Maharashtra Board Class 11 Physics Solution Chapter 7 – Thermal Properties of Matter
Balbharati Maharashtra Board Class 11 Physics Solution Chapter 7: Thermal Properties of Matter. Marathi or English Medium Students of Class 11 get here Thermal Properties of Matter full Exercise Solution.
Std 
Maharashtra Class 11 
Subject 
Physics 
Chapter 
7 
Chapter Name 
Thermal Properties of Matter 
1.) Choose the correct option.
i) Range of temperature in a clinical thermometer, which measures the temperature of human body, is
(A) 70 ºC to 100 ºC
(B) 34 ºC to 42 ºC
(C) 0 ºF to 100 ºF
(D) 34 ºF to 80 ºF
Answer B
It is standard range of temperature of 34 ºC to 42 ºC.
ii) A glass bottle completely filled with water is kept in the freezer. Why does it crack?
(A) Bottle gets contracted
(B) Bottle is expanded
(C) Water expands on freezing
(D) Water contracts on freezing
Answer C
We have property of water to expand on freezing which pushes the bottle boundaries lead to cracking of bottle.
iii) If two temperatures differ by 25 °C on Celsius scale, the difference in temperature on Fahrenheit scale is
(A) 65° (B) 45°
(C) 38° (D) 25°
AnswerB
T_{c}/100 = (TF32)/180
25/100 = (TF32)/180
T_{F}= 450
iv) If α, β and γ are coefficients of linear, areal and volume expansion of a solid then
(A) α:β:γ 1:3:2 (B) α:β:γ 1:2:3
(C) α:β:γ 2:3:1 (D) α:β:γ 3:1:2
AnswerB
Standard relation between α, β and γ is 1:2:3.
v) Consider the following statements
(I) the coefficient of linear expansion has dimension K ^{1}
(II) The coefficient of volume expansion has dimension K ^{1}
(A) I and II are both correct
(B) I is correct but II is wrong
(C) II is correct but I is wrong
(D) I and II are both wrong
AnswerA
The coefficient of linear expansion and volume expansion have dimension K ^{1}.
vi) Water falls from a height of 200 m. What is the difference in temperature between the water at the top and bottom of a water fall given that specific heat of water is 4200 J kg^{1} °C^{1}?
(A) 0.96°C (B) 1.02°C
(C) 0.46°C (D) 1.16°C
AnswerC
We know that all potential energy at top gets converted into kinetic energy at bottom. Then if we assume total kinetic energy is converted into heat. So we can conclude that total potential energy is converted into heat energy. Let ΔT be the rise in temperature of water then
Mgh= ms ΔT
ΔT= mgh/ms
ΔT = 9.8 x 200/4200
ΔT = 0.460c.
2.) Answer the following questions.
i) Clearly state the difference between heat and temperature?
Answer
Sr no.  heat  temperature 

It is form of energy  It is physical quantity 

it is derived quantity  It is fundamental quantity 

S.I unit is joule  S.I unit is kelvin 

Can be measured with the help of calorimeter.  Can be measured with the help of thermometer. 
ii) How a thermometer is calibrated?
Answer
For calibration purpose we require fixed points or called reference points .we have melting point of ice(0^{0}) and boiling point of water(100^{0}).
The temperature interval between these two standard fixed temperatures is divided into equal parts called degree. This sub division is based on some physical properties which changes with temperature like pressure volume etc. then temperature scale are created.
iii) What are different scales of temperature? What is the relation between them?
Answer basically there is tree temperature scales
 Celsius scale
 Fahrenheit scale
 Kelvin scale
(TC0)/100 = (TF32)/180 = (TK273.15)/100
Where,
T_{C}= Temperature in Celsius scale
T_{F}= temperature in Fahrenheit scale
T_{K}= Temperature in Kelvin scale
iv) What is absolute zero?
Answer the temperature at which pressure of gas become zero is called absolute zero temperature. This is lowest possible temperature that can be practically achieved and it is given as 273.15 ^{0}c.
v) Derive the relation between three coefficients of thermal expansion.
Answer
Relation between β and α:
Consider a square plate of side L_{0} at 0 °C and L_{T }at T °C.
L_{T }= L_{0} (1+αT)
If area of plate at 0 °C is A_{0}Then A_{0} = L_{0}^{2}.
If area of plate at T °C is A_{T }Then
A_{T}= L_{T}^{2} = L_{0}^{2} (1+αT)^{ 2}
A_{T} = A_{0}^{2 }(1+αT)^{ 2}……1
Also we know that expression for coefficient of area expansion
A_{T} = A0 (1+βT)………2
From 1 and 2 we get
A_{0}^{2 }(1+αT)^{ 2}= A0 (1+βT)
1+ 2αT +α^{2}T^{2 }=1+βT
Since the values of α are very small, the term α^{2}T^{2} is very small and may be neglected.
∴ βT = 2αT
∴ β = 2α
Relation between γ and α:
Consider a square plate of side L_{0} at 0 °C and L_{T }at T °C.
L_{T }= L_{0} (1+αT)
If volume of the cube at 0 °C is V_{0}, V_{0} =L_{0}^{3}.
If volume of the cube at T °C is V_{T},
V_{T} = L_{T}^{3} =L_{0}^{3} (1+αT)^{3}
V_{T} = V_{0} (1+αT)^{3}…………..3
V_{T} = V_{0} (1+γT) ……………4
From equation 3 and 4 we get
V_{0} (1+αT)^{3}= V_{T} = V_{0} (1+γT)
or 1+ 3αT +3α^{2}T^{2} + α^{3}T^{3} =1+γT
Since the values of α are very small, the terms with higher powers of α may be neglected.
∴ γ = 3α
Again the result is general because any solid can be regarded as a collection of small cubes.
Finally, the relation between α, β and γ is
α= β/2= γ/3.
vi) State applications of thermal expansion.
Answer
 When boiling water is poured in glass bottle it cracks due to thermal expansion.
 In railway tracks a small gap is left between rails to compensate for the expansion of length due to heat in summer.
vii) Why do we generally consider two specific heats for a gas?
Answer
In case of gas there is noticeable change in pressure and volume of gas due to small change in temperature of gas hence two specific heats are defined specific heats at constant pressure and specific heats constant volume.
viii) Are freezing point and melting point same with respect to change of state? Comment.
Answer
No. freezing point and melting point are not same with respect to change of state as at melting point there is phase or state change of solid into liquid while at freezing point there is state change of liquid into solids.
x) Explain the term ‘steady state’.
Answer steady state temperature means at each point or cross section temperature remains the same.
Let’s consider example of rod heated at one end. Heat flows from hot end to other part of rod thereby increasing the temperature. After some time temperature at each point become same this state is called as steady state.
xi) Define coefficient of thermal conductivity. Derive its expression.
Answer
Coefficient of thermal conductivity of a material may also be defined as the rate of flow of heat per unit area per unit temperature gradient when the heat flow is at right angles to the faces of slab of given material.
Consider a cube of each side x and each face of Crosssectional area A. Suppose its opposite faces are maintained at temperatures T1 and T2 (T1 > T2). Experiments show that under steady state condition the quantity of heat ‘Q’ that flows from the hot face
to the cold face is
 Directly proportional to the crosssectional area A of the face. i.e., Q ∝ A.
 Directly proportional to the temperature difference between the two faces i.e., Q ∝ (T1 T2).
 Directly proportional to time t (in seconds) for which heat flows i.e. Q ∝ t.
 Inversely proportional to the perpendicular distance x between hot and cold faces.
Where k is a constant of proportionality and is called coefficient of thermal conductivity.
xii) Give any four applications of thermal conductivity in every day life.
Answer
 Cooking utensils are made of metals but are provided with handles of bad conductor.
 Thick walls made up of bricks are used in the construction of cold storage rooms. Brick is a bad conductor of heat so it helps in reducing the flow of heat from the surroundings to the rooms hence keeps the cold room at desired temperature.
 A gunny bag is a poor conductor of heat hence it is used to store ice to prevent ice from melting.
 Heat transfer from iron to cloths while pressing.
xiii) Explain the term thermal resistance. State its SI unit and dimensions.
Answer
We know the expression for Conduction rate P_{cond} is the amount of energy transferred per unit time through a slab of area A and thickness x, the two sides of the slab being at temperatures T1 and T2 and is given by
Pcond= Q/t= (kA(T_{1}T_{2}))/x
We know that when a current flows through a conductor, the ratio V/I is called the electrical
Resistance of the conductor where V is the electrical potential difference between the ends of the conductor and I is the current. Using analogy (T1T2) is analogous to potential difference V and Pcond is analogous to rate of flow of current therefore the ratio (T1T2)/p_{cond} is called thermal resistance in analogous to electric resistance and given by
(T_{1}T_{2}))/Pcond= kA/x
R_{THERMAL}= kA/x .
The SI unit of thermal resistance is °C s/ kcal OR °C s/J and dimensions are [M^{1} L^{2} T3 K^{1}].
xiv) How heat transfer occurs through radiation in absence of a medium?
Answer
Thermal energy is a result of temperature. The body having temperature posses thermal energy. The rapidly moving molecules of a hot body emit EM waves called thermal radiations travelling with the velocity of light. These molecules carry energy with them and transfer it to the lowspeed molecules of a cold body on which they fall. This results in an increase in the molecular motion of the cold body and its temperature rises. Thus transfer of heat by radiation is a twofold process the conversion of thermal energy into waves and reconversion of waves into thermal energy by the body on which they fall.
xv) State Newton’s law of cooling and explain how it can be experimentally verified.
Answer
According to Newton’s law of cooling the rate of loss of heat dT/dt of the body is directly proportional to the difference of temperature (T T_{0}) of the body and the surroundings provided the difference in temperatures is small.
Mathematically this may be expressed as
dT/dt ∝ (T T_{0})
dT/dt = c (T T_{0})
Experiment verification:
 Fill the calorimeter to two third of its capacity and cover it with a cover having hole for passing the thermometer.
 Insert the thermometer through hole in hot water.
 note down the temperature on thermometer at every one minute interval until temperature drops to 25^{0}
 Graph of temperature versus time is plotted. Tangent to the curve dT/dt gives rate of fall of temperature at that temperature.
 Now graph of dT/dt against TT_{0} IS Plotted. Obtain graph is straight line which verifies Newton law of cooling.