By | September 22, 2020

CBSE Class 11 PHYSICS

Revision Notes
CHAPTER 11
THERMAL PROPERTIES OF MATTER

1. Temperature, heat and its measurement
2. Thermal Expansion
3. Heat capacity and Calorimetry, Change of state
4. Heat transfer

The branch dealing with measurement of temperature is called thremometry and the devices used to measure temperature are called thermometers.

Heat

Heat is a form of energy called thermal energy which flows from a higher temperature body to a lower temperature body when they are placed in contact.

Heat or thermal energy of a body is the sum of kinetic energies of all its constituent particles, on account of translational, vibrational and rotational motion.

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.
Mechanical energy or work (W) can be converted into heat (Q) by 1 W = JQ
where J = Joule’s mechanical equivalent of heat.
J is a conversion factor (not a physical quantity) and its value is 4.186 J/cal.

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.

Highest possible temperature achieved in laboratory is about 108 while lowest possible temperature attained is 10-8 K.

Branch of Physics dealing with production and measurement temperature close to 0 K is known as cryagenics, while that deaf with the measurement of very high temperature is called pyromettemperature of the core of the sun is 107 K while that of its surface 6000 K.

Different Scale of Temperature

1. Celsius Scale In this scale of temperature, the melting point ice is taken as 0°C and the boiling point of water as ${100}^{\circ }C$ and space between these two points is divided into 100 equal parts

2. Fahrenheit Scale In this scale of temperature, the melt point of ice is taken as 32°F and the boiling point of water as 211 and the space between these two points is divided into 180 equal parts.

3. Kelvin Scale In this scale of temperature, the melting pouxl ice is taken as 273 K and the boiling point of water as 373 K the space between these two points is divided into 100 equal pss

Relation between Different Scales of Temperatures

$\frac{C}{100}=\frac{F-32}{180}=\frac{K-273}{100}=\frac{R}{80}$

Thermometric Property
The property of an object which changes with temperature, is call thermometric property. Different thermometric properties thermometers have been given below

(i) Pressure of a Gas at Constant Volume

$\frac{P}{{T}_{1}}=\frac{{P}_{2}}{{T}_{2}}$

${P}_{t}={P}_{0}\left(1+\frac{t}{273}\right)$

$t=\left(\frac{{P}_{1}-{P}_{0}}{{P}_{100}-{P}_{0}}×100\right){\phantom{\rule{thinmathspace}{0ex}}}^{o}C$

whereand pt, are pressure of a gas at constant volume.A constant volume gas thermometer can measure tempera from

(ii) Electrical Resistance of Metals

where α and β are constants for a metal.
As β is too small therefore we can take

where, α = temperature coefficient of resistance and R0 and Rt, are electrical resistances at 0°C and t°C.
$\alpha =\frac{{R}_{2}-{R}_{1}}{{R}_{1}{t}_{2}-{R}_{2}{t}_{1}}$
Platinum resistance thermometer can measure temperature from

(iii) Length of Mercury Column in a Capillary Tube

ThermoElectro Motive Force

When two junctions of a thermocouple are kept at different temperatures, then a thermo-emf is produced between the junctions, which changes with temperature difference between the junctions. Thermo-emf

where a and b are constants for the pair of metals.
Unknown temperature of hot junction when cold junction is at ${0}^{\circ }\text{C}$.
Where ${\text{E}}_{100}$ is the thermo-emf when hot junction is at $\text{1}{00}^{\circ }\text{C}$.
A thermo-couple thermometer can measure temperature from

Thermal Equilibrium

When there is no transfer of heat between two bodies in contact, the the bodies are called in thermal equilibrium.

Zeroth Law of Thermodynamics

If two bodies A and B are separately in thermal equilibrium with thirtli body C, then bodies A and B will be in thermal equilibrium with each other.

Triple Point of Water

The values of pressure and temperature at which water coexists in equilibrium in all three states of matter, i.e., ice, water and vapour called triple point of water.
Triple point of water is 273 K temperature and 0.46 cm of mere pressure.

Specific Heat

The amount of heat required to raise the temperature of unit mass the substance through 1°C is called its specific heat.
It is denoted by c or s.
Its SI unit is joule/kilogram ${}^{\circ }{C}^{\prime }\left(J/Kg{-}^{\circ }C\right)$. Its dimensions is $\left[{L}^{2}{T}^{-2}{\theta }^{-1}\right]$
The specific heat of water is , which high compared with most other substances.

Gases have two types of specific heat

1. The specific heat capacity at constant volume (Cv).
2. The specific heat capacity at constant pressure (Cr).

Specific heat at constant pressure (Cp) is greater than specific heat constant volume

.

For molar specific heats

where R = gas constant and this relation is called Mayer’s formula.

The ratio of two principal sepecific heats of a gas is represented by γ.

$\gamma =\frac{{C}_{p}}{{C}_{v}}$

The value of y depends on atomicity of the gas.
Amount of heat energy required to change the temperature of any substance is given by Q = mcΔt

• where, m = mass of the substance,
• c = specific heat of the substance and
• Δt = change in temperature.

Thermal (Heat) Capacity
Heat capacity of any body is equal to the amount of heat energy required to increase its temperature through 1°C.
Heat capacity = me
where c = specific heat of the substance of the body and m = mass of the body.
Its SI unit is joule/kelvin (J/K).

Water Equivalent

It is the quantity of water whose thermal capacity is same as the heat capacity of the body. It is denoted by W.
W = ms = heat capacity of the body.

Latent Heat

The heat energy absorbed or released at constant temperature per unit mass for change of state is called latent heat.
Heat energy absorbed or released during change of state is given by Q = mL
where m = mass of the substance and L = latent heat.
Its unit is cal/g or J/kg and its dimension is ${L}^{2}{T}^{-2}$

For water at its normal boiling point or condensation temperature (100°C), the latent heat of vaporisation is
L = 540 cal/g
= 40.8 kJ/ mol
= 2260 kJ/kg

For water at its normal freezing temperature or melting point (0°C), the latent heat of fusion is
L = 80 cal/ g = 60 kJ/mol
= 336 kJ/kg

It is more painful to get burnt by steam rather than by boiling was 100°C gets converted to water at 100°C, then it gives out 536 heat. So, it is clear that steam at 100°C has more heat than wat 100°C (i.e., boiling of water).

After snow falls, the temperature of the atmosphere becomes very This is because the snow absorbs the heat from the atmosphere to down. So, in the mountains, when snow falls, one does not feel too but when ice melts, he feels too cold.

There is more shivering effect of ice cream on teeth as compare that of water (obtained from ice). This is because when ice cream down, it absorbs large amount of heat from teeth.

Melting

Conversion of solid into liquid state at constant temperature is melting.

Evaporation

Conversion of liquid into vapour at all temperatures (even below boiling point) is called evaporation.

Boilding

When a liquid is heated gradually, at a particular temperature saturated vapour pressure of the liquid becomes equal to atmospheric pressure, now bubbles of vapour rise to the surface d liquid. This process is called boiling of the liquid.

The temperature at which a liquid boils, is called boiling point The boiling point of water increases with increase in pre sure decreases with decrease in pressure.

Sublimation

The conversion of a solid into vapour state is called sublimation.

Hoar Frost

The conversion of vapours into solid state is called hoar fr.

Calorimetry

This is the branch of heat transfer that deals with the measorette heat. The heat is usually measured in calories or kilo calories.

Principle of Calorimetry

When a hot body is mixed with a cold body, then heat lost by ha is equal to the heat gained by cold body.
Heat lost = Heat gain

Thermal Expansion

Increase in size on heating is called thermal expansion. There are three types of thermal expansion.
1. Expansion of solids
2. Expansion of liquids
3. Expansion of gases

Expansion of Solids

Three types of expansion -takes place in solid.
Linear Expansion Expansion in length on heating is called linear expansion.
Increase in length

where, ll and l2 are initial and final lengths, Δt = change in temperature and α = coefficient of linear expansion.
Coefficient of linear expansion

where 1= real length and Δl = change in length and
Δt= change in temperature.

Superficial Expansion Expansion in area on heating is called superficial expansion. Increase in area A2 = A1(1 + β Δt)
where, A1 and A2 are initial and final areas and β is a coefficient of superficial expansion.
Coefficient of superficial expansion

where. A = area, AA = change in area and At = change in temperature.

Cubical ExpansionExpansion in volume on heating is called cubical expansion. Increase in volume

where V1 and V2 are initial and final volumes and γ is a coefficient of cubical expansion.

Coefficient of cubical expansion

$\gamma =\frac{\mathrm{\Delta }V}{V×\mathrm{\Delta }t}$

where V = real volume, AV =change in volume and Δt = change in temperature.

Relation between coefficients of linear, superficial and cubical expansions

2. Expansion of Liquids

In liquids only expansion in volume takes place on heating.

(i) Apparent Expansion of Liquids When expansion of th container containing liquid, on heating is not taken into accoun then observed expansion is called apparent expansion of liquids.

Coefficient of apparent expansion of a liquid

$\left({\gamma }_{r}\right)=\frac{real\phantom{\rule{thinmathspace}{0ex}}increase\phantom{\rule{thickmathspace}{0ex}}in\phantom{\rule{thickmathspace}{0ex}}volume}{original\phantom{\rule{thinmathspace}{0ex}}volume\phantom{\rule{thickmathspace}{0ex}}×\phantom{\rule{thickmathspace}{0ex}}rise\phantom{\rule{thickmathspace}{0ex}}in\phantom{\rule{thickmathspace}{0ex}}temperture}$

(ii) Real Expansion of Liquids When expansion of the container, containing liquid, on heating

is also taken into account, then observed expansion is called real expansion of liquids.

Coefficient of real expansion of a liquid

$\left({\gamma }_{r}\right)=\frac{real\phantom{\rule{thinmathspace}{0ex}}increase\phantom{\rule{thickmathspace}{0ex}}in\phantom{\rule{thickmathspace}{0ex}}volume}{original\phantom{\rule{thinmathspace}{0ex}}volume\phantom{\rule{thickmathspace}{0ex}}×\phantom{\rule{thickmathspace}{0ex}}rise\phantom{\rule{thickmathspace}{0ex}}in\phantom{\rule{thickmathspace}{0ex}}temperture}$

Both, yr, and ya are measured in °C-1

We can show that

where,  are coefficient of real and apparent expansion of liquids and yg is coefficient of cubical expansion of the container.

Anamalous Expansion of Water

When temperature of water is increased from 0°C, then its vol decreases upto 4°C, becomes minimumat 4°C and then increases. behaviour of water around 4°C is called, anamalous expansion water.

3. Expansion of Gases

There are two types of coefficient of expansion in gases

1. Volume Coefficient (&gamma;v) At constant pressure, the change in volume per unit volume per degree celsius is called volume coefficient. $\gamma v=\frac{{V}_{2}-{V}_{1}}{{V}_{o}\left({t}_{2}-{t}_{1}\right)}$ where V0, V1, and V2 are volumes of the gas
2. Pressure Coefficient (γp) At constant volume, the change in pressure per unit pressure per degree celsius is called pressure coefficient. ${\gamma }_{p}=\frac{{p}_{2}-{P}_{1}}{{P}_{o}\left({t}_{2}-{t}_{1}\right)}$ Where  are pressure of the gas at

Practical Applications of Expansion

1. When rails are laid down on the ground, space is left between the end of two rails.
2. The transmission cables are not tightly fixed to the poles.
3. The iron rim to be put on a cart wheel is always of slightly smaller diameter than that of wheel.
4. A glass stopper jammed in the neck of a glass bottle can be taken out by warming the neck of the bottles.

Important Points

• Due to increment in its time period a pendulum clock becomes slow in summer and will lose time. Loss of time in a time period
• Loss of time in any given time interval t can be given by ΔT =(1/2)α Δθt
• At some higher temperature a scale will expand and scale reading will be lesser than true values, so that

Here, Δt is the temperature difference.

• However, at lower temperature scale reading will be more or true value will be less.

Newton’s Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature (i.e. the temperature of its surroundings).

Newton’s Law makes a statement about an instantaneous rate of change of the temperature. We will see that when we translate this verbal statement into a differential equation, we arrive at a differential equation. The solution to this equation will then be a function that tracks the complete record of the temperature over time. Newton’s Law would enable us to solve the following problem.

Example 1: The Big Pot of Soup As part of his summer job at a resturant, Jim learned to cook up a big pot of soup late at night, just before closing time, so that there would be plenty of soup to feed customers the next day. He also found out that, while refrigeration was essential to preserve the soup overnight, the soup was too hot to be put directly into the fridge when it was ready. (The soup had just boiled at 100 degrees C, and the fridge was not powerful enough to accomodate a big pot of soup if it was any warmer than 20 degrees C). Jim discovered that by cooling the pot in a sink full of cold water, (kept running, so that its temperature was roughly constant at 5 degrees C) and stirring occasionally, he could bring tht temperature of the soup to 60 degrees C in ten minutes. How long before closing time should the soup be ready so that Jim could put it in the fridge and leave on time ?

Solution: Let us summarize the information briefly and define notation for this problem.
Let T(t)  = Temperature of the soup at time t (in min).
T(0) = T0 = Initial Temperature of the soup =100 deg.
Ta = Ambient temperature (temp of water in sink) = 5 deg .

Given: The rate of change of the temperature dT/dt, is (by Newton’s Law of Cooling) proportional to the difference between the temperature of the soup T(t) and the ambient temperature Ta This means that:
$\frac{dT}{dt}$ is proportional to (T-Ta).

Here a bit of care is needed: Clearly if the soup is hotter than the water in the sink T(t) – Ta > 0, then the soup is cooling down which means that the derivative dT/dt should be negative. (Remember the connection between a decreasing function and the sign of the derivative ?). This means that the equation we need has to have the following sign pattern:
$\frac{dT}{dt}$ = -k (T – Ta)

where k is a positive constant.
This equation is another example of a differential equation. The independent variable is t for time, the function we want to find is T(t) , and the quantities Ta, k are constants. In fact, from Jim’s measurements, we know that Ta = 5, but we still don’t know what value to put in for the constant k. We will discuss this further below.