Maharashtra Board Class 11 Physics Chapter 7 Thermal Properties of Matter Solution

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°

Answer-B

Tc/100  = (TF-32)/180

25/100  = (TF-32)/180

TF= 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

Answer-B

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

Answer-A

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

Answer-C

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(00) and boiling point of water(1000).

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

(TC-0)/100  = (TF-32)/180   = (TK-273.15)/100

Where,

TC= Temperature in Celsius scale

TF= temperature in Fahrenheit scale

TK= 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 0c.

v) Derive the relation between three coefficients of thermal expansion.

Answer-

Relation between β and α:

Consider a square plate of side L0 at 0 °C and Lat T °C.

LT = L0 (1+αT)

If area of plate at 0 °C is A0Then A0 = L02.

If area of plate at T °C is AT Then

AT= LT2 = L02 (1+αT) 2

AT = A02 (1+αT) 2……1

Also we know that expression for coefficient of area expansion

AT = A0 (1+βT)………2

From 1 and 2 we get

A02 (1+αT) 2= A0 (1+βT)

1+ 2αT +α2T2 =1+βT

Since the values of α are very small, the term α2T2 is very small and may be neglected.

∴ βT = 2αT

∴ β = 2α

 

Relation between γ and α:

Consider a square plate of side L0 at 0 °C and Lat T °C.

LT = L0 (1+αT)

If volume of the cube at 0 °C is V0, V0 =L03.

If volume of the cube at T °C is VT,

VT = LT3 =L03 (1+αT)3

VT = V0 (1+αT)3…………..3

VT = V0 (1+γT) ……………4

From equation 3 and 4 we get

 

V0 (1+αT)3= VT = V0 (1+γT)

or 1+ 3αT +3α2T2 + α3T3 =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-

  1. When boiling water is poured in glass bottle it cracks due to thermal expansion.
  2. 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 Cross-sectional 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 cross-sectional 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-

  1. Cooking utensils are made of metals but are provided with handles of bad conductor.
  2. 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.
  3. A gunny bag is a poor conductor of heat hence it is used to store ice to prevent ice from melting.
  4. 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 Pcond 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(T1-T2))/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 (T1-T2) is analogous to potential difference V and Pcond is analogous to rate of flow of current therefore the ratio (T1-T2)/pcond is called thermal resistance in analogous to electric resistance and given by

(T1-T2))/Pcond=  kA/x

 RTHERMAL= kA/x .

The SI unit of thermal resistance is °C s/ kcal OR °C s/J and dimensions are [M-1 L-2 T3 K1].

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 low-speed 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 two-fold 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 -T0) of the body and the surroundings provided the difference in temperatures is small.

Mathematically this may be expressed as

dT/dt ∝ (T -T0)

dT/dt  = c (T -T0)

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 250
  • 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 T-T0 IS Plotted. Obtain graph is straight line which verifies Newton law of cooling.


Updated: April 23, 2022 — 11:47 pm

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