Thermal Properties of Matter Class 11 Notes: CBSE Physics Chapter 10

1. Thermal Properties of Matter:

This topic discusses various thermal phenomena and how matter behaves when subjected to the flow of thermal energy. We are specifically concerned in 

• Thermal Expansion.

• Heat and Calorimetry

• Transfer of Heat

1.1 Temperature and Heat 

Temperature: 

• Temperature is a relative measure of a body's hotness or coldness.

• SI Unit: Kelvin (K) 

• Commonly used unit: \[\text{ }\!\!{}^\circ\!\!\text{ C or  }\!\!{}^\circ\!\!\text{ F}\]

• Conversion: \[\text{t}\left( \text{K} \right)\text{ = t}\left( \text{ }\!\!{}^\circ\!\!\text{ C} \right)\text{ + 273}\text{.15}\]

Heat: 

• Heat is a type of energy flow that occurs, 

i. between two bodies or 

ii. between a body and its surroundings as a result of a temperature difference.

• SI unit: Joule (J) 

• Commonly used unit: Calorie (Cal) 

• Conversion: \[\text{1 cal = 4}\text{.186 J}\]

• Heat is always transferred from a higher temperature system to a lower temperature system. 

1.2 Measurement of Temperature: 

Principle: 

• Thermometric properties are observed as temperature changes and compared to certain reference situations.

• Generally, the reference situation is ice point or steam point.

1.2.1 Celsius and Fahrenheit Temperature Scale:

It implies that 100 divisions in Celsius are equivalent to 180 divisions in Fahrenheit.

Hence, \[\dfrac{{{\text{t}}_{\text{f}}}\text{ }-\text{ }32}{180}\text{ = }\dfrac{{{\text{t}}_{\text{c}}}}{100}\]

1.2.2 Absolute Temperature Scale: 

• It is kelvin scale 

$\text{Ice point }\to \text{ 273}\text{.15 K}$

$\text{Steam point }\to \text{ 373}\text{.15 K}$  

• When compared to the Celsius scale, the number of scale divisions in both scales is the same.

\[\dfrac{{{\text{t}}_{\text{c}}}\text{ }-\text{ 0}{}^\circ \text{C}}{100}\text{ = }\dfrac{{{\text{t}}_{\text{k}}}-\text{ }273.15}{100}\]

• The Kelvin scale is known as an absolute scale because it is nearly impossible to go beyond 0 K on the negative side.

1.2.3 Thermometers: 

• A thermometer is a device used to measure the temperature of any system. The liquid in glass thermometers, Platinum Resistance Thermometers, and Constant Volume Gas Thermometers are a few examples.

• The Liquid in Glass thermometer and the Platinum Resistance thermometer provide uniform readings for ice point and steam point but vary for different liquids and materials.

• A constant-volume gas thermometer provides the same readings regardless of which gas is used. It is based on the fact that at low pressures and constant volume, \[\text{P  }\!\!\times\!\!\text{  T}\] for a gas

  

• All gases collide to absolute zero at zero pressure. 

Type of expansion	\[\dfrac{\text{ }\!\!\Delta\!\!\text{ x}}{\text{x}}=\text{ K}\Delta \text{T}\]	Constant (k)
Linear Expansion	\[\dfrac{\text{ }\!\!\Delta\!\!\text{ L}}{\text{L}}=\text{  }\!\!\alpha\!\!\text{ }\Delta \text{T}\]	Coefficient of linear expansion: It rises in length per unit length per degree increase in temperature.
Area expansion	\[\dfrac{\text{ }\!\!\Delta\!\!\text{ A}}{\text{A}}=\text{  }\!\!\beta\!\!\text{ }\Delta \text{T}\]	Coefficient of area expansion: It rises in area per unit area per degree increase in temperature.
Volume Expansion	\[\dfrac{\text{ }\!\!\Delta\!\!\text{ V}}{\text{V}}=\text{  }\!\!\gamma\!\!\text{ }\Delta \text{T}\]	Coefficient of volume expansion: It rises in volume per unit volume per degree increase in temperature.

Units of \[\text{ }\!\!\alpha\!\!\text{ ,  }\!\!\beta\!\!\text{ ,  }\!\!\gamma\!\!\text{  }=\text{ }/\text{ }\!\!{}^\circ\!\!\text{ C or }/\text{K}\]

In general, as the volume changes, the density changes as well.

• \[\text{ }\!\!\alpha\!\!\text{ }\] is generally higher for metals than \[\text{ }\!\!\alpha\!\!\text{ }\] for nonmetals 

• \[\text{ }\!\!\gamma\!\!\text{ }\] is nearly constant at high temperatures but varies with temperature at low temperatures.

1.3 Thermal Expansion: 

Most materials, it has been observed, expand when heated and contract when cooled. This expansion is multidimensional.

It has been demonstrated experimentally that a fractional change in any dimension is proportional to a change in temperature.

Coefficient of Volume Expansion of Cu as a Function of Temperature:

• For ideal gases, $\gamma$ is inversely related to the temperature at constant pressure.

$\text{V = }\dfrac{\text{nRT}}{\text{P}} $

$\Rightarrow \dfrac{\Delta \text{V}}{\text{V}}\text{ = }\dfrac{\Delta \text{T}}{\text{T}} $

$\Rightarrow \text{ }\!\!\gamma\!\!\text{  = }\dfrac{\bot }{\text{T}}$ 

Water, on the other hand, contracts when heated from \[\text{0 }\!\!{}^\circ\!\!\text{ C to 4 }\!\!{}^\circ\!\!\text{ C}\] and thus its density rises from \[\text{0 }\!\!{}^\circ\!\!\text{ C to 4 }\!\!{}^\circ\!\!\text{ C}\]. This is known as anomalous expansion.

• In general \[\text{ }\!\!\gamma\!\!\text{  }=\text{ }3\text{ }\!\!\alpha\!\!\text{   = }\dfrac{\text{3}}{2}\text{ }\!\!\beta\!\!\text{ }\]

Proof: Consider a cube of length l that expands equally in all directions when its temperature rises by a small \[\text{ }\!\!\Delta\!\!\text{ T;}\]

We have 

\[\Delta l\text{ }=\text{ }\!\!\alpha\!\!\text{ }l\Delta \text{T}\]

Also, 

$\Delta \text{V}=\text{ }{{(l\Delta l)}^{3}}\text{ }\text{ }{{l}^{3}}\text{ } $

$=\text{ }{{l}^{3}}\text{ }+\text{ }3{{l}^{2}}\Delta l\text{ }+\text{ }3l\Delta {{l}^{2}}\text{ }+\text{ }\Delta {{l}^{2}}\text{ }\text{ }{{l}^{3}}\text{ }$ 

$ =\text{ }3{{l}^{2}}\Delta l\text{ }...\left( 1 \right)  $

In Equation (1) we ignore \[3l\Delta {{l}^{2}}\text{ }\And \Delta {{l}^{3}}\text{ as }\Delta l\] is very small as compared to l. 

So, 

$\Delta \text{V = }\dfrac{3\text{V}}{l}\Delta l $

$=3\text{V }\!\!\alpha\!\!\text{  }\!\!\Delta\!\!\text{ T              }\left[ \text{Using }\dfrac{\text{V}}{l}=\text{ }{{l}^{\text{2}}} \right]\cdots \cdots \left( 2 \right) $ 

$\dfrac{\Delta \text{V}}{\text{V}}\text{ = 3 }\!\!\alpha\!\!\text{  }\!\!\Delta\!\!\text{ T} $

Similarly, we can prove that the area expansion coefficient of thermal expansion is prevented inside the rod by rigidly fixing its ends, and then the rod acquires a compressive strain due to external fones at the ends. The corresponding stress set up in the rod is called thermal stress.

Hence, thermal stress $=\frac{F}{A}=Y\left(L-L_{0}\right) / L_{0}$ where Y is Young's modulus of the given material.

This can be simplified into $Y(\alpha \Delta T) / L_{0}$.

Practical applications include railway tracks, metal tires on cartwheels, bridges, and a variety of other structures.

1.4 Heat and Calorimetry: 

When two systems at different temperatures are linked together, heat flows from the higher temperature to the lower temperature until their temperatures become the same.

The principle of calorimetry states that heat lost by a body at a higher temperature equals heat gained by a body at a lower temperature, ignoring heat loss to surroundings.

When heat is applied to anybody, either its temperature or its state changes.

1.4.1 Change in Temperature:

On heating, when the temp changes, then 

$\text{Heat supplied }\propto \text{ change in temp (}\Delta \text{T)} $

$\text{                       }\propto \text{amount of substance }\left( \text{m / n} \right)  $

$\text{                       }\propto \text{nature of substance }\left( \text{s / C} \right)$

\[\text{ }\!\!\Delta\!\!\text{ H = ms }\!\!\Delta\!\!\text{ T}\]

Here, \[\text{m}\] is the mass of body, \[\text{s}\] is specific heat capacity per kg, \[\text{ }\!\!\Delta\!\!\text{ T}\] is change in temp 

or \[\text{ }\!\!\Delta\!\!\text{ H = nC }\!\!\Delta\!\!\text{ T}\] 

Here, n is the number of moles, C is the Specific/Molar heat Capacity per mole, \[\text{ }\!\!\Delta\!\!\text{ T}\] is the change in temp. 

Specific Heat Capacity: 

• The amount of heat needed to increase the temperature of a substance's unit mass by one degree.

• Units:

i. $\text{SI }\to \text{ J / KgK           }{{\text{S}}_{{{\text{H}}_{\text{2}}}\text{O}\left( \text{e} \right)}}\text{ = 1 cal / g}{}^\circ \text{C} $ 

ii. $\text{Common }\to \text{ Cal / g}{}^\circ \text{C           }{{\text{S}}_{{{\text{H}}_{\text{2}}}\text{O}\left( \text{ice} \right)}}\text{ = 0}\text{.5 cal / g}{}^\circ \text{C} $

Molar Heat Capacity: 

• The amount of heat required to raise the temperature of a unit mole of a substance by one degree.

• Units: 

i. $\text{SI }\to \text{ J / mol K}  $

ii. $\text{Common }\to \text{ Cal / gc }\!\!{}^\circ\!\!\text{ }$

Heat Capacity: 

• The amount of heat needed to raise a system's temperature by one degree.

\[\Rightarrow \Delta \text{H = S}\Delta \text{T}\] where S is heat capacity. Units 

$\text{SI }\to \text{ J / K } $ 

$\text{Common }\to \text{ Cal / C }\!\!{}^\circ\!\!\text{ } $

• For \[{{\text{H}}_{\text{2}}}\text{O}\], the specific heat capacity varies, but only slightly.

• Materials with a higher specific heat capacity require a large amount of heat for a given temperature.

1.4.2 Change in State: 

On heating, when the phase changes then the heat supplied is directly related to the amount of substance that changes the state (M) and the nature of the substance (L). 

\[\Rightarrow \Delta \text{H = mL}\]

Where L is the latent heat of the process. 

• Latent Heat: The amount of heat required to change the state of any substance per mass. 

Units: 

$\text{SI }\to \text{ J/Kg}$

$\text{Common }\to \text{ Cal/g}$

• The change in state always occurs at a constant temperature. 

For example:   

$\text{Solid }\rightleftharpoons \text{ Liq      }{{\text{L}}_{\text{f}}} $

$\text{Liq }\rightleftharpoons \text{ Gas        }{{\text{L}}_{\text{v}}}  $

${{\text{L}}_{\text{f}}}\text{ =  Latent Heat of fusion}  $

${{\text{L}}_{\text{v}}}\text{ = Latent heat of evaporation} $

• If a material is not at its B.P. or M.P, heating will cause the temperature to change until a specific state change temperature is reached. 

For Example:  If water is initially at \[50\text{ }\!\!{}^\circ\!\!\text{ C}\] at 1 atm pressure in its solid state, then on heating

Step 1: Make a temperature change to the first to \[\text{0 }\!\!{}^\circ\!\!\text{ C}\].

Step 2: The ice melts to \[{{\text{H}}_{\text{2}}}\text{O }\left( \text{l} \right)\] while the temperature remains constant.

Step 3: The temperature is inverted to \[\text{100 }\!\!{}^\circ\!\!\text{ C}\text{.}\]

Step 4: \[{{\text{H}}_{\text{2}}}\text{O }\left( \text{l} \right)\] boils to steam while the temperature remains constant.

Step 5: Increase the temperature further.

• The slope is inversely related to the heat capacity. 

• The length of the horizontal line totally depends upon mL for the process. 

1.4.3 Pressure dependence on melting point and boiling point: 

• The melting point decreases with increasing pressure for some substances and increases for others. 

• The melting point increases with increasing temperature. The above results can be seen using phaser diagrams.

Line AO is the sublimation curve, line OB is the fusion curve, line OC is the evaporation curve, point O is the triple point, point C is a critical temperature 

Triple Point: 

The pressure and temperature combination at which all three states of matter (solids, liquids, and gases) coexist. Its value is 273.16 K and 0.006 atm for \[{{\text{H}}_{\text{2}}}\text{O}\text{.}\]

The combination of pressure and temperature beyond which vapour cannot be liquified is referred to as the critical point. The corresponding temperature and pressure are referred to as the critical temperature and critical pressure.

• The phasor diagram shows that the melting point of H2O decreases as pressure increases. The concept of regulation is based on this.

• Regulation: The phenomenon of refreezing water that has melted below its normal melting point due to the addition of pressure. Cooking on mountains is difficult due to the pressure effect on melting point, whereas cooking in a pressure cooker is easier.

1.5 Heat Transfer: 

There are three modes of heat transfer. 

• Conduction 

• Convection 

• Radiation 

1.5.1 Conduction 

Thermal conduction is the process by which thermal energy is transferred from the hotter to the colder part of a body or from a hot body to a cold body in contact with it without the transfer of material particles.

The rate of heat energy flowing through the rod becomes constant at a steady state. It is given by,

\[\text{Q = kA}\dfrac{\left( {{\text{T}}_{\text{C }}}\text{- }{{\text{T}}_{\text{D}}} \right)}{\text{L}}\]

This is the rate for rods with a uniform cross-section.

Here, Q is the rate of heat energy flow, A is an area of cross-section, \[{{\text{T}}_{\text{C}}}\text{ and }{{\text{T}}_{\text{D}}}\] are the temperature of the hot end and cold end respectively, L is the length of the rod, K is coefficient of thermal conductivity.

Some of the examples of conduction that we experience in our day-to-day life are being burnt after touching a stove. Your hand is being cooled with ice. By putting a red-hot piece of iron into the water, it is brought to a boil.

Coefficient of Thermal Conductivity: 

• It is defined as the amount of heat conducted in unit time during a steady state through a unit area of any cross-section of a substance under a unit temperature gradient, with the heat flow being normal to the area. 

• Units: 

\[\text{SI }\to \text{ J / mSk or W/mK}\]

• The greater the thermal conductivity, the faster heat energy flows for a given temperature difference.

• Metals have higher thermal conductivity than nonmetals.

• Insulators have a very low thermal conductivity. As a result, heat energy cannot be easily conducted through air.

• The concept of equivalent thermal conductivity of the composite rod can be used for combinations of rods between two ends kept at different temperatures.

\[{{\text{K}}_{\text{eq}}}\] is the equivalent thermal conductivity of the composite. 

• The term \[\dfrac{\left( {{\text{T}}_{\text{C }}}\text{- }{{\text{T}}_{\text{D}}} \right)}{\text{L}}\] is known as temperature Gradient. 

Temperature Gradient: 

• Temperature Gradient refers to the decrease in temperature per unit length in the direction of heat energy flow. 

• Units: 

\[\text{SI }\to \text{ K/m}\] 

• The term Q is the rate of flow of heat energy can also be named as heat current 

• The term \[\left( \text{L / KA} \right)\]  is known as the thermal resistance of any conducting rod. 

Thermal Resistance: 

• It is defined as the medium's obstruction of the flow of heat current.

• Units: 

\[\text{SI }\to \text{ K/W}\]

1.5.2 Convection: 

Thermal convection is the process by which heat is transferred from one point to another by the actual movement of heated material particles from a higher-temperature location to a lower-temperature location.

• Forced convection occurs when the medium is forced to move using a fan or a pump. Natural or free convection occurs when material moves due to differences in density in the medium.

• Examples of forced convection are the circulatory system, cooling system, and heat connector of an automobile

• Examples of natural convection are trade winds, sea breezes/land breezes, monsoons, and tea burning.

1.5.3 Radiation: 

It is a method of heat transmission in which heat travels directly from one location to another without the use of an intermediary medium.

• This radiation of heat energy takes the form of EM waves. 

• These radiators are emitted as a result of their temperature, similar to how a red-hot iron or a filament lamp emits light. 

• Everybody both radiates and absorbs energy from their surroundings. The amount of energy absorbed is proportional to the colour of the body.

• Black-body radiation is the thermal electromagnetic radiation emitted by a black body within or surrounding a body in thermodynamic equilibrium with its environment (an idealised opaque, non-reflective body). It has a specific spectrum of wavelengths that are inversely related to intensity and are only affected by the body's temperature, which is assumed to be uniform and constant for the sake of calculations and theory.

• Stefan’s Law of Radiation: 

Stefan's Law states that the radiated power density of a black body is directly related to its absolute temperature T raised to the fourth power.

Newton's Law of Cooling: 

According to Newton’s Law of cooling, the rate of loss of heat, that is, \[-\dfrac{\text{d}}{\text{dt}}\] of the body is directly related to the temp difference.

Now,\[\text{- }\dfrac{\text{ds}}{\text{dt}}\text{= k }\left( {{\text{T}}_{\text{2}}}\text{ - }{{\text{T}}_{\text{1}}} \right)\text{      }\cdots \cdots \left( 4 \right)\] 

Here, k is a positive constant which depends on the area and nature of the surface of the body. 

Suppose a body of mass m, specific heat capacity s is at temperature \[{{\text{T}}_{\text{2}}}\text{  }\!\!\And\!\!\text{  }{{\text{T}}_{\text{1}}}\] be the temperature of surroundings if \[\text{d}{{\text{T}}_{\text{2}}}\] the fall of temperature in time dt. 

The amount of heat lost is given by,

\[\text{dcs = msd}{{\text{T}}_{\text{2}}}\]

 Therefore, the rate of loss of heat is given by

\[\dfrac{\text{dcs}}{\text{dt}}\text{= ms }\dfrac{\text{d}{{\text{T}}_{\text{2}}}}{\text{dt}}\text{      }\cdots \cdots \left( 5 \right)\]

From Equations 4 and 5

$-\text{ms}\dfrac{\text{d}{{\text{T}}_{\text{2}}}}{\text{dt}}\text{= k }\left( {{\text{T}}_{\text{2}}}-{{\text{T}}_{\text{1}}} \right) $

$\dfrac{\text{d}{{\text{T}}_{\text{2}}}}{\left( {{\text{T}}_{\text{2}}}-{{\text{T}}_{\text{1}}} \right)}\text{ = -}\dfrac{\text{k}}{\text{ms}}\text{ dt } $

$\text{               = -Kdt }  $

Here, \[\text{K}=\dfrac{\text{k}}{\text{ms}}\]

On integrating,

$\log \left( {{\text{T}}_{\text{2}}}-{{\text{T}}_{\text{1}}} \right)\text{ =  -Kt + }\text{C}$

${{\text{T}}_{\text{2}}}\text{  = }{{\text{T}}_{\text{1}}}\text{ + }{{\text{C}}_{\text{1}}}{{\text{e}}^{-\text{kt}}}\text{  }\cdots \cdots \left( 6 \right)  $

The equation (6) allows us to calculate the time it takes a body to cool through a given temperature range.

For small temperature differences, the rate of cooling due to conduction, convection, and radiation combined is proportional to the temperature difference.

Approximation: 

If a body cools from \[{{\text{T}}_{\text{a}}}\text{ to }{{\text{T}}_{\text{b}}}\] in t times in a medium with a temperature of \[{{\text{T}}_{\text{0}}}\text{,}\]then

\[\left( {{\text{T}}_{\text{a}}}-{{\text{T}}_{\text{b}}} \right)\text{ = K }\left( \dfrac{{{\text{T}}_{\text{a}}}+{{\text{T}}_{\text{b}}}}{2}-{{\text{T}}_{\text{0}}} \right)\]

Newton's law of cooling can be studied experimentally.

Configuration: 

• A double-walled vessel (v) with water contained between two walls.

• Inside the double-walled vessel is a copper calorimeter (c) containing hot water. 

• Two thermometers threaded through the carbs are used to measure the temperature \[{{\text{T}}_{\text{2}}}\]of \[{{\text{H}}_{\text{2}}}\text{O}\] in the calorimeter T of water in between the double walls.

Experiment: 

• The temperature of hot water in the calorimeter at equal time intervals.

• As a result, A line graph is drawn between log \[\left( {{\text{T}}_{2}}\text{ }{{\text{T}}_{1}} \right)\] and time (t). 

• The graph is observed to be a straight line, as predicted by Newton's law of cooling.

Thermal Properties of Matter Class 11 Notes Physics - Basic Subjective Questions

Section – A (1 Mark Questions)

1. Three bodies are having temperature $T_{A}=-42^{\circ}F,T_{B}=-10^{\circ}C,T_{C}=200K$ . Which body among these is most warm?

Ans. Let us calculate all body temperatures on the same scale

Let choose Celsius

C=K-273.15

C=(5F-160)/9

$T_{A}=-42^{\circ}F=(5\times -42-160)/9=-41^{\circ}C$

$T_{B}=-10^{\circ}C$

$T_{C}=200K=200-273\cdot 15=-73\cdot 1^{\circ}C$

So the most warm body is TB.

2. If a thermometer reads the freezing point of water as 20°C and the boiling point as 150°C, how much does the thermometer read, when the actual temperature is 60°C?

Ans. 150C-20C=100x

1x=1.3C

thermometer reading = 1.3C + 20

$=1\cdot 3(60)+20=98^{\circ}C$

3. Burns from steam are usually more serious than those from boiling water. Why?

Ans. Steam will produce more severe burns than boiling water because steam has more heat energy than water due to its latent heat of evaporation.

For water, the latent heat of fusion and vaporisation is, $L_{f}=3\cdot 33\times 10^{5}Jkg^{-1}$ and $L_{V}=22\cdot 6\times 10^{5}Jkg^{-1}$ respectively. It means $22\cdot 6\times 10^{5}Jkg^{-1}$ of heat is needed to convert 1 kg of water to steam at 100°C. So, steam at 100°C carries more heat than water at 100°C. This is why burns from steam are usually more serious than those from boiling water.

4. A bimetallic strip consists of brass and iron when it is heated it bends into an arc with brass on the convex and iron on the concave side of the arc. Why does this happen?

Ans. Two strips of equal lengths but of different materials (different coefficient of linear expansion $\varpropto$ ) when joined together, are called a bimetallic strip. This strip has the characteristic property of bending on heating due to the unequal linear expansion of the two metals. The brass side bends on the outer side (convex side) due to greater $\varpropto$ and the iron bends on the inner side (concave side) due to smaller $\varpropto$.

5. Why are birds often seen to swell their feathers in winter?

Ans. When birds swell their feathers, they trap air in the feathers. Air being a poor conductor prevents loss of heat and keeps the bird warm.

Section – B (2 Marks Questions)

6. The reading of the Centigrade thermometer coincides with that of the Fahrenheit thermometer in a liquid. Then find the temperature of the liquid.

Ans. $\dfrac{C}{5}=\dfrac{F-32}{9}4$

C=F=x

$\dfrac{x}{5}=\dfrac{x-32}{9}$

$9x=5x-32\times 5$

$4x=-32\times 5$

$x=\dfrac{-32\times 5}{4}=-40^{\circ}C$

7. Two metal rods A and B have their initial lengths in the ratio 2 : 3 and the coefficient of linear expansion in the ratio 3: 4. When they are heated through the same temperature difference, then find the ratio of their change in length or expansion.

Ans. $ \dfrac{lA}{lB}=\dfrac{2}{3};\dfrac{\varpropto_{A}}{\varpropto_{B}}=\dfrac{3}{4}$

$\dfrac{\Delta lA}{\Delta lB}=\dfrac{l_{A}\varpropto_{A}\Delta T}{l_{b}\varpropto_{B}\Delta T}=\left ( \dfrac{l_{A}}{l_{B}} \right )\left ( \dfrac{\varpropto_{A}}{\varpropto_{B}} \right )=\dfrac{2}{3}\times \dfrac{3}{4}=1:2$

8. The surface temperature of the sun which has maximum energy emission at 500 nm is 6000 K. Now find the temperature of a star which has maximum energy emission at 400 nm.

Ans. By using Wien’s displacement law,

$(\lambda _{m})_{1}T_{1}=(\lambda _{m})_{2}T_{2}$

$\therefore 500\times 6000=400\times T_{2}$

$\Rightarrow T_{2}=\dfrac{500\times 6000}{400}$

=7500K

9. Ice starts forming in a lake with water at 0°C when atmospheric temperature is –10°C. If the time taken for the 5mm thickness of ice to be formed is 2h, then find the time taken for the thickness of ice to change from 5mm to 20 mm.

Ans. $t=\dfrac{\rho L}{2K\theta }\left ( x_{1}^{2}-x_{2}^{2} \right )\;t\varpropto\left ( x_{1}^{2}-x_{2}^{2} \right )$

$\dfrac{{t}'}{t}=\dfrac{(x'{_{1}^{2}}-x'{_{2}^{2}})}{(x_{1}^{2}-x_{2}^{2})}$

$\dfrac{{t}'}{2}=\dfrac{20^{2}-5^{2}}{5^{2}-0^{2}}$

${t}'=30h$

10. The volume of mercury in the bulb of a thermometer is 10-6 m3. The area of the cross-section of the capillary tube is $2\times 10^{-7}m^{2}$. If the temperature is raised by $100^{\circ}C$, then find the increase in the length of the mercury column. $\left ( Take,\gamma _{Hg}=18\times 10^{-5}/^{\circ}C \right )$

Ans. By cubical expansion relation,

$\Delta V=V\times _{\gamma }\Delta T$

Where, $\gamma =$ coefficient cubical expansion,

$V=10^{-6}m^{3}$

 = initial volume

$\gamma =18\times 10^{-5}/^{\circ}C$

$\Delta T=100^{\circ}C$

$\Delta V=10^{-6}\times 18\times 10^{-5}\times 10^{2}$

$=18\times 10^{-9}m^{3}$ .

Since, $\Delta V=A\times \Delta l$

$\therefore 18\times 10^{-9}=2\times 10^{-7}\times \Delta l$

or $9\times 10^{-2}=\Delta l$

or $\Delta l=9cm$ .

Important Topics of Class 11 Physics Chapter 10 Thermal Properties of Matter

Important formula in Class 11 Physics Chapter 10 Thermal Properties of Matter

1. Heat (Q) absorbed or released:

$Q = mc\Delta T$
Where Q is the heat absorbed or released, m is the mass of the substance, ccc is the specific heat capacity, and $\Delta T$ is the temperature change.

2. Thermal Expansion:

• Linear Expansion: $\Delta L = \alpha L_0 \Delta T$ Where $\Delta L$ is the change in length, $\alpha$ is the coefficient of linear expansion, $L_0$​ is the original length, and $\Delta T$ is the temperature change.

• Area Expansion: $\Delta A = \beta A_0 \Delta T$ Where $\Delta A$ is the change in area, $\beta$ is the coefficient of area expansion, and $A_0$​ is the initial area.

• Volume Expansion: $\Delta V = \gamma V_0 \Delta T$ Where $\Delta V$ is the volume change, $\gamma $is the coefficient of volume expansion, and $V_0$​ is the initial volume.

3. Specific Heat Capacity (c):

$c = \frac{Q}{m\Delta T}$
Where c is the specific heat capacity, QQQ is the heat energy, mmm is the mass, and $\Delta T$ is the temperature change.

4. Latent Heat (L):
Q=mL
Where Q is the heat required, mmm is the mass, and L is the latent heat of the material.

5. Heat Conduction (Fourier’s Law):
$Q = \frac{kA(T_1 - T_2)t}{d}$

Where Q is the heat transferred, k is the thermal conductivity, A is the area, ttt is the time, $T_1$​ and $T_2$​ are the temperatures, and ddd is the thickness of the material.

6. Stefan-Boltzmann Law (for radiation):
$P = \sigma A T^4$
Where P is the power radiated, $\sigma$ is the Stefan-Boltzmann constant, A is the surface area, and T is the temperature in kelvins.

7. Newton’s Law of Cooling:
$\frac{dT}{dt} = -k(T - T_{\text{ambient}})$
Where $\frac{dT}{dt}$​ is the rate of cooling, k is a constant, and $T_{\text{ambient}}$​ is the ambient temperature.

Importance of Class 11 Physics Chapter 10 Thermal Properties of Matter Revision Notes 

• They simplify key concepts such as heat transfer, thermal expansion, and specific heat, making them easier to understand and apply in solving problems.

• These notes provide a structured way to revise important formulas and laws, helping students during last-minute exam preparation.

• The notes also highlight practical applications of thermal properties, allowing students to see the relevance of the topic in everyday life and making the subject more engaging.

• By summarising complex topics, these notes enable quick revision, making it easier for students to focus on the essential points for their exams.

• The notes are especially helpful in understanding derivations and theory, which are key areas for scoring well in both school and competitive exams.

Tips for Learning the Physics Class 11 Chapter 10 Thermal Properties of Matter:

• Begin by understanding the basic concepts of heat, temperature, and thermal expansion. These fundamentals form the foundation of the chapter and help in understanding more advanced topics.

• Focus on the three methods of heat transfer: conduction, convection, and radiation. Know how each process works and where they are applied in real-life situations.

• Practice solving numerical problems related to specific heat, latent heat, and thermal conductivity. These are common exam questions and mastering them can help improve your problem-solving skills.

• Break down complex formulas and learn their derivations step by step. This will help you understand why certain formulas are used for specific situations.

• Use visual aids like diagrams and graphs to better understand concepts such as the expansion of materials with temperature and the heat flow process.

• Revise regularly using concise notes and summaries. This will help in retaining key concepts and formulas for exams.

• Apply what you learn to practical examples, like understanding why metal expands when heated or how insulation works to retain heat, to make the concepts more relatable.

Conclusion 

The Class 11 Physics Chapter 10: Thermal Properties of Matter notes summarise important concepts like heat transfer, thermal expansion, and specific heat in an easy-to-understand way. These notes help students grasp the fundamental ideas of how different materials behave under temperature changes. By breaking down complex formulas and concepts, the notes make it easier to revise and prepare for exams. They focus on practical applications, such as how heat flows in everyday objects, making learning more relatable. Overall, these notes offer a clear and structured approach to mastering the topic.

Related Study Materials for Class 12 Physics Chapter 10

Chapter-wise Links for Physics Notes for Class 11 PDF FREE Download

Related Study Materials Links for Class 11 Physics

Along with this, students can also download additional study materials provided by Vedantu for Physics Class 11–