**CBSE NOTES CLASS 11 CHAPTER 11**

**THERMAL PROPERTIES OF MATTER**

**Heat**

Heat is the form of energy transferred between two (or more) systems or a system and its surroundings by virtue of temperature difference.

The SI unit of heat energy is joule (J).

The practical unit of heat energy is calorie. 1 cal = 4.18 J

1 calorie is the quantity of heat required to raise the temperature of 1 g of water by 1°C.

**Temperature**

Temperature of a body is the degree of hotness or coldness of the body.

A device which is used to measure the temperature is called a thermometer.

Temperature of the core of the sun is 10^{7}** **K while that of its surface is 6000 K.

NTP or STP implies 273.15 K (= 0°C = 32°F).

**Different Scale of Temperature**

**Celsius Scale**The melting point of ice is taken as 0 °C and the boiling point of water as 100°C and the gap between these two points is divided into 100 equal parts, each representing 1 °C.

**Fahrenheit Scale**The melting point of ice is taken as 32 °F and the boiling point of water as 212 °F and the gap between these two points is divided into 180 equal parts, each representing 1 °F.

**Kelvin Scale**The melting point of ice is taken as 273.16 K and the boiling point of water as 373.16 K the space between these two points is divided into 100 equal parts.

**Conversion formulae**

$$\frac{\mathrm{F}-32}{180}=\frac{\mathrm{C}}{100}\mathrm{}$$

And

K = C + 273.16

**Ideal gas equation**

PV = nRT

Where

P = pressure,

V = volume,

T = temperature in K

n = number of moles of gas

and R = universal gas constant = 8.31 J mol^{–1} K^{–1}

**For adiabatic expansion**

PV^{γ} = constant, where γ = $\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{v}}}$

**Thermal Expansion**

The increase in the dimensions of a body due to the increase in its temperature is called thermal expansion.

**Linear Expansion**

The expansion in length is called **linear expansion**.

If the substance is in the form of a long rod, then for small change in temperature, ΔT, the fractional change in length, Δl/l, is directly proportional to ΔT.

$$\frac{\mathrm{\Delta}\mathrm{l}}{\mathrm{l}}=\mathrm{}\mathrm{}\mathrm{\alpha}\mathrm{}\mathrm{\Delta}\mathrm{T}\mathrm{}$$

where *α *is known as the **coefficient of linear expansion **and is characteristic of the material of the rod.

**Area Expansion**

The expansion in area is called **area expansion**.

If the substance is in the form of a sheet, then for small change in temperature, ΔT, the fractional change in area, $\frac{\mathrm{\Delta}\mathrm{A}}{\mathrm{A}}$, is directly proportional to ΔT.

$$\frac{\mathrm{\Delta}\mathrm{A}}{\mathrm{A}}=\mathrm{}\mathrm{}\mathrm{\Delta}\mathrm{T}\mathrm{}$$

where β is known as the **coefficient of area expansion **and is characteristic of the material of the sheet.

**Volume Expansion**

The expansion in volume is called **volume expansion**.

If the substance is in the form 3d object, then for small change in temperature, ΔT, the fractional change in volume, $\frac{\mathrm{\Delta}\mathrm{V}}{\mathrm{V}}$, is directly proportional to ΔT.

$$\frac{\mathrm{\Delta}\mathrm{V}}{\mathrm{V}}=\mathrm{}\mathrm{\gamma}\mathrm{}\mathrm{\Delta}\mathrm{T}\mathrm{}$$

Where γ** **is known as the **coefficient of volume or bulk expansion **and is characteristic of the material of the object.

For an ideal gas PV = nRT

At constant pressure

PΔV = nR ΔT

$$\frac{\mathrm{\Delta}\mathrm{V}}{\mathrm{V}}=\frac{\mathrm{\Delta}\mathrm{T}}{\mathrm{T}}$$

Or

$$\mathrm{\gamma}\mathrm{}\mathrm{\Delta}\mathrm{T}=\mathrm{}\frac{\mathrm{\Delta}\mathrm{T}}{\mathrm{T}}$$

Or coefficent of volume expansion for gases γ = $\frac{1}{\mathrm{T}}$

**Relationship between different coefficients of expansion**

Solids have three coefficients α, β and γ**,** liquids and gases have only γ**.**

γ_{ }= 3α and β = 2α

**Proof **

**Linear vs Area Coefficients**

Consider a rectangular sheet of the solid material of length ‘*a*’ and breadth ‘*b*’.

When the temperature increases by ΔT, *a* increases by

Δ*a* = α_{ }*a* ΔT

and *b *increases by

Δ*b* = α_{ }*b* ΔT.

Now, ΔA = ΔA_{1} +ΔA_{2} + ΔA_{3}

Or, ΔA = *a Δb + b *Δ*a + (*Δ*a) (*Δ*b)*

* = a αb *ΔT* + b α a *ΔT* + (α)*^{2}* **ab (*ΔT*)*^{2}

* = α ab *ΔT* (*2* + α *Δ*T) = α *A* *ΔT* (*2* + α*_{ }ΔT*)*

Since α ≈10^{–5} K^{–1}, the product α ΔT for fractional temperature is small in comparision with 2 and may be neglected.

$$\frac{\mathrm{\Delta}\mathrm{A}}{\mathrm{A}}\mathrm{=}2\mathrm{\alpha \; \Delta T}\mathrm{}$$

⇒ β = 2α

**Linear vs Volume Coefficients**

Let us cbonsider *a* cube of side = *l*

Then orginal volume V_{ }= *l*^{3}

On raising the temperature by ΔT,

new side will be = *l* + α*l*ΔT

and new volume = (*l *+ α*l*ΔT)^{3}

So, ΔV = (*l *+ Δ*l*)^{3} – *l*^{3} = 3*l*^{2}Δ*l *

terms in (Δ*l*)^{2} and (Δ*l*)^{3} have been neglected since Δ*l* is small compared to l.

Therefore,

$$\frac{\mathrm{\Delta}\mathrm{V}}{\mathrm{V}}=\frac{3\mathrm{\alpha}\mathrm{l}\mathrm{\Delta}\mathrm{T}}{\mathrm{l}}=\mathrm{}3\mathrm{\alpha}\mathrm{\Delta}\mathrm{T}$$

⇒ γ = 3α

**Thermal Expansion of Rod Fixed Rigidly at Both Ends**

If the rod is fixed rigidly at both the ends, it acquires a compressive strain due to the external forces provided by the rigid support at the ends. The corresponding stress set up in the rod is called **thermal stress**.

The compressive strain is,

$$\frac{\mathrm{\Delta}\mathrm{l}}{\mathrm{l}}=\mathrm{}\mathrm{\alpha}\mathrm{}\mathrm{\Delta}\mathrm{T}$$

Thermal stress

$$\frac{\mathrm{F}}{\mathrm{A}}=\mathrm{}\mathrm{Y}\frac{\mathrm{\Delta}\mathrm{l}}{\mathrm{l}}=\mathrm{}\mathrm{\alpha}\mathrm{}\mathrm{\Delta}\mathrm{T}$$

Or the force developed due to this is

$$\mathrm{F}\mathrm{}=\mathrm{}\mathrm{Y}\mathrm{}\mathrm{A}\frac{\mathrm{\Delta}\mathrm{l}}{\mathrm{l}}$$

Here Y is the Young’s modulus of the material.

If we consider steel, the force comes in the range of 10^{7} N, which is very strong force. If the iron rails are placed end to end touching each other, this force will easily bend the rails. That is the reason, gap is left between the two rails.

**Heat Capacity of a System**

Heat capacity (S) of a system is defined as the amount of heat required to raise the temperature of a system by 1°C.

q = S × ΔT

⇒ S = $\frac{\mathrm{q}}{\mathrm{\Delta}\mathrm{T}}$

- Heat capacity is an extensive property.

**Molar Heat Capacity**

Molar heat capacity (C_{m}) of a system is defined as the amount of heat required to raise the temperature of 1 mole of substance by 1°C.

$${\mathrm{C}}_{\mathrm{m}}\mathrm{}\mathrm{}=\frac{\mathrm{S}}{\mathrm{n}}=\frac{\mathrm{q}}{\mathrm{\Delta}\mathrm{T}}\times \frac{\mathrm{M}}{\mathrm{m}}$$

where n = number of moles, m = given mass, M = molar mass

**Specific Heat Capacity**

Specific heat capacity (s) of a system is defined as the amount of heat required to raise the temperature of 1gram of substance by 1°C.

$$\mathrm{s}\mathrm{}=\frac{\mathrm{S}}{\mathrm{m}}=\mathrm{}\frac{\mathrm{q}}{\mathrm{m}\mathrm{\Delta}\mathrm{T}}\mathrm{}\mathrm{}$$

where, m = mass of substance, ΔT = change in temperature.

**The Relationship between C**_{p}** and C**_{V}** for an Ideal Gas**

At constant volume, the heat capacity, C is denoted by C_{V }and at constant pressure, this is denoted by C_{p}.

C_{p} - C_{V} = R

C_{P}/C_{V} = γ

**Measurement of ΔU And ΔH**

**Calorimetry**

An experimental technique used to measure energy changes associated with chemical or physical processes is called calorimetry. In calorimetry, the process is carried out in a vessel called calorimeter, which is immersed in a known volume of a liquid.

Measurements are made under two different conditions:

i) at constant volume, q_{V, }ΔU

ii) at constant pressure, q_{p,} ΔH

**(a) ΔU measurements**

Heat absorbed at constant volume, is measured in a bomb calorimeter. A steel vessel (the bomb) is immersed in a water bath. The whole device is called calorimeter. The steel vessel is immersed in water bath to ensure that no heat is lost to the surroundings. A combustible substance is burnt in pure dioxygen supplied in the steel bomb. Heat evolved during the reaction is transferred to the water around the bomb and its temperature is monitored. Since the bomb calorimeter is sealed, its volume does not change i.e., the energy changes associated with reactions are measured at constant volume. Hence no work is done.

$$\mathrm{\Delta}{\mathrm{U}}_{\mathrm{M}}\mathrm{}=\mathrm{}{\mathrm{q}}_{\mathrm{V}}\mathrm{}=\mathrm{}\mathrm{}\mathrm{C}\mathrm{\Delta}\mathrm{T}\mathrm{}\times \frac{\mathrm{M}}{\mathrm{m}}$$

Where

C = heat capacity of the calorimeter.

M = molar mass of the substance,

m = given mass of the substance,

ΔT = change in temperature

U_{M}= Internal energy per mole of the substance.

**(b) ΔH measurements**

Measurement of heat change at constant pressure (generally under atmospheric pressure) can be done.

ΔH = q_{p }= Δ_{r}H

= (Mass of reactants + Solvent) × specifc heat of the solvent × change in temperature

= (m_{r }+ m_{s}) × c ΔT

Or ΔH_{M }= (m_{r}+ m_{s}) × c $\frac{\mathrm{\Delta}\mathrm{T}}{\mathrm{M}}$ ……[per mole]

**Change of State**

**Change of State from Solid to Liquid**

The change of state from solid to liquid is called **melting **and from liquid to solid is called **fusion**.

The temperature remains constant until the entire amount of the solid substance melts.

Both the solid and liquid states of the substance coexist in thermal equilibrium during the change of state from solid to liquid**. **

The temperature at which the solid and the liquid states of the substance in thermal equilibrium with each other is called its **melting point**.

The melting point of a substance at standard atmspheric pressure is called its **normal melting point**.

- Melting point decreases with increase in pressure and increases with decrease in pressure.
- Melting point decreases with addition of soluble impurities

**Regelation** is the phenomenon of melting of ice under pressure and freezing again when the pressure is reduced.

When we loop a fine wire around a block of ice, with a heavy weight attached to it. The pressure exerted on the ice slowly melts it locally, permitting the wire to pass through the entire block. The wire's track will refill as soon as pressure is relieved, so the ice block will remain solid even after wire passes completely through.

Skating is possible on snow due to the formation of water below the skates. Water is formed due to the increase of pressure and it acts as a lubricant.

**Change of State from Liquid to Gas**

The change of state from liquid to vapour (or gas) is called **vaporisation**.

The temperature remains constant until the entire amount of the liquid is converted into vapour.

Both the liquid and vapour states of the substance coexist in thermal equilibrium, during the change of state from liquid to vapour.

The temperature at which the liquid and the vapour states of the substance coexist is called its **boiling point**.

The boiling point of a substance at standard atmospheric pressure is called its **normal boiling point**.

- Boiling point increases with increase in pressure and decreases with decrease in pressure.
- Boiling point increases with addition of soluble impurities

**Sublimation **

The change from solid state to vapour state without passing through the liquid state is called **sublimation**, and the substance is said to sublime.

Dry ice (solid CO_{2}) sublimes, so also iodine. During the sublimation process both the solid and vapour states of a substance coexist in thermal equilibrium.

**Latent Heat**

The amount of heat per unit mass transferred during change of state of the substance is called latent heat of the substance for the process.

At this temperature, the addition of more heat does not increase the temperature but causes the change of state at the melting or boiling point.

The latent heat for a solid-liquid state change is called the **latent heat of fusion **(L_{f}), and that for a liquid-gas state change is called the **latent heat of vaporisation **(L_{v}).

If mass m of a substance undergoes a change from one state to the other, then the quantity of heat required is given by

Q = m L or L = Q/m

where L is known as latent heat.

**Triple Point of Water**

The values of pressure and temperature at which water coexists in equilibrium in all three states, i.e., ice, water and vapour is called triple point of water.

The triple point of pure water is at 0.01°C (273.16K, 32.01°F) and 4.58 mm (611.2 Pa) of mercury

**Anmolous Behaviour of Water **

Water contracts on heating between 0°C and 4°C. Also the volume of a given amount of water decreases as it is cooled from room temperature, until its temperature reaches 4°C. Below 4°C, the volume increases, and therefore the density decreases.

This means that water has a maximum density at 4°C.

Lakes and ponds freeze at the top first. As a lake cools toward 4°C, water near the surface loses energy to the atmosphere, becomes denser, and sinks; the warmer, less dense water near the bottom rises. However, once the colder water on top reaches temperature below 4°C, it becomes less dense and remains at the surface, where it freezes.

If water did not have this property, lakes and ponds would freeze from the bottom up, which would destroy much of aquatic life.

Thermal expansion of solids < liquids < gases.

**Heat Transfer**

Heat is energy transfer from one system to another or from one part of a system to another part, arising due to temperature difference.

Different ways of heat transfer are

**Conduction**

Conduction is the mechanism of transfer of heat between two adjacent parts of a body because of their temperature difference.

Heat conduction may be described quantitatively as the time rate of heat flow in a material for a given temperature difference.

In the steady state, **the rate of flow of heat** (or heat current) through a rod is proportional to the temperature difference (T_{C} – T_{D}) and the area of cross section A and is inversely proportional to the length L:

$$\mathrm{Q}\mathrm{}=\mathrm{K}\mathrm{}\mathrm{A}\mathrm{}\frac{{\mathrm{T}}_{\mathrm{C}}-{\mathrm{T}}_{\mathrm{D}}\mathrm{}}{\mathrm{L}}\mathrm{\Delta}\mathrm{t}$$

The constant of proportionality K is called the **thermal conductivity **of the material. The greater the value of K for a material, the more rapidly will it conduct heat.

The SI unit of K is J s^{–1} m^{–1} K^{–1} or Watt m^{–1} K^{–1}.

**Convection**

Convection is heat transfer by mass motion of a fluid such as air or water when the heated fluid is caused to move away from the source of heat, carrying energy with it. It is possible only in fluids.

Convection above a hot surface occurs because hot air expands, becomes less dense, and rises.

Example – Sea breeze and land breeze

**Sea Breeze**

During the day, the ground heats up more quickly than large bodies of water do. This occurs because the water has a greater specific heat and because mixing currents disperse the absorbed heat throughout the great volume of water.

The air in contact with the warm ground is heated by conduction. It expands, becoming less dense than the surrounding cooler air. As a result, the warm air rises (air currents) and colder air from the sea moves to fill the space-creating a sea breeze near a large body of water. Cooler air descends, and a thermal convection cycle is set up, which transfers heat away from the land.

**Land Breeze**

At night, the ground loses its heat more quickly, and the water surface is warmer than the land. The air in contact with the warm water is heated by conduction. It expands, becoming less dense than the surrounding cooler air. As a result, the warm air rises (air currents) and colder air from the land moves to fill the space-creating a sea breeze near a large body of water. Cooler air descends, and a thermal convection cycle is set up, which transfers heat away from the water. This is called land breeze.

**Trade Wind**

The steady surface wind on the earth blowing in from north-east towards the equator is called trade wind.

The equatorial and polar regions of the earth receive unequal solar heat. Air at the earth’s surface near the equator is hot while the air in the upper atmosphere of the poles is cool. In the absence of any other factor, a convection current would be set up, with the air at the equatorial surface rising and moving out towards the poles, descending and streaming in towards the equator. The rotation of the earth, however, modifies this convection current. Because of this, air close to the equator has an eastward speed of 1600 km/h, while it is zero close to the poles. As a result, the air descends not at the poles but at 30° N (North) latitude and returns to the equator.

**Radiation**

Heat transfer due to emission of electromagnetic waves is known as thermal radiation. Heat transfer through **radiation** takes place in form of electromagnetic waves mainly in the infrared region. Radiation emitted by a body is a consequence of thermal agitation of its composing molecules.

A **black body** is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence.

We wear white or light coloured clothes in summer so that they absorb the least heat from the sun. However, during winter, we use dark coloured clothes which absorb heat from the sun and keep our body warm.

The bottoms of the utensils for cooking food are blackened so that they absorb maximum heat from the fire and give it to the vegetables to be cooked.

**Working of thermos flask**

It consists of a double-walled glass vessel with the inner and outer walls coated with silver. Radiation from the inner wall is reflected back into the contents of the bottle. The outer wall similarly reflects back any incoming radiation. The space between the walls is evacuted to reduce conduction and convection losses and the flask is supported on an insulator like cork.

**Newton’s Law of Cooling**

According to Newton’s law of cooling, the rate of loss of heat, – dQ/dt of the body is directly proportional to the difference of temperature ΔT = (T_{2}–T_{1}) of the body and the surroundings. The law holds good only for small difference of temperature.

$$-\frac{\mathrm{d}\mathrm{Q}}{\mathrm{d}\mathrm{t}}=\mathrm{k}\mathrm{}({\mathrm{T}}_{2}-{\mathrm{T}}_{1})$$

Also, the loss of heat by radiation depends upon the nature of the surface of the body and the area of the exposed surface.

Suppose a body of mass m and specific heat capacity s is at temperature T_{2}. Let T_{1} be the temperature of the surroundings. If the temperature falls by a small amount dT_{2} in time dt, then the amount of heat lost is

dQ = ms dT_{2}

Rate of loss of heat is given by

$$-\frac{\mathrm{d}\mathrm{Q}}{\mathrm{d}\mathrm{t}}=-\mathrm{m}\mathrm{s}\mathrm{}\frac{\mathrm{d}{\mathrm{T}}_{2}}{\mathrm{d}\mathrm{t}}$$

$$\Rightarrow \mathrm{}\mathrm{m}\mathrm{s}\mathrm{}\frac{\mathrm{d}{\mathrm{T}}_{2}}{\mathrm{d}\mathrm{t}}=\mathrm{k}\mathrm{}({\mathrm{T}}_{2}-{\mathrm{T}}_{1})\mathrm{}$$

$$\Rightarrow \mathrm{}\mathrm{}\frac{\mathrm{d}{\mathrm{T}}_{2}}{{\mathrm{T}}_{2}-{\mathrm{T}}_{1}}=-\frac{\mathrm{k}}{\mathrm{m}\mathrm{s}}\mathrm{}\mathrm{}\mathrm{d}\mathrm{t}=\mathrm{}-\mathrm{K}\mathrm{}\mathrm{d}\mathrm{t}$$

Where $\frac{\mathrm{k}}{\mathrm{m}\mathrm{s}}$ = K

Integrating both sides we get,

*ln* (T_{2}-T_{1}) = - Kt + c

⇒ T_{2}-T_{1} = e^{-Kt }e^{c}

⇒ T_{2} = T_{1} + C’ e^{-Kt }

Where C’ = e

We can plot these realtions as follows.