Properties Of Solids And Liquids

Solids maintain a definite shape and volume, with particles that are closely packed together and only vibrate in place, demonstrating rigidity and resistance to deformation. Key concepts include modulus of elasticity and stress-strain behavior (Hooke’s Law). Liquids, on the other hand, have a fixed volume but no fixed shape, adapting to their container, with particles that move past each other. Liquids expand with temperature (thermal expansion) and can be studied using calorimetry to measure heat changes. Both states are essential in applications like materials engineering and thermodynamics.

1.0Elasticity Key Points

1. Deforming force: An external force applied to a body which changes its size or shape or both is called deforming force.
2. Elasticity: The property of the body of virtue by which it tends to regain its original size and shape when the applied forces are removed, is known as elasticity.
3. Cause of Elasticity: Elasticity in solids is caused by intermolecular forces. When deformed, molecules move, and restoring forces return them to their original positions, returning the solid to its shape.
4. Internal Restoring Force: When an external force acts on any substance then due to the intermolecular force there is an internal resistance produced into the substance called internal restoring force.
5. Restoring Mechanism: The restoring mechanism in solids can be explained using a spring-ball model, where atoms are represented by balls and interatomic forces by springs. When a ball is displaced, the spring restores it to its original position. 

2.0Stress

• Stress is the internal restoring force applied per unit area of the cross-section of a deformed body.
• $Stress=\frac{InternalRestoringForce}{Areaofcross-section}=\frac{F_{internal}}{A}=\frac{F_{External}}{A}$
• SI Unit : $\frac{N}{m^2}$
• Dimensions : $[M^1{L}^{-1}{T}^{-2}]$

Types of Stress

1. Tensile Stress: The longitudinal stress, generated due to increase in length of a body.

2. Compressive Stress: The longitudinal stress, generated due to decrease in length of a body.

3. Tangential Stress or Shear Stress:

Shear stress occurs when an external force is applied tangentially or parallel to the surface of a body. This stress causes the body to change shape or twist.

Shear Stress=$\frac{F}{Area}$

4. Volume Stress or Bulk Stress or Hydraulic Stress: Fluid pressure compresses a solid, reducing its volume. Pressure, defined as force per unit area, acts as volume stress on the object and shares the same units as other types of stress.

Volume Stress=Pressure=$\frac{F}{A}=P$

Note:

$\sigma =\frac{F}{A}\text{as }F\propto A$

3.0Strain

Strain is the ratio of the change in configuration (such as shape, length, or volume) to the body's original configuration.

 $Strain(ϵ)=\frac{changeinconfiguration}{originalconfiguration}$

• It has no unit. It is dimensionless.

Types of Strain

1. Longitudinal Strain :This type of strain occurs when the deforming force results in a change in the body's length.

 $LongitudinalStrain,ϵ=\frac{\text{change in Length}}{\text{original Length}}$        

$LongitudinalStrain,ϵ=\frac{\Delta L}{L}$

2. Volume Strain or Hydraulic Strain : This type of strain is produced when the deforming force produces a change in volume of the body.

$\text{Volume Strain}={ϵ}_v=\frac{\Delta V}{V}=\frac{\text{change in Volume}}{\text{original Volume}}$

3. Shear Strain : This type of strain occurs when the deforming force causes a change in the shape of the body.

$\tan \varphi \text{or}\varphi =\frac{x}{l}(\because \tan \varphi \approx \varphi \text{for small deformation})$

4.0Hooke’s Law and Modulus of Elasticity

According to this law, within the elastic limit, stress is proportional to the strain.

 $\text{stress}\propto \text{strain}$

 $\text{stress}=\text{Constant}\times \text{strain}$

 $\frac{Stress}{Strain}=\text{Modulus of Elasticity}$

• Modulus of elasticity depends on the nature of the material of the body and is independent of its dimensions (i.e. length, volume etc.).
• Unit : The Sl unit of modulus of elasticity is $\frac{N}{m^2}$ or Pascal (Pa).

Types of Modulus of Elasticity

1. Young's modulus of elasticity

 $\text{Young’s modulus (Y)}=\frac{\text{Longitudinal stress}}{\text{Longitudinal strain}}$

$\text{Young’s modulus (Y)}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}=\frac{FL}{A\Delta L}$

2. Bulk Modulus:  It is defined as the ratio of the volume stress to the volume strain

$B=\frac{\text{Pressure}}{\text{Volumetric Strain}}=\frac{p}{-\frac{\Delta V}{V}}=-\frac{pV}{\Delta V}$

Negative sign shows that increase in pressure (p) causes decrease in volume (V).

$B=\frac{p}{-\frac{\Delta V}{V}}=-V\frac{\Delta p}{\Delta V}=-V\frac{dp}{dV}$

Compressibility : Compressibility is the inverse of the bulk modulus of elasticity.

$\text{SI Unit: }{N}^{-1}{m}^2\text{ or }Pasca{l}^{-1}$

$B_{\text{Solids}}>B_{\text{Liquids}}>B_{\text{Gases}}$

Isothermal bulk modulus of elasticity of gas B = P (pressure of gas)

Adiabatic bulk modulus of elasticity of gas B=  ✕ P Where =CpCv

$\text{B}=\gamma \times P\text{where}\gamma =\frac{C_p}{C_v}$

3. Modulus of Rigidity: Within the elastic limit it is the ratio of shearing stress to shearing strain.

$\eta =\frac{\text{shearing stress}}{\text{shearing strain}}=\frac{\frac{F_{\text{Tangential}}}{A}}{\varphi }=\frac{F_{\text{Tangential}}}{A\varphi }$

$\eta =\frac{F}{bc\varphi }$

Increment of Length Due to Own Weight   

$\Delta l=\frac{MgL}{2AY}$

Variation of Strain with Stress

• Elastic limit: The maximum stress where the wire returns to its original length if the force is removed.
• Beyond elastic limit: Stress is no longer proportional to strain.
• Yield point (C): The wire begins to flow like a fluid.
• Breaking point (D): The wire breaks under further stress.
• Ductile materials: Have a large plastic range (CD), deform before breaking.
• Brittle materials: Have a small plastic range, break quickly after exceeding the elastic limit.

Elastic Fatigue

It refers to the weakening of a material due to repeated strain. This is why bridges are considered unsafe after extended periods of use.

Analogy of Rod as a Spring

$Y=\frac{\text{Stress}}{\text{Strain}}⟹Y=\frac{FL}{A\Delta L}$

$F=\frac{AY}{l}\Delta l$

$\frac{AY}{l}=\text{Constant, depends on type of material and geometry of rod.}$

F=kl where k=AYl=equivalent spring Constant

$F=k\Delta l\text{where}k=\frac{AY}{l}\text{is the equivalent spring constant}$

Case 1: Springs in Series

$K_1=\frac{A_1{Y}_1}{l_1},K_2=\frac{A_2{Y}_2}{l_2}$

$K_{\text{eq}}=\frac{K_1{K}_2}{K_1+K_2}$

Case2: Springs in Parallel

$K_1=\frac{A_1{Y}_1}{l_1},K_2=\frac{A_2{Y}_2}{l_2},K_3=\frac{A_3{Y}_3}{l_3}$

$K_{\text{eq}}=K_1+K_2+K_3$

5.0Elastic Potential Energy Stored in a Stretched Wire or in a Rod

$U=\frac{1}{2}\times \text{Stress}\times \text{Strain}\times \text{Volume}$

$U=\frac{1}{2}Y(\text{Strain}{)}^2\times \text{Volume}$

$\text{Strain Energy Density}=\frac{\text{Strain Energy}}{\text{Volume}}$

$\text{Strain Energy Density}=\frac{1}{2}\frac{(\text{Stress}{)}^2}{Y}=\frac{1}{2}Y(\text{Strain}{)}^2=\frac{1}{2}\times \text{Stress}\times \text{Strain}$

6.0Heat

• Heat is the energy delivered between two bodies or between adjacent parts of a body due to a temperature difference.
• 1 calorie = 4.18 joules
• 1 calorie : The amount of heat needed to increase the temperature of 1 gm of water from 14.5°C to 15.5 °C at one atmospheric pressure is 1 calorie.

Mechanical Equivalent of Heat

• The mechanical equivalent of heat shows that heat is a form of mechanical energy. Experiments demonstrated that mechanical work can raise temperature.
• The relationship is given by W=JH, where:
• W is mechanical work.
• H is the heat.
• J  is the mechanical equivalent of heat.
• J is measured in joules per calorie.
• The value of J is 4.18 J/cal, representing the work needed to raise 1g of water by 1°C.

Specific Heat: It is the amount of heat gained or released by the substance to increase or decrease its temperature by 1°C per unit mass.

$Q=\int msdT$

$s=\frac{Q}{m\delta T}$

S.I Unit -Joule/kg-Kelvin

C.G.S Unit-cal./gm-°C.

Heat Capacity or Thermal Capacity

• It is the amount of heat needed to raise its temperature by 1°C. 
• Heat capacity = ms

     SI  Units -is, JK–1

    CGS system is, cal °C–1;

Relation between Specific Heat and Water Equivalent

It is the amount of water that requires the same quantity of heat to achieve the same temperature increase as the object.

$ms\Delta T=m_w{s}_w\Delta T⟹m_w=\frac{ms}{s_w}$

In calorie Sw  = 1    mw=ms

In calories,

$S_w=1m_w=ms$

$m_w$ is also represented by W 

W=ms

Law of Mixture

When two substances at different temperatures are combined, heat exchange occurs until their temperatures become equal. This resulting temperature is known as the final temperature of the mixture.

                 Heat taken by one substance = Heat given by another substance

$⟹m_1{s}_1\left(T_1-T_m\right)=m_2{s}_2\left(T_m-T_2\right)$

Phase Change 

The heat required for a phase or state change, Q = mL, L = latent heat.

1. Latent heat (L): The heat supplied to a substance of unit mass that causes a change in its state at a constant temperature.
2. Latent heat of Fusion $(L_f)$: The heat supplied to a substance that changes it from solid to liquid at its melting point and 1 atm pressure is called the latent heat of fusion. For ice, the latent heat of fusion is 80 kcal/kg.
3. Latent heat of vaporization $(L_v)$:The heat supplied to a substance that changes it from liquid to vapor at its boiling point and 1 atm pressure is called the latent heat of vaporization. For water, the latent heat of vaporization is 540 kcal/kg.

Heating Curve

7.0Calorimeter

• A calorimeter measures heat.
• It comprises a metallic vessel and a stirrer, usually made of copper or aluminum.
• The vessel is placed inside a heat-insulated wooden jacket.
• A thermometer is used to measure temperature.
• It is based on the principle that the heat gained is equal to the heat lost.
• The water equivalent of a calorimeter is the heat required to raise the temperature of an equivalent amount of water by 1°C.

8.0Potential Energy v/s Distance Curve

• Heating matter usually causes it to expand.
• According to atomic theory, thermal expansion is due to asymmetry in the potential energy curve.
• As temperature rises  the amplitude of atomic vibrations and their energy increases
• This results in an increase in the average distance between atoms .
• If the potential energy curve were symmetrical, no thermal expansion would occur.

           $T_2>T_1,E_2>E_1,r_{2\text{avg.}}>r_{1\text{avg.}}$

1. Linear Expansion: When the rod is heated, its increase in length ΔL is proportional to its original length $(L_0)$ and change in temperature ΔT where ΔT is in °C or K.

L=$L_0(1+\alpha \Delta T)$where L is the length after heating the rod.

Note: Thermal expansion is typically a 3-D phenomenon. However, when the other two dimensions of an object are negligible compared to one, the expansion is significant only in that dimension, and this is referred to as linear expansion.

2. Superficial (Areal) Expansion

 $A=A_0(1+\beta \Delta T)$

$\beta =2\alpha$

3. Volume Expansion

$V=V_0(1+\gamma \Delta T)$

$\gamma =3\alpha$

$\alpha :\beta :\gamma =1:2:3$

$\text{Unit of }\alpha :\beta :\gamma \text{ is }\frac{1}{{}^{\circ }\text{C}}\text{ or }1/K\text{ or }{K}^{-1}$

9.0Variation of Time Period of Pendulum Clocks                 

• A pendulum clock's time is based on the pendulum's oscillations.
• The second hand advances by one second when the pendulum reaches its extreme position.
• The second hand moves by two seconds for each complete oscillation.
• A pendulum with a 2-second time period is called a "seconds pendulum."

Therefore change (loss or gain) in time per unit time lapsed is $\frac{T^{\prime }-T}{T}=\frac{1}{2}\alpha \Delta \theta$

Gain or loss in time in duration of 't' in $\frac{1}{2}\alpha \Delta \theta T$ if T is the correct time then

1. $(\theta <{\theta }_0,T<T^{\prime })\Rightarrow$clock becomes fast and gains time 
2. $\theta >{\theta }_0,T^{\prime }>T\Rightarrow$ clock becomes slow and loses time

10.0Measurement of Length by Metallic Scale

Case (i) : When object is expanded only; No expansion of scale

$l_2=l_1\left\{1+{\alpha }_0\left({\theta }_2-{\theta }_1\right)\right\}$

Case (ii) : When only the measurement instrument is expanded, the actual length of the object will not change but the measured value (MV) decreases.

$MV=l_1\left\{1+{\alpha }_s\left({\theta }_2-{\theta }_1\right)\right\}$

Case (iii): If both expanded simultaneously

$MV=l_1\left\{1+\left({\alpha }_0-{\alpha }_s\right)\left({\theta }_2-{\theta }_1\right)\right\}$

11.0Thermal Stress of a Material

When the temperature of the rod is decreased or increased under constrained conditions, compressive or tensile stresses are developed in the rod. These stresses are known as thermal stresses.

$\text{Strain}=\frac{\Delta L}{L_0}=\frac{\text{Final Length}-\text{Original Length}}{\text{Original Length}}=\alpha \Delta T$

original length refers to the natural length of rod at new temperature.

12.0The Bimetallic Strip

• A bimetallic strip is made of two metals with different coefficients of linear expansion (e.g., brass and steel).
• Brass ( $\alpha =19\times 10^{-6}$) has a higher expansion coefficient than steel (α = 12 × 10⁻⁶ /°C).
• The metals are joined together by welding or riveting.
• When heated up , the brass expands more than the steel, causing the strip to bend.
• The brass side of the strip has a larger radius than the steel side.
• Upon cooling, the strip bends in the opposite direction.

13.0Thermal Expansion in Liquids

• Liquids experience volume expansion instead of linear or superficial expansion.
• Initially, as both the liquid and vessel are heated, the liquid level falls due to the vessel expanding more than the liquid.
• Later, the liquid level rises as the liquid expands faster than the vessel.
• The actual increase in the liquid's volume is the sum of the apparent volume increase and the vessel's volume increase.

Co–Efficient of Real Expansion $({\gamma }_r)$

${\gamma }_r=\frac{\text{Real increase in volume}}{\text{Initial Volume}\times \Delta \theta }=\frac{\Delta V}{V\times \Delta \theta }$

Co-Efficient of expansion of flask

${\gamma }_{\text{Vessel}}=\frac{{\left(\Delta V\right)}_{\text{vessel}}}{V\times \Delta \theta }$

${\gamma }_{\text{real}}={\gamma }_{\text{Apparent}}+{\gamma }_{\text{Vessel}}$

1. Variation of Density with Temperature:

• Mass of substance does not change with change in temperature so with increase of temperature, volume increases so density decreases and vice-versa.

$d=\frac{d_0}{\left(1+\gamma \Delta T\right)}$

• For solids values of  are generally small 

$d=d_0\left(1-\gamma \Delta T\right)$

Anomalous expansion of water

• Water's density increases from 0°C to 4°C, making $\gamma$ negative; from 4°C onward, $\gamma$ is positive.
• At 4°C, water's density is maximum.
• This causes ice to form at the surface of lakes in cold weather.
• As winter approaches, surface water cools, becomes denser, and sinks, leading to the formation of ice at 0°C.
• Ice on the surface slows further freezing, as ice is a poor heat conductor.
• Aquatic life survives winter because the lake bottom remains at around 4°C, unfrozen.

2. Variation of Force of Buoyancy with Temperature:

$F_B^{\prime }=F_B\left[1+({\gamma }_s-{\gamma }_L)\Delta \theta \right]$

Measurement of Temperature

Thermometry is the study of temperature measurement. A thermometer uses thermometric substances, like liquids or gases, to measure a body's temperature.

Different Scales of Temperature :A thermometer can be graduated into following scales.

(a) The Centigrade or Celsius scale (°C)

(b) The Fahrenheit scale (°F)

(c) Kelvin scale of temperature (K)

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

Modes of Heat Transfer

14.0Conduction

$\frac{Q}{t}=-KA\frac{dT}{dx}$

$\frac{dT}{dx}\to \text{Temperature Gradient}$

SI Unit : J s-1m-1K-1Dimensions : M1L1T-3K-1

$\text{SI Unit:}(J{s}^{-1}{m}^{-1}{K}^{-1})$

Dimensions: $[M^1{L}^1{T}^{-3}{K}^{-1}$

Steady State

$\frac{Q}{t}=KA\left(\frac{T_B-T_C}{L}\right)$

In this case, Q denotes the amount of heat passing through a cross-section of the slab at a specific position over a time period t.

1. Heat Flow from a Uniform Rod at Steady State

$\frac{dQ}{dt}=KA\left(\frac{\Delta T}{\Delta x}\right)$

Thermal Resistance to Conduction

$R=\frac{L}{KA}$

$\frac{Q}{t}=\frac{(T_B-T_C)}{R}$

$\frac{Q}{t}\text{ as thermal Current }{i}_r$

$i_r=\frac{(T_B-T_C)}{R}$

2. Heat Flow Through Slabs in Series

$R_{eq}=R_1+R_2$

$\frac{L_1+L_2}{K_{eq}A}=\frac{L_1}{K_{1}A}+\frac{L_2}{K_{2}A}$

Equivalent thermal conductivity of the system is

$K_{\text{eq}}=\frac{L_1+L_2}{\frac{L_1}{K_1}+\frac{L_2}{K_1}}=\frac{{\sum }_i{l}_i}{{\sum }_i\frac{l_i}{K_i}}$

3. Heat Flow Through Slabs in Parallel

$\frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2},R=\frac{L}{KA}$

$\frac{K_{\text{eq}}}{L}\left(A_1+A_2\right)=\frac{K_1{A}_1}{L}+\frac{K_2{A}_2}{L}$

Equivalent thermal conductivity

$K_{\text{eq}}=\frac{K_1{A}_1+K_2{A}_2}{A_1+A_2}=\frac{\sum K_i{A}_i}{\sum A_i}$

15.0Radiation

Radiation is the process in which energy is transferred by means of electromagnetic waves. All bodies continuously radiate energy in the form of electromagnetic waves.

Spectral, Emissive, Absorptive and Transmitive Power of a Given Body Surface

Due to incident radiations on the surface of a body following phenomena occur by which the radiation is divided into three parts. (a) Reflection (b) Absorption (c) Transmission

From Energy Conservation

$Q=Q_r+Q_a+Q_t\Rightarrow \frac{Q_r}{Q}+\frac{Q_a}{Q}+\frac{Q_t}{Q}=1\Rightarrow r+a+t=1$

$\text{Reflection Coefficient }r=\frac{Q_r}{Q}$

$\text{Absorptive Coefficient }a=\frac{Q_a}{Q}$

$\text{Transmittive Coefficient }t=\frac{Q_t}{Q}$

r = 1 and a = 0 , t = 0 Perfect reflector

a = 1 and r = 0, t = 0  Ideal absorber (ideal black body)

t = 1 and a = 0, r = 0  Perfect transmitter

$\text{Reflection power }(r)=\left[\frac{Q_r}{Q}\times 100\right]\%$

$\text{Absorption power }(a)=\left[\frac{Q_a}{Q}\times 100\right]\%$

$\text{Transmission power }(t)=\left[\frac{Q_t}{Q}\times 100\right]\%$

16.0Prevost's Theory of Heat Energy Exchange

• All bodies absorb and emit radiation at all temperatures.
• Emission rate depends on the surface nature and temperature of the body.
• Absorption rate depends on the surface nature and temperature of the surroundings.

$Tb>Ts{T}_b>T_s$ then rate of emission > Rate of absorbing and $T_b\downarrow T_b$

$T_b<Ts{T}_b<T_s$ then rate of emission < Rate of absorbing and $T_b\uparrow T_b$

$T_b=Ts{T}_b=T_s$ then rate of emission = Rate of absorbing and No Change

• According to Prevost, energy exchange between a body and its surroundings occurs continuously at all temperatures.
• The temperature difference determines whether the body experiences heating or cooling:
• Cold body in hot surroundings: Absorbs more energy, radiates less, and its temperature increases.
• Hot body in cooler surroundings: Radiates more energy, absorbs less, and its temperature decreases.
• Body temperature equal to surroundings: Energy radiated and absorbed are equal, so temperature remains constant.
• At absolute zero (0 K), atoms remain in their ground state, preventing radiation emission. Hence, Prevost's heat energy exchange theory doesn't apply at this temperature.
• This is considered the limitation of Prevost's theory.
• Using Prevost's theory, the rate of cooling of a body can be determined with respect to its surroundings, as applied in Stefan-Boltzmann law and Newton's law of cooling.

Absorptive power or absorptive coefficient: The ratio of the radiation absorbed by a surface to the radiation incident on it is known as the absorptance.

$a=\frac{Q_a}{Q}=\text{Unitless}$

For Black Body a=1;0 ≤a ≤1

Emissive power (E) :The amount of heat radiation emitted by unit area of the surface in one second at a particular temperature or It is radiation intensity at a particular temperature.

$E=\frac{\Delta Q}{\Delta tA}=\frac{W}{m^2}\text{or}\frac{J}{m^{2}s}$

Spectral Emissive power $(E_{\lambda })$: Emissive power per unit wavelength range at wavelength  is known as spectral emissive power.

$E={\int }_0^{\infty }{E}_{\lambda }d\lambda \text{and}\frac{dE}{d\lambda }=E_{\lambda }$

Emissivity, $e=\frac{\text{Emissive power of a body at temperature }T}{\text{Emissive power of a black body at same temperature }T}=\frac{E}{E_0}$

17.0Kirchhoff's Law

At a given temperature, the ratio of a body's spectral emissive power  to its spectral absorptive power is constant. This constant is equal to the spectral emissive power of an ideal black body at the same temperature:

$\frac{e_{\lambda }}{a_{\lambda }}=\text{Constant}=E_{\lambda }$

Thus, good absorbers are also good emitters, and bad absorbers are bad emitters. According to Kirchhoff's law, the absorptivity of a surface is equal to its emissivity (e=a).

Practical confirmation of Kirchhoff's law was carried out using the Rishi apparatus, with the Leslie container as the primary base.

18.0Stefan's Law

The amount of radiation emitted per second per unit area by a black body is directly proportional to the fourth power of its absolute temperature.

$E=\sigma T^4$

$\sigma =\text{Stefen’s Constant=}5.67\times 10^{-8}\text{Watt}/m^2{K}^4$

Stefan’s Radiation Law

• An ideal blackbody absorbs all radiation that falls on it, including visible, infrared, ultraviolet, and all other wavelengths.
• A blackbody is both a perfect absorber and a perfect emitter of radiation.
• It emits more radiant power per unit area than any real object at the same temperature.
• The rate of radiation emission per unit area is directly proportional to the fourth power of the absolute temperature.

$P=\sigma A{T}^4$

Rate of Emission of Radiation

$\frac{dQ}{dt}=eA\left(T_b^4-T_s^4\right)=ms\frac{d{T}_b}{dt}$

19.0Newton's Law of Cooling

• The temperature difference between the body and surroundings should be small.
• Heat loss from the body should occur only through radiation.
• The surrounding temperature must remain constant during the cooling process.

According to Newton's Law of Cooling, for a small temperature difference between the body and its surroundings, the rate of cooling is directly proportional to both the temperature difference and the surface area exposed.

${\theta }_f={\theta }_0+({\theta }_i-{\theta }_0)e^{-kt}$

• Solar Constant (S):
  The Sun emits radiant energy continuously in space, but only a small portion reaches the Earth. The solar constant is the amount of solar radiant energy received per unit area per unit time by a black surface placed at right angles to the Sun's rays at the Earth's mean distance, assuming no atmospheric interference. The value of the solar constant is approximately 1340 watts/m² or 1.937 Cal/cm²·minute.
• Temperature of the Sun is 5732 K

20.0Spectral Emissive Power

${\int }_0^{\infty }{E}_{\lambda }d\lambda =\text{total emissive power}$

Total area of graph = Total emissive power = $e\sigma T^4$(here e = emissivity of surface)

Spectral Energy Distribution Curve of Black Body Radiations

Spectral energy distribution curves are continuous, indicating that radiation is emitted at all wavelengths (from 0 to ∞) at any given temperature, though the intensity varies across different wavelengths.

• As the wavelength increases, the amount of emitted radiation initially increases, reaches a maximum, and then decreases.
• The area under the spectral energy curve at a specific temperature represents the body's spectral emissive power.

21.0Wein's Displacement Law

The wavelength corresponding to maximum emission of radiation decreases with increasing temperature.

$[{\lambda }_m\propto \frac{1}{T}]$

${\lambda }_m.T=b\Rightarrow {\lambda }_{m1}{T}_1={\lambda }_{m2}{T}_2=b$

Wein’s Constant= $2.89\times 10^{-3}mK.$

Relation between frequency and temperature

$v_m=\frac{c}{b}T$

22.0Sample Questions on Properties of Solids and Liquids

Q-1. Find out longitudinal stress and tangential stress on a fixed block. Force is uniformly applied on the upper surface of the block.

Solution:

Longitudinal or normal stress          

${\sigma }_1=\frac{100Sin30^{\circ }}{5\times 2}=5N/m2$

Tangential stress,

${\sigma }_2=\frac{100Cos30^{\circ }}{5\times 2}=5\sqrt{3}N/m2$

Q-2 Find $\frac{\Delta V}{V}$

Solution:

$\Delta P=\frac{M_g}{A}$

$|\frac{\Delta V}{V}|=\frac{\Delta P}{B}=\frac{M_g}{AB}$

Q-3. What is the change in potential energy (in calories) of a 10 kg mass after 10 m fall ?

Solution:

$\Delta U=mgh=10\times 10\times 10=1000J=\frac{1000}{4.186}cal$

Q-4. Find heat required to increase the temperature of 1 kg water by 20°C.

Solution:

Heat required = $Q=ms\Delta T$

s=1cal/gm°C== 1 K cal/kg°C

$\Delta Q=1\times 20=20Kcal$

Q-5. A Calorimeter of heat capacity 100 J/K is at room temperature of 30°C. 100 g of water at 40°C of specific heat 4200 J/kg–K is poured into the calorimeter. What is the temperature of water in the calorimeter?

Solution:

Let the temperature of water in the calorimeter is t. Then heat lost by water = heat gained by calorimeter

$0.1\times 4200\times (40-t)=100(t-30)$

$420\times 40-420t=100t-3000$

$\Rightarrow t=38.07°C$