two types:
1.
Linear thermal expansion
2.
Volume thermal expansion
Q No.9 What is linear thermal expansion and derive its equation.
Answer: The length of
solid changes with the change in the temperature is called linear thermal
expansion.
Derivation: Consider a
metal rod with,
Initial length=Lo
Initial Temperature=To
Final length=L
Final
Temperature=T
Increase in
length=ΔL=L-Lo
Increase in temperature=ΔT=T-To
ΔL
Lo and ΔL
T
ΔL
Lo ΔT
ΔL = α Lo ΔT
OR
L-Lo= α Lo ΔT
L=Lo (α ΔT+1)
Thus, we can define the
coefficient of linear expansion α of a substance
as the fractional increase in its length per Kelvin rise in temperature.
Q No.10 What is Volume thermal expansion and derive
its equation.
Answer: The Volume of solid changes with the
change in the temperature is called Volume thermal expansion or cubical
thermal expansion.
Derivation: Consider a solid with,
Initial
volume=Vo
Initial Temperature=To
Final Volume=V
Final Temperature=T
Change in Volume=V-Vo
Change in Temperature=T-To
ΔV
Vo and ΔV
T
ΔV =
Vo ΔT
OR
V-Vo=
Vo ΔT
V=
Vo ΔT+ Vo
Where
is the temperature coefficient of
volume expansion. Using equation 1, we get
Thus, we can define the temperature coefficient of volume expansion
as the fractional change in its
volume per Kelvin change in temperature the coefficients of linear expansion
and volume expansion are related by the equation:
Q No. 11 Describe some examples of expansion of
solids?
Answer: We use expansions of solids by heating:
1)
The iron rim round a Tonga wooden wheel is made
slightly smaller in diameter than the wheel. The iron is heated to high
temperature and thus it expands. And while still hot it is mounted to the
wheel. The cold water is poured over it, due to which the iron rim contracts
and firmly binds the wooden wheel.
2)
While laying rail tracks gaps are left at joints so as
to avoid damages caused by expansion or contraction.
3)
Pipes passing through the deserts and plains are
curved to allow expansion or contraction due to the change of season.
4)
Sometimes a glass stopper sticks into the neck of a
glass bottle. Then the neck of the bottle is slightly heated, due to which it expands
and the stopper is easily opened.
Q No. 12 What do you mean by bimetal strip?
Answer: In a bimetal strip, to thin strips of different metals
such as brass and iron are bounded together as shown
On heating the strip, brass expands more
than iron. This unequal expansion causes bending of the strip as shown.
Bimetal strips are
used in to measure temperatures especially in furnaces and ovens. They are also
used in thermostats to control temperature. A thermostats switch used to
control the temperature of heater coil in an electric iron.
Q No. 13 What is thermal expansion of
liquids?
Answer: Liquids
have no definite shape of their own. A liquid always attains shape of the
container in which it is poured. Therefore, when a liquid is heated, both
liquid and the container undergo a change in their volume. Thus, there are two
types of thermal volume expansion for liquid.
·
Apparent volume expansion
·
Real volume expansion
Activity:
Take a
long-necked flask. Fill it with some coloured liquid up to the mark A on its neck
as shown. Now
start heating the flask from bottom. The liquid level first falls to B and then
rises to C.
The heat first reaches the
flask which expands and its volume increases. As a result liquid descends in
the flask and its level falls to B. after sometimes, the liquid begins to rise
above B on getting hot. At certain temperature it reaches at C. The rise in
level from A to C is due to the apparent expansion in the volume of liquid.
Actual expansion of the liquid is greater than that due to the expansion
because of the expansion of the glass flask. Thus real expansion of the liquid
is equal to the volume difference between A to C in addition to the volume
expansion of flask.
Real expansion
of the liquid =apparent expansion of the liquid +expansion of the flask -
OR BC =
AC + AB
Examples of
chapter 08
Example 1: What will be the temperature on
Kelvin scale of temperature when it is 20°C on Celsius scale?
Solution:
From Celsius to Kelvin Scale: T(K)=273+C
C=20°C
T(K)=?
= 273+C
=273+20
=293K
Example 2: Change 300K on Kelvin scale
into Celsius scale of temperature.
Solution:
From Kelvin to Celsius scale: C=T(K)-273
T=300K
C=?
C=T(K)-273
C=300-273
C=27°C
Example 3: Convert 50°C on Celsius
scale into Fahrenheit temperature scale.
Solution:
From Celsius to Fahrenheit scale: F=1.8C+32
C=50°C
F=?
F=1.8C+32
F=1.8×50+32
F=90+32
F=122°F
Example 4: Convert 100°F into
temperature on Celsius scale.
Solution: F=100°F
F=1.8C+32
100=1.8C+32
100-32=1.8C
68=1.8C
68 ̷ 1.8=C
37.8°C=C
C=37.8°C
Example 5: A container has 2.5 liters
of water at 20° C. How much heat is required to boil the water?
Solution:
Volume of water =2.5 L
Mass of water m=2.5 kg
Specific heat of water C=4200J kg-1K-1
Initial temperature t1= 20° C
Final temperature t2=
100°C
Change in temperature ΔT=t2-t1
=100-20
=80°C OR 80K
Heat required ΔQ =?
ΔQ= c m Δ T
=
4200×2.5×80
= 840000J OR
840 KJ
Example 6: A brass rod is one meter
long at 0°C. Find its length at 30°C. (Coefficient of linear expansion of
brass= 1.9×10-5K-1).
Solution: Lo=1m
L=?
T1=0°C + 273=273K
T2=30°C+ 273=303K
α=1.9×10-5K-1
ΔT=T2-T1
ΔT=303-273
ΔT=
30K
L=Lo(1+α ΔT)
L=1(1+1.9×10-5×30
L=(1+57×10-5)
L=
L=(1+
)
)
L=(1+0.00057)
L=1.00057m
Example 7: Find the volume of a brass
cube at 100°C whose side is 10 cm at 0°C. (Coefficient of linear thermal
expansion of brass =1.9×10-5K-1).
Solution:
Lo=10
cm =0.1m
Vo=0.1×0.1×0.1 =1×10-3m3 =10-3m3
To=0°C =(0+273)K
T=100+273 =373K
ΔT=T-To
=373-273
=100K
α=1.9×10-5K-1
=3α=3×1.9×10-5=5.7×10-5
V=?
V=Vo(1+
ΔT)
ΔT)
V=10-3(1+5.7×10-5×100)
V=10-3(1+5.7×10-5×102)
V=10-3(1+5.7×10-3)
V=10-3(
)
)
V=10-3(
)
)
V=10-3(1+0.0057)
V=10-3(1.0057)
V=1.0057×10-3m3
Problems Chapter No. 08
Problem 1: Temperature of a water in
a beaker is 50° C. what is its value in Fahrenheit scale?
Solution: C=50°C
F°=?
F=1.8C+32
F=1.8×50+32
F=122F°
Problem 2: Normal human body
temperature is 98.6° F. Convert it into Celsius scale and Kelvin scale.
Solution: F=98.6°F
C=?
F=1.8C+32
98.6=1.8C+32
98.6- 32=1.8C
66.6=1.8C
37°C=C
TK=310K
Problem 3: Calculate the increase in the
length of an aluminum bar 2 m long when heated from 0°C to 20°C. the thermal
coefficient of linear expansion of
aluminum is 2.5x 10-5K-1.
Solution: Lo=2m
T1=0°C
+273=273K
T2 = 20°C+273=293
ΔL=?
ΔT=T2-T1
= 293-273
= 20K
ΔL=
L0ΔT
ΔL=2.5X10-5×2×20K
ΔL=100x10-5
= 102x 10-5
=102-5
=10-3
=
=
=0.001m
ΔL=0.001x100
Problem 4: A balloon
contains 1.2 m3 air at 15°C. Find its volume at 40°C. Thermal coefficient of volume expansion of
air is 3.67x10-3K-1.
Solution:
Vo=1.2m3
T1=15°C+273=288K
T2 =40°C +273= 313K
ΔT=T2-T1
=313-288
=25K
V=?
V=Vo(1+
ΔT)
V=1.2(1+3.67x10-3x25)
V=1.2(1+91.75x10-3)
V=1.2+110.1x10-3
V=1.2+
V=1.2+
V=1.3m3
Problem 5: How much
heat is required to increase the temperature of 0.5 kg of water from 10°C to
65°C?
Solution:
m=0.5kg
T1=10°C+273=283K
T2=65°C+273=338K
ΔT=T2-T1
=338-283=55K
C=4200Jkg-1K-1
ΔQ=?
ΔQ=c m Δ T
ΔQ=115500J
Problem 6: An electric
heater supplies heat at the rate of 1000 Joule per second. How much time is
required to raise the temperature of 200 g of water from 20°C to 90°C.
Solution:
P=1000Js-1
m=200g =
=0.2kg
T1=20°C+273=293K
T2=90°C+273=363K
ΔT=T2-T1
ΔT=363-293
ΔT=70K
t=?
P=
P=
P=
P=
Problem 7: How much ice
will melt by 50000J of heat ? Latent heat of fusion of ice=336000Jkg-1.
Solution:
ΔQf=50000J
Hf=336000Jkg-1
Hf=
m=
m=
m=
m=0.15kg
m=0.15×1000g
Problem 9: How much
heat is required to change 100 g of water
at 100°C into steam?(Latent heat of vaporization of water is 2.26×106Jkg-1?
Solution:
m=100g=
=0.1kg
Hv=2.26×106Jkg-1
ΔQv=?
Hv=
m×Hv=ΔQv
0.1×2.26×106=ΔQv
2.26×105J=ΔQv
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