# Thermal Properties of Matter NCERT Textbook PDF

NCERT Solutions for Class 11 Physics Chapter 11′ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 11 Physics Chapter 11 Exercise Solution’ using the download button.

### Chapter 11: Thermal Properties of Matter

#### 11.1 Introduction

We all have common-sense notions of heat and temperature. Temperature is a measure of the ‘hotness’ of a body.

A kettle with boiling water is hotter than a box containing ice. In physics, we need to define the notion of heat, temperature, etc., more carefully.

In this chapter, you will learn what heat is and how it is measured, and study the various processes by which heat flows from one body to another.

Along the way, you will find out why blacksmiths heat the iron ring before fitting on the rim of a wooden wheel of a bullock cart and why the wind at the beach often reverses direction after the
sun goes down.

You will also learn what happens when water boils or freezes, and its temperature does not change during these processes even though a great deal of heat is flowing into or out of it.

#### 11.2 Temperature And Heat

We can begin studying the thermal properties of matter with definitions of temperature and heat. Temperature is a relative measure or indication of hotness or coldness.

A hot utensil is said to have a high temperature, and an ice cube to have a low temperature. An object that has a higher temperature than another object is said to be hotter.

Note that hot and cold are relative terms, like tall and short. We can perceive temperature by touch.

However, this temperature sense is somewhat unreliable and its range is too limited to be useful for scientific purposes.

We know from experience that a glass of ice-cold water left on a table on a hot summer day eventually warms up whereas a cup of hot tea on the same table cools down.

It means that when the temperature of the body, ice-cold water, or hot tea in this case, and its surrounding medium are different, heat the transfer takes place between the system and the surrounding medium, until the body and the surrounding medium, are at the same temperature.

We also know that in the case of glass a tumbler of ice-cold water.

#### 11.3 Measurement of Temperature

A measure of temperature is obtained using a thermometer.

Many physical properties of materials change sufficiently with temperature to be used as the basis for constructing thermometers.

The commonly used property is the variation of the volume of a liquid with temperature.

For example, a common thermometer (the liquid-in-glass type) with which you are familiar. Mercury and alcohol are the liquids used in most liquid-in-glass thermometers.

### NCERT Solutions Class 11 Physics Chapter 11 Thermal Properties of Matter

Q.1: The triple points of neon and carbon dioxide are 24.57 K and 216.55 K, respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Ans:

Given: Kelvin and Celsius scales are related as:

TC = TK – 273.15  . . . . . . . . . . . . . . (1)

We know :

T= \left( \frac{ 9 }{ 5 } \right ) T _{ c }(59​)Tc​ + 32 . . . . . . . . . . . . (2)

For neon:

TK = 24.57 K\Rightarrow⇒ TC = 24.57 – 273.15 = – 248.580 C

TF = \left( \frac{ 9 }{ 5 } \right ) T _{ c } + 32(59​)Tc​+32

= \left( \frac{ 9 }{ 5 } \right ) \times \left ( -248.58 \right ) + 32(59​)×(−248.58)+32  = – 415.440F

For carbon dioxide :

TK = 216.55 K\Rightarrow⇒  TC = 216.55 – 273.15 = – 56.600 C

TF = \left( \frac{ 9 }{ 5 } \right ) T _{ c } + 32(59​)Tc​+32

= \left( \frac{ 9 }{ 5 } \right ) \times \left ( -56.60 \right ) + 32(59​)×(−56.60)+32 = -69.88F

Q.2: Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between TA and TB?

Ans:

Given:

The triple point of water on absolute scale B, T2 = 400 B

The triple point of water on absolute scale A, T1 = 200 A

The triple point of water on the Kelvin scale, TK = 273.15 K

273.15 K on the Kelvin scale is equivalent to 200 A on the absolute scale A.\Rightarrow⇒ T1 = TK

200 A = 273.15 K

Thus, A = \frac{ 273.15 }{ 200 }200273.15​

273.15 K on the Kelvin scale is equivalent to 350 B on the absolute scale B.\Rightarrow⇒ T2 = TK

350 B = 273.15 K
Thus, B = \frac{ 273.15 }{ 350 }350273.15​

Let, TA and Tbe the triple point of water on scales A and B respectively.

Thus, we have :
273.15 \times \frac{ T_{ A } }{ 200 } = 273.15 \times \frac{ T_{ B } }{ 200 }273.15×200TA​​=273.15×200TB​​

Therefore, T: T= 4 : 7

Q.3: The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law :
R = Ro[1 + α (T – To )]
The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?

Ans:

Triple point temperature, T0 =273.16 K

Resistance at the triple point, R0=101.6 Ω

Normal melting point of lead, T1= 600.5 K
Resistance at normal melting point, R=165.5 Ω
According to approximate law

R1=R0[1+α(T1−T0)]

165.5=101.6[1+α(600.5 – 273.16)]

165.5=101.6[1+α(327.34)]

165.5=101.6+α(101.6)(327.34)

165.5=101.6+α(101.6 x 327.34)

α = (165.5 – 101.6) /(101.6 x 327.34)

α = 63.9/(101.6 x 327.34)​

Now when resistance is 123.4Ω then temperature T2 is:

R2=R0[1+α(T2−T0)]
123.4=101.6[1+α(T2 – 273.16)]

123.4= 101.6[1 +(63.9/(101.6 x 327.34)) (T2 – 273.16)]

123.4 = 101.6 + (63.9/327.34)(T2 – 273.16)

123.4  =101.6 +  (0.195)(T2) – (0.195) (273.16)

123.4  =101.6 +  (0.195)(T2) – 53.32

T= (123.4 – 101.6 + 53.32) /0.195 = 75.12/0.195  = 385.23