This article is formulated according to the 11th chapter in the NCERT Syllabus of 11th Std Physics OR 1st PUC Physics.
Temperature is a relative measure or indication of the hotness or coldness of a body. The temperature of a body determines the direction of the flow of energy. The SI unit of temperature is kelvin (K). °C is a commonly used unit of temperature.
Heat is the form of energy transferred between two systems or a system and its surroundings by virtue of temperature difference. Heat energy flows from body of higher temperature to a body of lower temperature. Heat gained by a body is taken as positive while heat lost by a body is taken as negative. SI Unit of heat energy is joule (J).
The branch of science which deals with the measurement of temperature of a substance is known as thermometry.
A device used to measure the temperature of a body is called thermometer. Commonly used thermometers are liquid-in-glass type. Mercury and Alcohol are the liquids in most of this type of thermometer.
Principle of Thermometer
The principle of thermometer is, when a substance is heated, some of its physical properties change. The commonly used property is variation of the volume of a liquid with temperature.
To measure temperature of a body different temperature scales are defined. Each has two fixed points. The ice (freezing) point and steam (boiling) point of water are the two convenient fixed points.
Celsius scale (°C)
Celsius scale of temperature was invented by Andres Celsius. In this scale, the melting point of pure ice at standard atmospheric pressure is 0°C and marked as lower fixed point. The boiling point of pure water at standard atmospheric pressure is 100°C and marked as upper fixed point. The interval between these two points is divided into 100 equal parts. Each part is taken as “One degree Celsius”.
Fahrenheit scale (°F)
Fahrenheit scale was invented by Gabriel Fahrenheit. In this scale, the melting point of pure ice at standard atmospheric pressure is 32°F and marked as lower fixed point. The boiling point of pure water at standard atmospheric pressure is 212°F and marked as the upper fixed point. The interval between these two points is divided into 180 equal parts and each part is known as “One degree Fahrenheit”.
Relation between Celsius and Fahrenheit scale
A plot of Fahrenheit temperature (𝑡𝐹) versus Celsius temperature (𝑡𝐶) is as shown. The equation of the straight line is given by, 𝑦 = 𝑚𝑥 + 𝐶 or
𝑡𝐹 = (180/100)𝑡𝑐 + 32
𝑡𝐹 − 32 = (180/100)𝑡𝑐
(𝒕𝑭 – 𝟑𝟐)/𝟏𝟖𝟎 = 𝒕𝒄/𝟏𝟎𝟎
Drawbacks of liquid-in-glass thermometers
Liquid-in-glass thermometers show different readings for temperatures other than the fixed points. This is because of the different expansion properties of liquids. This problem can be removed if a thermometer uses a gas. They give the same reading regardless of which gas is used and experiments show that all gases at low densities exhibit the same expansion behavior.
The laws which describe the behaviour of a gas at different conditions are called gas laws and the behaviour of a given quantity of gas is explained using the variables such as pressure, volume and temperature.
When temperature is held constant, the volume of a given mass of gas is inversely proportional to its pressure.
𝑉 ∝ 1/𝑃
𝑷𝑽 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
At constant pressure, the volume of a given mass of an ideal gas is directly proportional to its absolute temperature.
𝑉 ∝ 𝑇
𝑽/𝑻 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
Ideal Gas equation
For low-density gases, we can combine Boyle’s law and Charles’s law into a single relationship as,
𝑃𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑉/𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑃𝑉/𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑃𝑉/𝑇 = 𝑅 where 𝑅 → 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
For 𝜇 moles of gas, 𝑃𝑉/𝑇 = 𝜇𝑅
𝑷𝑽 = 𝝁𝑹𝑻 This equation is called Ideal gas equation. Where R is a constant called universal gas constant and 𝑅 = 8.31𝐽𝑚𝑜𝑙−1𝐾−1
From the Ideal gas equation, we have 𝑃𝑉 ∝ 𝑇. This relationship allows a gas to be used to measure temperature in constant volume gas thermometer. At constant volume, P versus T graph is as shown. The relationship is linear over a large temperature range. It looks as pressure might reach zero, with decreasing temperature and gas remains to be a gas. Thus, pressure of a gas becomes zero at −273.15°𝐶.
The temperature at which the pressure of an ideal gas becomes zero is termed as absolute zero.
Kelvin’s scale of temperature
This scale was suggested by Kelvin. Absolute zero is foundation of the Kelvin temperature scale. The zero of the absolute scale of temperature is denoted by 0 𝐾 and known as absolute zero. Hence, 𝑡𝐾 = 𝑡𝐶 + 273.15. Ice point in Kelvin scale is 273.15K. Boiling point in Kelvin scale is 100 + 273.15 = 373.15K. Practically, 𝒕𝑲 = 𝒕𝑪 + 𝟐𝟕𝟑
Effects of Heat
The common effects of Heat are,
- Thermal expansion
- Rise in temperature
- Change in state
The increase in the dimensions of a body due to an increase in its temperature is called thermal expansion. There are three types of thermal expansion.
- Linear expansion
- Area expansion
- Volume expansion
The expansion in length is called linear expansion. When a rod-like solid is heated its length increases. The increase in length (∆𝑙) is directly proportional to, (i) its original length (𝑙0) (ii) the change in temperature (∆𝑇).
Mathematically, ∆𝑙 ∝ 𝑙0∆𝑇
∆𝑙/𝑙0 ∝ ∆𝑇
∆𝑙/𝑙0 = 𝛼𝑙∆𝑇
Where 𝛼𝑙 → constant and called co-efficient of linear expansion.
Co-efficient of linear expansion
It is defined as the increase in length per unit length per degree increase in temperature. SI unit of Co-efficient of linear expansion is 𝐾−1. It is also expressed in °𝐶−1.
(i) The increase in length (∆𝑙) is also depends on the material of the solid.
(ii) Normally metals expand more and have relatively high values of 𝛼𝑙
Consider a metallic rod whose ends are fixed rigidly. When the temperature of the rod increases, its length increases. Since there is no space left to increase its length, so it bends. If the rod is not allowed to bend, then it will be under a great stress. This is known as thermal stress.
Expression for thermal stress
Let 𝑙 be the length of the rod and 𝐴 be its cross-sectional area. 𝛼𝑙 be the co-efficient of linear expansion of the material of the rod. Let ∆𝑇 be the increase in temperature and ∆𝑙 be the increase in its length. Then, ∆𝑙/𝑙 = 𝛼𝑙∆𝑇.
But ∆𝑙/𝑙 = 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛
𝑠𝑡𝑟𝑎𝑖𝑛 = 𝛼𝑙∆𝑇
Young’s modulus, 𝑌 = 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠/𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠/𝛼𝑙∆𝑇
𝑻𝒉𝒆𝒓𝒎𝒂𝒍 𝒔𝒕𝒓𝒆𝒔𝒔 = 𝒀𝜶𝒍∆𝑻
This is the expression for thermal stress developed in the rod.
Force developed in the rod due to Thermal stress
We have, 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝐹𝑜𝑟𝑐𝑒 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑡𝑟𝑒𝑠𝑠/𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛
Force, 𝐹 = 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 × 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛
𝐹 = 𝑌𝛼𝑙∆𝑇 × 𝐴
𝐹 = 𝑌𝐴𝛼𝑙∆𝑇
𝑭 = 𝒀𝑨∆𝒍/𝒍
This is the expression for force developed in the rod.
Area (Superficial) expansion
The expansion in the area is called area expansion. When a planar body like solid is heated its area increases. The increases in area (∆𝐴) is directly proportional to, (i) its original area (𝐴0) (ii) its change in temperature (∆𝑇) Mathematically, ∆𝐴 ∝ 𝐴0∆𝑇
∆𝐴/𝐴0 ∝ ∆𝑇
∆𝐴/𝐴0 = 𝛼𝐴∆𝑇 Where 𝛼𝐴 → constant and called co-efficient of area expansion.
Co-efficient of area expansion
It is defined as the increase in area per unit area per degree increase in temperature. SI unit of Co-efficient of area expansion is 𝐾−1. It is also expressed in °𝐶−1.
Volume (Cubical) expansion
The expansion in the volume is called volume expansion. When a substance is heated its volume increases. The increases in volume (∆𝑉) is directly proportional to, (i) its original volume (𝑉0) (ii) its change in temperature (∆𝑇) Mathematically, ∆𝑉 ∝ 𝑉0∆𝑇
∆𝑉/𝑉0 ∝ ∆𝑇
∆𝑉/𝑉0 = 𝛼𝑉∆𝑇 Where 𝛼𝑉 → constant and called co-efficient of volume expansion.
Co-efficient of volume expansion
It is defined as the increase in volume per unit volume per degree increase in temperature. SI unit of Co-efficient of volume expansion is 𝐾−1. It is also expressed in °𝐶−1.
Note: Co-efficient of volume expansion is not strictly a constant. It depends on the temperature. It is initially small and rises sharply and becomes a constant at high temperatures.
Variation of Co-efficient of volume expansion (𝜶𝑽) with temperature – For SOLIDS and LIQUIDS
At ordinary temperatures, Solids and Liquids expand less compared to gases. For liquids, the coefficient of volume expansion is relatively independent of the temperature.
Variation of Co-efficient of volume expansion (𝜶𝑽) with temperature – For GASES
For gases 𝛼𝑉 is dependent on temperature, which can be obtained as follows, The ideal gas equation is given by, 𝑃𝑉 = 𝜇𝑅𝑇 At constant pressure, the equation becomes, 𝑃∆𝑉 = 𝜇𝑅∆𝑇
Taking, 𝑃∆𝑉/𝑃𝑉 = 𝜇𝑅∆𝑇/𝜇𝑅𝑇 we have, ∆𝑉/𝑉 = ∆𝑇/𝑇
∆𝑉/𝑉∆𝑇 = 1/𝑇
𝜶𝑽 = 𝟏/𝑻
It shows that 𝛼𝑉 decreases with an increase in temperature.
Anomalous expansion of water
Water contracts on heating from 0°C to 40°C. This is known as anomalous expansion of water. In the temperature range of 0°C to 40°C, the volume of water decreases as temperature increases. Hence, the Coefficient of cubical expansion of water is negative. Hence water has the greatest density at 40°C. The graph below shows the variation in volume and density of water with temperature.
Relation between 𝜶𝒍, 𝜶𝑨, and 𝜶𝑽
(i)We have, ∆𝑙/𝑙0 = 𝛼𝑙∆𝑇
∆𝑙 = 𝑙0𝛼𝑙∆𝑇
𝑙 − 𝑙0 = 𝑙0𝛼𝑙∆𝑇
𝑙 = 𝑙0 + 𝑙0𝛼𝑙∆𝑇
𝑙 = 𝑙0(1 + 𝛼𝑙∆𝑇)
Squaring on both sides,
𝑙2 = 𝑙02(1 + 𝛼𝑙∆𝑇)2
𝐴 = 𝐴0(1 + 2𝛼𝑙∆𝑇 + 𝛼𝑙2∆𝑇2)
Since 𝛼𝑙 is very small, 𝛼𝑙2∆𝑇2 can be neglected.
𝐴 = 𝐴0(1 + 2𝛼𝑙∆𝑇)
Comparing with the equation, 𝐴 = 𝐴0(1 + 𝛼𝐴∆𝑇)
We have, 𝜶𝑨 = 𝟐𝜶𝒍
(ii) We have, 𝑙 = 𝑙0(1 + 𝛼𝑙∆𝑇)
Cubing on both sides, 𝑙3 = 𝑙03(1 + 𝛼𝑙∆𝑇)3
𝑉 = 𝑉0(1 + 3𝛼𝑙∆𝑇 + 3𝛼𝑙2∆𝑇2 + 𝛼𝑙3∆𝑇3)
Since 𝛼𝑙 is very small, 𝛼𝑙2∆𝑇2 and 𝛼𝑙3∆𝑇3 can be neglected.
𝑉 = 𝑉0(1 + 3𝛼𝑙∆𝑇)
Comparing with the equation, 𝑉 = 𝑉0(1 + 𝛼𝑉∆𝑇)
We have, 𝜶𝑽 = 𝟑𝜶𝒍
Further, 𝜶𝒍:𝜶𝑨:𝜶𝑽 = 𝜶𝒍:𝟐𝜶𝒍:𝟑𝜶𝒍 = 𝟏: 𝟐: 𝟑
Heat Capacity (S)
The heat capacity of a substance is defined as the amount of heat required to raise the temperature of the substance through 1°𝐶. The SI unit is 𝑗𝑜𝑢𝑙𝑒/°𝐶 (𝐽−°𝐶−1 𝑜𝑟 𝐽−𝐾−1). The quantity of heat required to warm a given substance depends upon its change in temperature That is, ∆𝑄 ∝ ∆𝑇
∆𝑄/∆𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑺 = ∆𝑸/∆𝑻 where 𝑆 is called Heat capacity
Note: The quantity of heat required to warm a given substance also depends upon (i) its mass (𝑚) and (ii) the nature of the material of the substance
Specific heat capacity (s)
The specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of the unit mass of a substance through 1°𝐶. Specific heat capacity is also defined as heat capacity per unit mass. The SI unit is 𝐽𝑘𝑔−1 °𝐶−1 𝑜𝑟 𝐽𝑘𝑔−1𝐾−1𝑠 = 𝑆/𝑚 = ∆𝑄/∆𝑇
𝒔 = (∆𝑸/∆𝑻)(𝟏/𝒎) where 𝑠 is called Specific heat capacity.
Note: Specific heat capacity of a substance depends on, (i) the nature of the material and (ii) the raise or fall in temperature (∆𝑇).
Molar specific heat capacity
It is defined as the amount of heat required to raise the temperature of one mole of substance through 1°𝐶. 𝑪 = (∆𝑸/∆𝑻)(𝟏/𝝁)
The SI unit is 𝐽𝑚𝑜𝑙−1°𝐶−1 𝑜𝑟 𝐽 𝑚𝑜𝑙−1𝐾−1
Specific heat of a gas
Solids and liquids have very small co-efficient of expansion. Therefore, amount of heat spent in their expansion is negligible and heat supplied is assumed to increase only the temperature. Gases have very large co-efficient of expansion. Therefore, amount of heat supplied to a gas is used in two parts, (i) to raise the temperature of gas and (ii) to do mechanical work by the gas.
When heat is supplied to a gas, the increase in temperature of the gas is accompanied either by increase in pressure or volume or both. Thus, a gas can be heated under two conditions: (i) at constant volume and (ii) at constant pressure. Therefore, we consider the specific heat of a gas at constant volume (𝐶𝑉) or the specific heat of a gas at constant pressure (𝐶𝑃).
Specific heat capacity at constant volume (𝑪𝑽)
It is defined as, the amount of heat required to raise the temperature of unit mass of gas through 1°𝐶 at constant volume.
Specific heat capacity at constant pressure (𝑪𝑷)
It is defined as, the amount of heat required to raise the temperature of unit mass of gas through 1°𝐶 at constant pressure.
Note: (i) 𝐶𝑃 is greater than 𝐶𝑉. (ii) 𝐶𝑃− 𝐶𝑉= 𝑅. This relation is called Mayer’s relation.
Change of state
Matter normally exists in three states: solid, liquid and gas. A transition from one state to another is called change of state.
The change of state from solid to liquid is called melting. During this change of state temperature remains constant. That is both the solid and liquid states of the substance co-exist in thermal equilibrium during this change of state.
The temperature at which the solid and liquid co-exist in thermal equilibrium during change of state from solid to liquid is called melting point.
The change of state from liquid to solid is called fusion or freezing. Effect of pressure on melting point; Regelation: When a metallic wire carrying two masses on either end is hung from an ice cube, the wire passes through the ice cube without splitting it. This phenomenon is called regelation. This is because ice melts at lower temperature due to increase in pressure. That is melting point decreases with increase in pressure on ice.
Note: Skating on ice is possible due to increase of pressure on it.
The change of state from liquid to vapour (gas) is called vaporisation. During vaporisation temperature remains constant and liquid and vapour state co-exist in thermal equilibrium.
The temperature at which the liquid and the vapour states of the substance co-exist is called boiling point.
Effect of pressure on boiling point
Boiling point of the substance increases with increase in pressure. This effect is used in the construction of pressure cooker.
Note: At high altitudes, atmospheric pressure is lower and reduces the boiling point of water. This is why cooking is difficult on hills.
The change from solid state to vapour state without passing through the liquid state is called sublimation. Ex: Dry ice (solid 𝐶O2), camphor etc.
The temperature at which the solid, liquid and vapour co-exist in thermal equilibrium is called triple point.
The amount of heat transferred per unit mass during the change of state of the substance is called latent heat. If 𝑚 is the mass of the substance that undergoes a change from one state to the other, then the quantity of heat required is given by, 𝑸 = 𝒎𝑳 where 𝐿 → latent heat. SI unit of latent heat is 𝐽 𝑘𝑔−1 (𝑗𝑜𝑢𝑙𝑒 𝑝𝑒𝑟 𝑘𝑔).
Types of latent heat
Latent heat of fusion (𝑳𝒇)
The amount of heat required to melt unit mass of solid completely at its melting point is called latent heat of fusion.
Latent heat of vaporisation (𝑳𝒗)
The amount of heat required to vaporise unit mass of liquid completely at its boiling point is called latent heat of vaporisation. The graph between temperature and amount of heat supplied for water is as shown.
The branch of science which deals with the measurement of heat is called Calorimetry.
Principle of Calorimetry
Heat lost by the hot body is equal to the heat gained by the colder body, when they are kept in contact with each other, provided no heat is allowed to escape to the surroundings.
A device in which heat measurement can be made is called Calorimeter.
Calorimeter is a hollow cylinder made of copper with a lid and a stirrer placed in it. The calorimeter is placed in an insulated enclosure so that there is no loss of heat by radiation.
Transfer of heat
Heat transfers from a body at higher temperature to a body at lower temperature. There are three modes of heat transfer, (i) Conduction (ii) Convection (iii) Radiation.
It is the mechanism of transfer of heat between two adjacent parts of a body because of their temperature difference. In this mode of heat transfer, transfer of heat takes place without any actual movement of the particles of the medium.
Law of Thermal conductivity
The amount of heat that flows from hotter part to colder part is,
- directly proportional to area of cross-section (𝐴) of the body.
- directly proportional to the temperature difference (∆𝑇 = 𝑇1 − 𝑇2) between the hotter part and colder part.
- directly proportional to the time (𝑡) for which heat flows.
- inversely proportional to the distance (𝑥) between the parts.
Thus we can write, 𝑄 ∝ 𝐴 (𝑇1 − 𝑇2)𝑡/𝑥
𝑄/𝑡 = 𝐾[𝐴(𝑇1 − 𝑇2)/𝑥]
𝑯 = 𝑲[𝑨(𝑻𝟏 − 𝑻𝟐)/𝒙] where 𝐾 → constant, called Thermal conductivity.
It is defined as the rate of flow of heat per unit area of its surface normal to the direction of heat flow under unit temperature gradient. SI unit of thermal conductivity is 𝐽𝑠−1𝑚−1𝐾−1 𝑜𝑟 𝑊𝑚−1𝐾−1. Dimensions are 𝑀𝐿𝑇−3𝐾−1.
Classification of substance based on thermal conductivity
(a) Good conductors of heat: These substances have large values of thermal conductivity. Ex: Most metals. For Ideal conductor thermal conductivity, 𝐾 = ∞
(b) Bad conductors of heat: These substances have small values of thermal conductivity. Ex: wood, air, wool, etc. For ideal bad conductor thermal conductivity, 𝐾 = 0
Applications of thermal conduction
- In winter, the iron chairs appear to be colder than the wooden chairs.
- Cooking utensils are made of aluminium and brass whereas their handles are made of wood.
- We feel warm in woollen cloths.
- Houses made of concrete roofs get very hot during summer days.
- Steel utensils with copper bottom are good for uniform hearting of food.
It is the process in which heat is transferred by the actual movement of the particles of the medium. Convection is possible only in fluids. Convection can be natural or forced. Natural convection is responsible for many familiar phenomena such as sea breeze, land breeze, trade wind.
In daytime, earth is heated by the sun and hence air in contact with the earth gets heated up. This heated air, being lighter, rises up and is replaced by the cold and heavier air from large reservoir of water creating a sea breeze. At night this cycle is reversed forming land breeze.
The equatorial and polar regions of the earth receive unequal solar heat. Air at the earth surface near the equator is hot while the air in the upper atmosphere of the poles is cool. The cold air from the poles rushes towards the equator whose pressure is low. Thus, convection current of air starts between the equator and the poles. Due to rotation of earth from west to east, the convection current drifts towards the east. Convection current blows from North-east towards the equator, which is called Trade wind.
Radiation is the process in which heat is transferred from one region to another without the necessity of any intervening medium. Radiation also refers to the energy emitted by a body and energy is emitted in the form of electromagnetic waves. Energy so radiated/emitted is called Radiant energy.
Everybody emits energy in the form of waves due to its temperature. These waves are known as thermal radiations.
Properties of thermal radiations
- They travel along straight line at the seed of light.
- They can travel in vacuum.
- They do not heat intervening medium.
- They can be reflected and refracted.
- They exhibit the phenomenon of interference, diffraction and polarisation.
- They obey inverse square law that is their intensity varies inversely as the square of the distance from the source.
Note: Thermal radiations can be detected by thermopile, radiometer and bolometer.
A body that absorbs all the radiations falling on it is called a black body.
Black body radiation
Radiations emitted by a black body are called black body radiations. A black body at a given temperature emits all possible wavelengths at that temperature. The intensity and wavelength emitted are independent of the material of the black body but depend only on the temperature of the body. At low temperature, the wavelengths of the radiation emitted are in the infrared region. As the temperature of the black body is increased to about 1100𝐾, the emitted wavelength corresponds to the red region. At sufficiently high temperatures (3000𝐾), the emitted radiations contain shorter wavelengths. The black body radiation consists of a continuous distribution of wavelengths covering infrared, visible and ultraviolet portions of the electromagnetic waves.
Wien’s displacement law
According to this law, the wavelength (𝜆𝑚), corresponding to maximum intensity of emission of black body radiation is inversely proportional to absolute temperature of the black body. 𝜆𝑚 ∝ 1/𝑇
𝝀𝒎𝑻 = 𝒃
Where 𝑏 is constant, called Wien’s constant and 𝑏 = 2.898 × 10−3𝑚𝐾
(i) The colour of a piece of iron heated first becomes dull red, then reddish yellow and finally white hot. This can be explained using Wien’s displacement law.
(ii) Wien’s displacement law is used to find the temperature of Sun and stars.
Stefan’s Law (Stefan-Boltzmann law)
Energy emitted by a black body per unit time per unit area is directly proportional to the fourth power of the temperature. Mathematically,
𝑯 = 𝑨𝝈𝑻𝟒
Where 𝜎 is called Stefan-Boltzmann constant and 𝜎 = 5.67 × 10−8𝑊𝑚−2𝐾−4
This law is obtained experimentally by Stefan and later proved theoretically by Boltzmann. Therefore, it is also called as Stefan-Boltzmann law.
(i) For a body other than black body, the energy radiated per unit time is given by, 𝐻 = 𝑒𝐴𝜎𝑇4.
(ii) A body at temperature 𝑇, with surroundings at temperature 𝑇𝑆, emits as well as receives energy, the net rate of loss of radiant energy is,
𝐻 = 𝐴𝜎(𝑇4 − 𝑇𝑆4) Where 𝑒 → emissivity of black body
Emissivity of a body is defined as the ratio of the heat energy radiated per second per unit area by the body to the amount of heat energy radiated per second per unit area by a perfect black body at the same temperature. Emissivity of black body is one.
The earth surface is a source of thermal radiation. Large portion of this radiation is absorbed by greenhouse gases, namely carbon di oxide (𝐶𝑂2), methane (𝐶𝐻4), nitrous oxide (𝑁2𝑂), chlorofluorocarbon (𝐶𝐹𝑥𝐶𝑙𝑥) and atmospheric ozone (𝑂3). This heats up the atmosphere which in turn, gives more energy to earth resulting in warmer surface. This heating up of earth’s surface and atmosphere is known as greenhouse effect.
Consequences of Greenhouse effect
Greenhouse effect heats up the earth’s surface and atmosphere. Human activities have increased the greenhouse gases, resulting in increase in the average temperature of earth by 0.3°𝐶 to 0.6°𝐶. Without greenhouse effect, the temperature of the earth would have been −180𝐶. It has been estimated that, if such activities continue, then the temperature of the earth will increase by 1°𝐶 to 3°𝐶 after 50 to 60 years. This global warming may cause problem for human life, plants and animals. Because of global warming, ice caps are melting faster, sea level is rising, and weather pattern is changing. Many coastal cities are at the risk of getting submerged. The enhanced Greenhouse effect may also result in expansion of deserts.
Newton’s law of cooling
“The rate of loss of heat by a body is directly proportional to the temperature difference between the body and the surrounding.”
Explanation: Consider a body of mass 𝑚 and specific heat capacity 𝑠 at temperature 𝑇2. Let 𝑇1 be the temperature of the surroundings of the body. According to Newton’s law,
−𝑑𝑄/𝑑𝑡 ∝ (𝑇2 − 𝑇1)
−𝑑𝑄/𝑑𝑡 = 𝑘(𝑇2 − 𝑇1)
𝑑𝑄/𝑑𝑡 = −𝑘(𝑇2 − 𝑇1) − − − (1)
Let the temperature of the body decreases by 𝑑𝑇2 in time 𝑑𝑡. Heat lost by the body is, 𝑑𝑄 = 𝑚𝑠𝑑𝑇2
𝑑𝑄/𝑑𝑡 = 𝑚𝑠𝑑𝑇2/𝑑𝑡 − − − (2)
From equation (1)and (2), 𝑚𝑠(𝑑𝑇2/𝑑𝑡) = −𝑘(𝑇2 − 𝑇1)
𝑑𝑇2/(𝑇2 − 𝑇1) = −(𝑘/𝑚𝑠)𝑑𝑡
𝑑𝑇2/(𝑇2 − 𝑇1) = −𝐾𝑑𝑡 𝑤ℎ𝑒𝑟𝑒 𝐾 = 𝑘/𝑚𝑠
Integrating, ∫𝑑𝑇2/(𝑇2 − 𝑇1) = −𝐾∫𝑑𝑡
𝑙𝑜𝑔𝑒(𝑇2 − 𝑇1) = −𝐾𝑡 + 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑻𝟐 − 𝑻𝟏 = 𝒆−𝑲𝒕 + 𝒄′ where 𝑐′ = 𝑒𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
This equation shows a straight line having a negative slope. The graph between 𝑙𝑜𝑔𝑒(𝑇2 − 𝑇1) and time 𝑡 is as shown.
Important Questions for Exams.
1. State Charles law.
2. Mention the principle of calorimetry.
3. What is greenhouse effect?
4. What is anomalous expansion of water?
5. What is regelation?
6. What is Ideal gas?
7. Give an example for greenhouse gas.
1. State and explain Newton’s law of cooling.
2. Show that 𝑃𝑉 = 𝑅𝑇 for an ideal gas.
1. Plot a graph of temperature versus time showing the changes in the states of ice on heating at one atmospheric pressure. Indicate how much energy is absorbed in the two changes of states.
2. Mention the modes of heat transfer.
3. Define three types of thermal expansions.
4. Write any three properties of heat radiation.
5. Explain the greenhouse effect.
6. Derive 𝛼 = 1/𝑇 for an ideal gas, where the symbols have their usual meaning.
7. Explain the thermal conduction and hence define the coefficient of thermal conductivity.
8. State Stefan’s law and Draw the intensity distribution graph of black body radiation.
1. Define the specific heat of a gas at constant pressure and at constant volume. Give the relation between them. Define latent heat of fusion and vaporisation.2. State and explain the law of thermal conductivity. Define the coefficient of thermal conductivity. Mention its unit and dimension.
3. State and explain Newton’s law of cooling.
4. Write a note on Kelvin’s scale of temperature.
1. A brass boiler has a base of area 0.15𝑚2 and a thickness of 1𝑐𝑚. It boils water at the rate of 6 𝑘𝑔𝑚𝑖𝑛−1 when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Given that thermal conductivity of brass = 109𝐽𝑠−1𝑚−1𝐾−1 and latent heat of vaporization of water= 2256 × 103𝐽𝑘𝑔−1.
2. What is the temperature of the steel-copper junction in steady state? Length of steel rod = 0.15𝑚, length of copper rod = 0.1𝑚, temperature of free end of steel rod is 3000°𝐶, temperature of free and of copper rod is 300°. Area of cross-section of steel rod is twice that of copper rod. K of steel= 50.2𝑊𝑚−1𝐾−1. K of copper = 385𝑊𝑚−1𝐾−1.
3. A cubical ice box of thermocol has each side 30𝑐𝑚 and thickness of 5𝑐𝑚, 4𝑘𝑔 of ice is put in the box. If outside temperature is 450°𝐶 and co-efficient of thermal conductivity is 0.01𝐽𝑠−1𝑚−1𝐾−1. Calculate the mass of ice left after 6ℎ𝑟𝑠. Take latent heat of fusion of ice is 335 × 103 𝐽𝑘𝑔−1.
4. A 10𝑘𝑊 drilling mechine is used to drill a boar in a small aluminium block of mass 8 𝑘𝑔. How much is the raise in temperature of the block in 2.5 𝑚𝑖𝑛𝑢𝑡𝑒 assuming 50% of power is used up in heating. The machine itself or lost to the surroundings. Specific heat of aluminium is 0.91𝐽𝑔−1𝐶−1.
5. Two pieces of copper and aluminium of equal thickness and cross-sectional area are soldered together. The other end of copper is kept at 1000°𝐶 and aluminium at 0°𝐶. Find the temperature of the interface if the thermal conductivities of copper and aluminium are 385𝑊𝑚−1𝐾−1 and 180𝑊𝑚−1𝐾−1 respectively.
6. Two identical iron and brass bars are soldered end to end. The free ends of iron bar and brass bar are maintained at ends of iron bar and brass bar are maintained at 373 𝐾 respectively. Calculate the temperature of the junction. (Given that 𝐾𝑖 = 79𝑊𝑚−1𝐾−1).
7. How much heat is required to convert 10𝑔 of ice at −5°𝐶 into steam at 1000°𝐶. Given specific heat of ice 2.1 𝐽𝑔−1°𝐶−1. Latent heat of steam = 2268𝐽𝑔−1 and latent heat of fusion of ice is 336𝐽𝑔−1. Specific heat of water = 4.2𝐽𝑔−1°𝐶−1.