TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Telangana TSBIE TS Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter

Very Short Answer Type Questions

Question 1.
Distinguish between heat and temperature. [TS Mar. ’15]
Answer:
Differences between heat and temperature :

Heat Temperature
1. Heat is a form of energy. 1. It represents relative degree of hotness (or) Coldness of a body.
2. Unit: joule (or) calorie 2. Unit: °C or °F
3. It is cause 3. It is effect.
4. Heat is measured by calorimeter 4. It is measured by thermometer.
5. Quantity of heat supplied Q = m st 5. Change in temperature of a body ∆ t = \(\frac{Q}{ms}\)

Question 2.
Explain triple point of water.
Answer:
The temperature of a substance remains constant during its change of state.

A graph plotted between temperature (T) and pressure (P) of a substance during change of state is called “phase diagram”.
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 1

In phase diagram of water, the P – T plane is divided into three regions.

The line ‘AO’ is called fusion curve. It gives equilibrium temperature and pressure between solid and liquid states.

The line ‘CO’ is called vaporisation curve. It gives equilibrium temperature and pressure between liquid and vapour states.

The line ‘BO’ is called sublimation curve. It gives the relation between of temperature and pressure between solid and vapour states.

Triple Point:
At point ‘O’ the curves AO, BO and CO will intersect.

It gives the temperature and pressure at which solid, liquid and vapour states of water are in equilibrium.

Coordinate of triple point of water a temperature = 273.16 K, pressure = 0.006 atmos (or) 611 pascals.

Question 3.
What are the lower and upper fixing points in Celsius and Fahrenheit scales? [TS Mar. ’16; AP Mar. ’19, ’18 ’16, May ’14]
Answer:
Centigrade (Celsius) scale of temperature:
In centigrade scale of temperature lower fixed point is freezing point of water at one atmosphere pressure, as 0°C. The upper limit is boiling point of water at 1 atm pressure, as 100°C.

Fahrenheit scale of temperature :
In Fahrenheit scale, the lower fixed point is freezing point of water at one atmosphere pressure, as 32°F. The upper fixed point is boiling point of water at 1 atm pressure, as 212°F.

TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 4.
Do the values of coefficients of expansion differ, when the temperatures are measured on Centigrade scale or on Fahrenheit scale?
Answer:
Yes. Coefficients of expansion α, αa and αv are not same in Celsius scale and in Fahrenheit scale. In Fahrenheit scale values of α, αa and αv are less than those in Celsius scale.

Since magnitude of 1°C > magnitude of 1°F this change takes place. The values of Celsius scale are 1.8 times more than the values in Fahrenheit scale.
αF = 5/9 αc

Question 5.
Can a substance contact on heating? Give an example. [AP Mar. ’18, ’16, May ’16; TS May, ’18, ’16]
Answer:
Yes. Some substances will contact on heating. Ex: Leather, rubber, cast Iron type metal.

Question 6.
Why gaps are left between rails on a railway track? [TS Mar. ’19; AP Mar. ’19, ’17, ’16, ’09; May ’16; June ’15]
Answer:
To allow the linear expansion rails.

In summer temperature of atmosphere increases so rails will expand. If no gap is given between rails then the rails will bend it leads to accidents. If gap is given the rails will expand into that gap and that track is safe.

Question 7.
Why do liquids have no linear and areal expansions? [TS Mar. ’19]
Answer:
Liquids have only volume expansion. No linear expansion or areal expansion. Because liquids does not have any independent shape they must be taken in a container. So we will consider only volume of liquid.

Question 8.
What is latent heat of fusion? [AP & TS May ’17]
Answer:
Latent heat of fusion (melting):
It is defined as the amount of heat energy absorbed or rejected by unit mass of substance while converting from solid to liquid or from liquid to solid.

Question 9.
What is latent heat of vapourisation? [AP Mar. ’13]
Answer:
Latent heat of vapourisation :
It is defined as the amount of heat energy absorbed or rejected by unit mass of substance while converting from liquid to vapour or from vapour to liquid state.

Question 10.
Why are utensils coated black? Why is the bottom of the utensils made of copper? [AP May ’18; TS Mar. ’18]
Answer:
Lower portion of the utensils is in contact with fire. Black bodies are good heat absorbers. So, a black bottom will absorb more heat.

Copper is a good conductor of heat. So, copper is used at the bottom of cooking utensils.

TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 11.
What is triple point of Water? Mention the values of temperature and pressure at triple point of water. [TS June ’15]
Answer:
Triple point:
The temperature and pressure where a substance can coexist in all its three states is called the “triple point”.

i.e., The substance will exist as a solid, as liquid and as vapour at that particular temperature and pressure.

For water the triple point is at a temperature of 273.16 K and at a pressure of 6.11 × 10-3 atmospheres or nearly 610 pascals.

Question 12.
State Boyle’s law and Charles law. [AP June 15; TS Mar. 15]
Answer:
Boyle’s Law :
At constant temperature, the volume (V) of a given mass of a gas is inversely proportional to its pressure (P).
∴ V ∝ \(\frac{1}{P}\) ⇒ PV = constant = K.

Charles Law:
At constant pressure, the volume (V) of a given mass of a gas is directly proportional to its absolute temperature (T).
∴ V ∝ T ⇒ \(\frac{V}{T}\) = K (constant)

Question 13.
State Wein’s displacement law. [AP Mar. ’17]
Answer:
Wein’s Displacement Law :
The wavelength (Ain) of maximum intensity of emission of black body radiation is inversely proportional to absolute temperature (T) of the black body.
i.e., λm ∝ \(\frac{1}{T}\) (or) λm = \(\frac{b}{P}\)
where ‘b’ is called ‘Wein’s constant”.

Question 14.
Ventilators are provided in rooms just below the roof. Why?
Answer:
Density of hot air is less. So in a room hot air goes to top layers i.e., nearer to the roof.

When ventilators are provided nearer to the roof hot air will escape easily from room. So we feel that the room is cool and circulation of air will become easy.

Question 15.
Does a body radiate heat at 0 K? Does it radiate heat at 0°C?
Answer:
According to Precost’s theory, every body above zero Kelvin will radiate heat energy to the surroundings.
So, i) A body at O’ Kelvin does not radiate heat energy.
ii) A body at 0°C i.e., at 273Kwill radiate heat energy.

Question 16.
State the different modes of transmission of heat. Which of these modes require medium? [TS May ’18]
Answer:
Transmission of heat is of three types.
1) Conduction 2) Convection 3) Radiation

For propagation of heat energy medium is required in case of conduction and convection.

TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 17.
Define coefficient of thermal conductivity and temperature gradient.
Answer:
“The coefficient of thermal conductivity”
(K) it is the quantity of heat flowing normally per second through unit area of the substance per unit temperature gradient.
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 2

‘Temperature gradient” is defined as the change in temperature along the conductor per unit length.

Temperature gradient
Change in temperature
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 3

Question 18.
Define emissive power and emissivity.
Answer:
Emissive power:
The emissive power of a body is defined as the energy radiated by the body per second per unit area at a given temperature and wavelength.

Emissivity:
Emissivity is defined as the ratio of the emissive power of a body to that of a black body at the same temperature.

Question 19.
Is there any substance available in nature that contracts on heating? If so, give an example. [TS May ’16]
Answer:
Yes. Some substances will contact on heating.
Ex: Leather, rubber, cast Iron type metal.

Question 20.
What is greenhouse effect? Explain global warming. [AP Mar. ’15, ’13; TS Mar. & May ’16]
Answer:
Green house effect: Earth will absorb heat radiation and reradiate heat energy of longer wavelength. This longer wave length heat radiation is reflected back to earth due to green house gases such as Carbon dioxide [CO2], Methane (CH4) Chloroflurocarbons, Ozone (O3), etc. As a result temperature of earth’s atmosphere is gradually increasing. This is known as “green house effect.”

Global warming:
Earth receives heat energy during day time from sun. It reradiates heat energy in the form of longer electromagnetic waves.

But due to presence of green house gases the longer electromagnetic waves were reflected back to earth. As a result temperature of earth’s atmosphere is gradually increasing.

This process will increase with the increased content of green house gases in atmosphere. As a result temperature of earth’s atmosphere increases gradually.

Question 21.
Define absorptive power of a body. What is the absorptive power of a perfect black body?
Answer:
Absorptive power of a body is defined as the ratio of energy absorbed by the body within the wave length range of A and A + dA to the total energy flux following on the body.
Absorptive power,
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 4

Question 22.
State Newton’s law of cooling. [AP Mar. ’18, ’16, May ’18, ’17, June ’15; TS Mar. ’18, TS May ’16]
Answer:
Newton’s Law of cooling states that the rate of loss of heat of a hot body is directly pro-portional to the difference in temperature between the body and its surroundings pro-vided the difference in temperatures is small and the nature of the radiating surface remains same.
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 5
Where k is the proportionality constant

Question 23.
State the conditions under which Newton’s law of cooling is applicable. [AP May ’16; TS June ’15]
Answer:
Newton’s law of cooling is applicable

  1. loss of heat is negligible by conduction and only when it is due to convection.
  2. loss of heat occurs in a streamlined flow of air i.e., forced convection.
  3. temperature of the body is uniformly distributed over it.
  4. temperature difference between the body and surroundings is moderate i.e., upto 30 K.

TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 24.
The roofs of buildings are often painted white during summer. Why? [TS Mar. ’17, ’15; AP May ’16]
Answer:
When roofs of buildings are coated white we will feel relatively cold during summer.

Absorptive power of white surface is less. So roofs coated white will absorb less heat energy. So less quantity of heat is transmitted into the house. So we feel less hot or cold inside the house.

Question 25.
What is thermal expansion? [TS Mar. ’16]
Answer:
The increase in interatomic distance due to thermal energy is called “thermal expansion”.

As a result the length solids or volume of liquids or pressure of gases will increase.

Question 26.
Why is it easier to perform the skating on the snow? [TS Mar. ’16]
Answer:
Due to increase of pressure melting point decreases, So it is easier to perform the skating on the snow.

Short Answer Questions

Question 1.
Explain Celsius and Fahrenheit scales of temperature. Obtain the relation between Celsius and Fahrenheit scales of temperature.
Answer:
Celsius (Centigrade) scale of temperature :
In centigrade scale of temperature lower fixed point is freezing point of water at one atmosphere pressure, as 0°C. The upper limit is boiling point of water at 1 atm pressure, as 100°C.

The interval between lower limit and upper limit [100 – 0 = 100] is divided into 100 equal parts and each part is called 1°C.

Fahrenheit scale of temperature :
In Fahrenheit scale, the lower fixed point is freezing point of water at one atmosphere pressure, as 32°F. The upper fixed point is boiling point of water at 1 atm pressure, as 212°F.

The interval between upper fixed point and lower fixed point (212 -32 = 180) is divided into 180 equal parts and each part is called 1°F.

Relation between Fahrenheit and Celsius scale of temperatures:
In both scales, lower limit and upper limit are same. The only change is in numerical values of lower and upper limits.

In Fahrenheit lower limit = 32, upper limit = 212, difference of limits = 180

In Celsius scale lower limit = 0, upper limit = 100, difference of limits = 100
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 6
C = Temperature in Celsius scale.
F = Temperature in Fahrenheit scale.

Question 2.
Two identical rectangular strips one of copper and the other of steel, are riveted together to form a compound bar. What will happen on heating?
Answer:
When two dissimilar metals say copper and steel are riveted together that arrangement is called “bimetallic strip”.

When a bimetallic strip is heated copper strip will expand more than steel due to more expansion coefficient.
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 7

Since they are riveted they must expand as a common piece. As a result bimetallic strip will bend in the form of an arc. For the metallic strip with high a its length is more so it is on the outer side. For the strip with less a its length is less. It will be at the inner side of the arc.

TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 3.
Pendulum clocks generally go fast in winter and slow in summer. Why? [TS Mar. ’19, ’17]
Answer:
In summer due to increase in temperature of atmosphere length of pendulum will increase.

Time period of pendulum T = 2π\(\sqrt{\frac{l}{g}}\)

T ∝ √l. So it will make less number of oscillations per day. So clock will run slowly in summer.

In winter temperature of atmosphere decreases. So length of pendulum decreases.

Hence time period of oscillation will also decrease. As a result, pendulum will make more oscillations per day so clocks will run fast in winter.

Question 4.
In what way is the anomalous behaviour of water advantageous to aquatic animals? [AP Mar. ’18, 17, 14; May 18. 17, 14; TS May ’18]
Answer:
In cold countries and at polar region temperature falls below 0°C at winter. So surface of water will be frozen. Due to anomalous expansion of water even though the surface of lakes, and sea are frozen water will exist at bottom layers at 4°C.

Different layers in between ice and bottom will have different temperatures like 1°C, 2°C or 3°C. In these layers, aquatic animals are able to survive even in winter.

Anomalous expansion of water helps for the survival of aquatic life at polar region and in cold countries.

Question 5.
Explain conduction, convection and radiation with examples. [TS Mar. ’18, ’16, ’15, June ’15; AP Mar. ’19, ’15, May, ’16]
Answer:
Conduction :
It is a mode of transfer of heat from one part of the body to another, from particle to particle in the direction of fall of temperature without any actual movement of the heated particles.
Ex: When one end of a metal rod is heated, its other end becomes hot. Here, the heat goes from hot end of the metal rod towards cold end, by conduction.

Convection :
It is a mode of transfer of heat from one part of the medium to another part by the actual movement of the heated particles of the medium.
Ex : Seabreeze, Tradewind, etc.

Radiation :
It is a mode of transfer of heat from the source to the receiver without any actual movement of source or receiver and also without heating the intervening medium.
Ex : Heat from sun comes to us through radiation. On standing near fire, we feel hot as heat comes to us through radiation.

Long Answer Questions

Question 1.
Explain thermal conductivity and coefficient of thermal conductivity.
A copper bar of thermal conductivity 401 W (mK) has one end at 104°C and the other end at 24°C. The length of the bar is 0.10 m and the cross-sectional area is 1.0 × 10-6 m-2. What is the rate of heat conduction along the bar?
Answer:
The ability to conduct heat in solids is called ‘Thermal conductivity.”

Consider a bar with rectangular cross-section as shown in the figure. The faces ABCD and EFGH are maintained at θ1 and θ2 respectively (θ1 > θ2). Heat passes from one end to the other.
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 8

The amount of heat conducted (Q) depends on,

  1. Amount of heat conducted Q is proportional to area of cross-section A perpendicular to flow.
    ∴ Q ∝ A ………. (1)
  2. is proportional to temperature difference between the two ends.
    ∴ Q ∝ (θ2 – θ1) …………. (2)
  3. is proportional to the time of flow.
    Q ∝ t ………. (3) and
  4. is inversely proportional to the length of the conductor.
    Q ∝ \(\frac{1}{l}\) …………. (4)

TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 9
where k = constant called coefficient of thermal conductivity.

Coefficient of thermal conductivity (k) :
It is defined as the amount of heat conducted normally per sec per unit area of cross-section per unit temperature gradient.
S.I. Unit w m k-1
Dimensional formula = [ M¹L¹T-3θ-1]

Problem:
Thermal conductivity of copper,
Kc = 401 W/m-k
Temperature at one end, θ2 = 104°C
Temperature of 2nd end, θ1 = 24°C
Length of copper bar, l = 0.1 m; Area,
A = 1.0 × 10-6 m-2
Rate of conduction,
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 10

TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 2.
State the explain Newton’s law of cooling. State the conditions under which Newton’s law of cooling is applicable.
A body cools down from 60°C to 50°C in 5 minutes and to 40°C in another 8 minutes. Find the temperature of the surroundings. [TS May ’17, ’16; AP May ’13]
Answer:
Newtons’ Law of cooling :
The rate of loss of heat of the body is directly proportional to the difference of temperature of the body and the surroundings.

Explanation :
The law holds good only for small difference of temperature. Also, the loss of heat by radiation depends upon the nature of the surface of the body and the area of the exposed surface. We can write
– \(\frac{dQ}{dt}\) = k (T2 – T1) (sign indicates loss) …….. (1)

where k is a positive constant depending upon the area and nature of the surface of the body. Suppose a body of mass ‘m’ and specific heat capacity ‘s’ is at temperature T2. Let T1 be the temperature of the surroundings. If the temperature falls by a small amount dT2 in time dt, then the amount of heat lost is
dQ = ms dT2
∴ Rate of loss of hfeat is given by
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 11
where K = k/ms

Conditions (under which Newton’s law of cooling is applicable):
Newton’s law of cooling is applicable

  1. loss of heat is negligible by conduction and only when it is due to convection.
  2. loss of heat occurs in a streamlined flow of air i.e., forced convection.
  3. temperature of the body is uniformly distributed over it.
  4. temperature difference between the body and surroundings is moderate i.e., upto 30 K.

PROBLEM :
Let ‘θo‘ be the temperature of the surroundings.

In first case :
Initial temperature, θ1 = 60°C
Final temperature, θ2 = 50°C
Time of cooling, t = 5 minutes = 5 × 60 = 300s
From Newton’s law of cooling we can write,

In secoend case :
Initial temperature, θ1 = 60°C
Final temperature, θ2 = 40°C
Time of cooling, t = 13 minutes = 13 × 60 = 780s
Again from Newton’s law of cooling we can write,
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 13
On solving equations (1) and (2) we get, θo = 33.33°C

Problems

Question 1.
What is the temperature for which the readings on Kelvin Fahrenheit scales are same?
Solution:
On the Kelvin and Fahrenheit scales
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 14
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 15

Question 2.
Find the increase in temperature of aluminium rod if it’s length is to be increased by 1%. (α for aluminium = 25 × 10-6/°C) [AP Mar. ’15; June ’15]
Solution:
Coefficient of linear expansion of aluminium, α = 25 × 10-6/°C
We know that percentage increase in length
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 16

Question 3.
How much steam at 100°C is to be passed into water of mass 100g at 20°C to raise its temperature by 5°C? (Latent heat of steam is 540 cal / g and specific heat of water is 1 cal / g°C)
Solution:
Latent heat of steam, Ls = 540 cal/g
Specific heat of water, Lw = 1 cal / g°C
Mass of water, mw = 100g

According to method of mixture or from the principle of calorimetry we can write, Heat lost by steam = heat gained by water
i.e., msLs + msSw(100 – t) = mwSw (t – 20)
∴ ms × 540 + ms × 1(100 – 25)
⇒ 100 × 1 × (25 – 20)
⇒ 615ms = 500(or)ms = \(\frac{500}{615}\) = 0.813 g

TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 4.
2 kg of air is heated at constant volume. The temperature of air is increased from 293 K to 313 K. If the specific heat of air at constant volume is 0.718 kJ/kg K, find the amount of heat absorbed in kJ and kcal. (J = 4.2 kJ/kcal.)
Solution:
Mass of air, m = 2 kg
Change in temperature, ∆T = 313 – 293 = 20K.
Specific heat at constant volume, Cv = 0.718 k.J/kg – K.
Heat mechanical equivalent, J = 4.2 kJ/k.cal.
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 17
∴ Heat energy absorbed,
Q = 2 × 0.718 × 10³ × 20. = 28.72 kJ
= 6.838 k calories.

Question 5.
A clock, with a brass pendulum, keeps correct time at 20°C, but loses 8.212 s per day, when the temperature rises to 30°C. Calculate the coefficient of linear expansion of brass.
Solution:
Temperature of correct time, t1 = 20°C
Loss or gain of time in seconds per day = 8.212 sec.
Final temperature, t2 = 30°C
∴ ∆t = 30 – 20 = 10
α of pendulum material = ?

In pendulum loss or gain of time in seconds per day = 43,200. α ∆t
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 18

Question 6.
A body cools from 60°C to 40°C in 7 minutes. What will be its temperature after next 7 minutes if the temperature of its surroundings is 10°C? [AP May ’13]
Solution:
In first case :
Initial temperature, θ1 = 60°C
Final temperature, θ2 = 40°C
Time of cooling, t1 = 7 minutes
= 7 × 60 = 420s
Temperature of surroundings, θ0 =10°C
From. Newton’s law of cooling, we can write,
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 19

In second case:
Initial temperature, θ1 = 40°C
Time of cooling, t2 = 7 minutes = 420s
Again, from Newton’s law of cooling we can write,
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 20
on solving equations (1) & (2) we get, 0 = 28°C

TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 7.
If the maximum intensity of radiation for a black body is found at 2.65 µ m, what is the temperature of the radiating body? (Wein’s constant = 2.9 × 10-3 mK)
Solution:
Wavelength corresponding to maximum intensity, λmax = 2.65 µm = 2.65 × 10-6 m.
Wein’s constant, b = 2.90 × 10-3 mK.
From Wein s Law, T = \(\frac{\mathrm{b}}{\lambda_{\mathrm{m}}}=\frac{2.90 \times 10^{-3}}{2.65 \times 10^{-6}}\)
= 1094 K.

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