DP IB Physics: SL

B. The particulate nature of matter

B.1 Thermal energy transfers

DP IB Physics: SL B. The particulate nature of matter B.1 Thermal energy transfers Understandings Students should understand:
a)	Molecular theory in solids, liquids and gases
b)	Density ρ as given by [math]\rho = \frac{m}{v}[/math]
c)	That Kelvin and Celsius scales are used to express temperature
d)	That the change in temperature of a system is the same when expressed with the Kelvin or Celsius scales
e)	That Kelvin temperature is a measure of the average kinetic energy of particles as given by [math]\overline{E_k} = \frac{3}{2} k_B T[/math]
f)	That the internal energy of a system is the total intermolecular potential energy arising from the forces between the molecules plus the total random kinetic energy of the molecules arising from their random motion
g)	That temperature difference determines the direction of the resultant thermal energy transfer between bodies
h)	That a phase change represents a change in particle behavior arising from a change in energy at constant temperature
i)	Quantitative analysis of thermal energy transfers Q with the use of specific heat capacity c and specific latent heat of fusion and vaporization of substances L as given by [math]Q = mcΔT and Q = mL[/math]
j)	That conduction, convection and thermal radiation are the primary mechanisms for thermal energy transfer
k)	Conduction in terms of the difference in the kinetic energy of particles
l)	Quantitative analysis of rate of thermal energy transfer by conduction in terms of the type of material and cross-sectional area of the material and the temperature gradient as given by [math]\frac{\Delta Q}{\Delta t} = kA \frac{\Delta T}{\Delta x}[/math]
m)	Qualitative description of thermal energy transferred by convection due to fluid density differences
n)	Quantitative analysis of energy transferred by radiation as a result of the emission of electromagnetic waves from the surface of a body, which in the case of a black body can be modelled by the Stefan Boltzmann law as given by [math]L = σAT^4[/math] where L is the luminosity, A is the surface area and T is the absolute temperature of the body
o)	The concept of apparent brightness b
p)	Luminosity L of a body as given by [math]b = \frac{L}{4 \pi d^2}[/math]
q)	The emission spectrum of a black body and the determination of the temperature of the body using Wien’s displacement law as given by [math]\lambda_{\text{max}} \, T = 2.9 \times 10^{-9} \, \text{mK}[/math] where [math]\lambda_{\text{max}}[/math] is the peak wavelength emitted.

a) Molecular Theory in Solids, Liquids, and Gases
The molecular theory describes the behavior and properties of matter in three states: solids, liquids, and gases. It is based on the kinetic theory of matter, which states that all matter is made up of tiny particles (atoms or molecules) that are in constant motion.
Figure 1 State of matter
⇒ Molecular Structure and Properties of Solids, Liquids, and Gases

State of Matter	Molecular Arrangement	Intermolecular Forces	Motion of Particles	Properties
Solids	Particles are closely packed in a fixed, ordered arrangement	Strongest intermolecular forces	Particles vibrate about fixed positions but do not move freely	Fixed shape and volume, incompressible
Liquids	Particles are close together but not in a fixed position	Weaker than in solids but stronger than in gases	Particles move freely past each other but remain in contact	Fixed volume, takes shape of container, slightly compressible
Gases	Particles are far apart and move randomly	Very weak intermolecular forces	Particles move freely and rapidly in all directions	No fixed shape or volume, highly compressible

b) Density (ρ)
Density is defined as the mass per unit volume of a substance. It represents how much matter is packed into a given space. The formula for density is:
[math]\rho = \frac{m}{v}[/math]
Where:
ρ = Density (kg/m³ or g/cm³)
m = Mass (kg or g)
V = Volume (m³ or cm³)
Figure 2 Density of cork, wood, and metal
⇒ Units of Density
The SI unit of density is kilograms per cubic meter (kg/m³).
In smaller-scale measurements, density is often expressed in grams per cubic centimeter (g/cm³).
The relation between the two units:
[math]1g/cm^3 = 1000kg/m^3[/math]
⇒ Comparison of Density in Different States of Matter:
Solids: Highest density due to tightly packed molecules.
Liquids: Lower density than solids (except for water, where ice is less dense than liquid water).
Gases: Lowest density, as particles are widely spaced.
⇒ Factors Affecting Density
1. Temperature:
– For most substances, density decreases as temperature increases (thermal expansion).
– Example: Warm air is less dense than cold air, causing warm air to rise.
2. Pressure:
– Affects gases significantly (compressibility).
– Example: As pressure increases, gas molecules are forced closer together, increasing density
c) Temperature Scales: Celsius and Kelvin
Temperature is a measure of the average kinetic energy of particles in a substance. It determines how hot or cold a body is.
There are multiple temperature scales, but the most commonly used ones in science are:
Celsius (°C) – Based on the freezing and boiling points of water.
– 0°C = Freezing point of water
– 100°C = Boiling point of water at standard atmospheric pressure (1 atm)
It is commonly used in daily life and many scientific fields.
Kelvin Scale (K):
The absolute temperature scale where 0 K represents absolute zero, the point at which all molecular motion theoretically stops.
– 0 K (absolute zero) = -273.15°C
– 273.15 K = Freezing point of water
– 373.15 K = Boiling point of water
The Kelvin scale does not use the degree symbol (°).
d) Relationship Between Celsius and Kelvin
Since the Kelvin and Celsius scales have the same size units, the difference between temperatures in Kelvin and Celsius is a fixed value of 15.
⇒ Conversion Between Celsius and Kelvin
Celsius to Kelvin:
[math]T(K) = T(℃) + 273.15[/math]
Kelvin to Celsius:
[math]T(℃) = T(K) – 273.15[/math]
Example Calculations
1. Convert 25°C to Kelvin:
[math]T(K) = T(℃) + 273.15 \\ T(K) = 25 + 273.15 \\ T(K) = 298.15 K[/math]
2. Convert 310 K to Celsius:
[math]T(℃) = T(K) – 273.15 \\ T(℃) = 310 – 273.15 \\ T(℃) = 36.85 ℃[/math]
The Kelvin scale is used in scientific calculations because it is based on absolute zero, making it ideal for thermodynamics and gas laws.
The Kelvin temperature is always positive, eliminating negative values in energy calculations
Since the size of one Kelvin and one Celsius degree is the same, a change in temperature of ΔT is the same in both units:
[math]ΔT(K) = ΔT(°C)[/math]
Figure 3 Relationship between Kelvin and centigrade
e) Kelvin Temperature and Average Kinetic Energy
Temperature in Kelvin (K) is directly related to the average kinetic energy of the particles in a system. This relationship is expressed as:
[math]\overline{E_k} = \frac{3}{2} k_B T[/math]
Where:
– [math]\overline{E_k}[/math] = Average kinetic energy of a particle (J)
– [math]k_B[/math] = Boltzmann constant ([math]1.38 × 10^{-23} J/K[/math] )
– T = Temperature in Kelvin (K)
This equation shows that as temperature increases, the average kinetic energy of particles also increases. It explains why hotter substances have faster-moving molecules, while colder substances have slower-moving molecules.
f) Internal Energy of a System
The internal energy (U) of a system consists of two components:
1. Total random kinetic energy of molecules due to their motion.
2. Total intermolecular potential energy due to the forces between molecules.
⇒ Mathematically,
[math]U = \text{Kinetic Energy} + \text{Potential Energy}[/math]
⇒ Kinetic Energy Contribution
– Related to temperature (higher temperature → higher kinetic energy).
– Particles in gases have higher kinetic energy than in liquids or solids.
⇒ Potential Energy Contribution
– Related to intermolecular forces.
– Solids have the strongest intermolecular forces (lowest potential energy).
– Gases have the weakest intermolecular forces (highest potential energy).
– Heating a system increases its internal energy by increasing the kinetic and/or potential energy.
– Cooling a system decreases its internal energy by reducing the kinetic and/or potential energy.
Figure 4 Internal energy of a system
g) Temperature Difference and Thermal Energy Transfer
Heat transfer occurs due to a temperature difference between two bodies. Heat always flows from the hotter body to the colder body until thermal equilibrium is reached.
There are three main modes of heat transfer:
1. Conduction – Transfer through direct contact (e.g., metal rod heating up).
2. Convection – Transfer via fluid motion (e.g., boiling water).
3. Radiation – Transfer via electromagnetic waves (e.g., sunlight warming the Earth).
The greater the temperature difference, the faster the heat transfer.
Figure 5 Understand the thermal energy transfer
Example:
– A hot coffee cup transfers heat to your cold hands via conduction.
– A heater warms a room by convection.
– The Sun transfers heat to the Earth via radiation.
h) Phase Changes and Energy Transfer
A phase change occurs when a substance transitions between solid, liquid, and gas
⇒ Phase Changes
– Phase changes occur at a constant temperature.
– Heat energy is used to change particle behavior (not temperature).
– The kinetic energy remains constant, but the potential energy changes.
⇒ Types of Phase Changes

Phase Change	Description	Energy Change
Melting	Solid → Liquid	Absorbs energy
Freezing	Liquid → Solid	Releases energy
Vaporization	Liquid → Gas	Absorbs energy
Condensation	Gas → Liquid	Releases energy
Sublimation	Solid → Gas	Absorbs energy
Deposition	Gas → Solid	Releases energy

⇒ Latent Heat
The heat energy required for a phase change without changing temperature is called latent heat (L). The amount of heat required is:
[math]Q = mL[/math]
Where:
– Q = Heat energy (Joules)
– m = Mass of the substance (kg)
– L = Specific latent heat (J/kg)
There are two types:
Latent heat of fusion ([math]L_f[/math]) → Energy required for melting/freezing.
Latent heat of vaporization ([math]L_v[/math]) → Energy required for boiling/condensation.
⇒ Example: Boiling Water
When water boils at 100°C, the temperature remains constant even though heat is continuously supplied. The energy is used to break intermolecular bonds, allowing water to turn into steam.
Figure 6 Phase changes and energy transfer
i) Quantitative Analysis of Thermal Energy Transfers
⇒ Sensible Heat and Specific Heat Capacity
When the temperature of a substance changes without a phase change, the energy transferred as heat (sensible heat) is given by:
[math]Q = mcΔT[/math]
Where:
– Q = Thermal energy transferred (in Joules, J)
– m = Mass of the substance (in kilograms, kg)
– c = Specific heat capacity (in J/kg·°C or J/kg·K), a measure of the energy required to raise the temperature of 1 kg of the substance by 1°C (or 1 K)
– ΔT = Change in temperature (°C or K)
⇒ Example:
If you heat 2 kg of water (with [math]c ≈ 4186J/kgK[/math]) from 20°C to 80°C, the energy required is:
[math]Q = mcΔT \\ Q = (2)(4186)(80 – 20) \\ Q = 2 × 4186 × 60 \\ Q = 502,320 J[/math]
Latent Heat and Phase Changes
During a phase change (such as melting or vaporization), the temperature remains constant while energy is absorbed or released. The energy involved is calculated by:
[math]Q = mL[/math]
Where:
L = Specific latent heat (in J/kg)
– Latent heat of fusion ([math]L_f[/math]): Energy required to change a substance from solid to liquid (or vice versa) at its melting/freezing point.
– Latent heat of vaporization ([math]L_v[/math]): Energy required to change a substance from liquid to gas (or vice versa) at its boiling/condensation point.
⇒ Example:
To melt 1 kg of ice at 0°C, with [math]L_f ≈ 334,000J/kg[/math]:
[math]Q = mL_f \\ Q = (1)(334,000) \\ Q = 334,000 J[/math]
j) Primary Mechanisms of Thermal Energy Transfer
There are three main mechanisms by which thermal energy is transferred:
⇒ Conduction
Mechanism:
Energy is transferred through direct molecular collisions. In solids, particles vibrate and transfer kinetic energy from one molecule to another.
Quantitative Analysis:
The rate of heat transfer by conduction is given by Fourier’s Law:
[math]\frac{\Delta Q}{\Delta t} = kA \frac{\Delta T}{\Delta x}[/math]
Where:
– [math]\frac{\Delta Q}{\Delta t}[/math] = Rate of heat transfer (Watts, W)
– k = Thermal conductivity of the material (W/m·K)
– A = Cross-sectional area through which heat is transferred (m²)
– ΔT = Temperature difference between the two ends (K or °C)
– Δx = Distance between the regions at different temperatures (m)
⇒ Convection
Mechanism:
Thermal energy is carried by the movement of fluids (liquids or gases). Warmer, less dense regions of the fluid rise, and cooler, denser regions sink, setting up a convective current.
Example:
Heating of water in a pot, atmospheric circulation.
⇒ Thermal Radiation
Mechanism:
Energy is transferred by electromagnetic waves (infrared radiation) without requiring a medium.
Example:
The Sun’s energy reaching the Earth, heat from a fire felt across a room.
Figure 7 Heat transfer
k) Conduction in Terms of Kinetic Energy
In a solid, the molecular theory explains conduction as follows:
Molecules vibrate due to thermal energy.
When a molecule vibrates, it collides with neighboring molecules, transferring some of its kinetic energy.
A temperature difference means that molecules in the hotter region vibrate with higher energy than those in the cooler region.
The net effect is that thermal energy flows from the hotter region to the cooler region until thermal equilibrium is reached.
Figure 8 Conduction
l) Quantitative Analysis of Conduction Using Fourier’s Law
Fourier’s Law of conduction is expressed as:
[math]\frac{\Delta Q}{\Delta t} = kA \frac{\Delta T}{\Delta x}[/math]
⇒ Variables Explained:
Thermal Conductivity (k):
A property of the material that measures how well it conducts heat. For example, metals have high k values, while wood and plastics have low k values.
Cross-Sectional Area (A):
The area through which the heat is being conducted. Larger areas allow more heat to flow.
Temperature Gradient ([math]\frac{\Delta T}{\Delta x}[/math]):
The change in temperature per unit distance. A steeper gradient results in faster heat transfer.
⇒ Example Calculation:
Suppose you have a metal rod with:
k=200 W/mK
A=0.005 m²
ΔT=50 K (difference between the two ends)
Δx=0.2 m
Then the rate of heat transfer is:
[math]\frac{\Delta Q}{\Delta t} = 200 \times 0.005 \times \frac{50}{0.2} \\
\frac{\Delta Q}{\Delta t} = 200 \times 0.005 \times 250 \\
\frac{\Delta Q}{\Delta t} = 200 \times 1.25 \\
\frac{\Delta Q}{\Delta t} = 250 \, \text{W}[/math]
This means 250 Joules of thermal energy is transferred per second.
m) Convection: Qualitative Description of Thermal Energy Transfer
Convection is the process of thermal energy transfer in fluids (liquids and gases) due to differences in density caused by temperature variations.
⇒ Mechanism of Convection
1. Heating a fluid causes its particles to move faster, increasing the average kinetic energy.
2. The fluid expands, reducing its density.
3. The less dense, hotter fluid rises while the cooler, denser fluid sinks.
4. This movement creates a convection current, continuously transferring heat within the fluid.
Figure 9 Convection
⇒ Examples of Convection
Boiling Water: When water is heated in a pot, hot water at the bottom rises, and cooler water sinks, forming a circular convection current.
Sea and Land Breeze:
– Daytime: Land heats up faster than water, causing warm air over land to rise and cooler air from the sea to move in (sea breeze).
– Nighttime: Land cools faster than water, reversing the process (land breeze).
– Heating a Room: A radiator heats the air, which rises, and cooler air moves in to replace it, setting up a convection cycle.
n) Quantitative Analysis of Energy Transfer by Radiation
⇒ Thermal Radiation
Radiation is the transfer of thermal energy by electromagnetic waves, primarily in the infrared spectrum. Unlike conduction and convection, radiation does not require a medium and can occur in a vacuum.
⇒ Stefan-Boltzmann Law
For a black body (an ideal emitter and absorber of radiation), the total energy emitted per second (luminosity) is given by:
[math]L = σAT^4[/math]
Where:
– L = Luminosity (total energy radiated per second) in watts (W)
– σ = Stefan-Boltzmann constant ([math]5.67 × 10^{-8} W/m²K⁴[/math])
– A = Surface area of the emitting body (m²)
– T = Absolute temperature in Kelvin (K)
⇒ Implications
1. Higher Temperature → More Radiation
– Since [math]L ∝ T^4[/math] if the temperature doubles, the emitted energy increases by 2⁴=16 times.
2. Larger Surface Area → More Radiation
– A larger body emits more radiation because it has a greater emitting surface.
3. Black Bodies vs. Real Objects
– A perfect black body absorbs and emits all radiation, but real objects have an emissivity factor (ε) between 0 and 1.
– The actual emitted energy for a real object is:
[math]L = εσAT^4[/math]
o) Apparent Brightness (b)
Apparent brightness is the power per unit area received by an observer from a radiating source.
[math]b = \frac{L}{4πd^2}[/math]
Where:
b = Apparent brightness (W/m²)
L = Luminosity (W)
d = Distance between the observer and the source (m)
1. Inverse Square Law:
– If the distance from the source doubles, the brightness decreases by a factor of 1/4.
– If the distance triples, the brightness decreases by 1/9.
2. Stars and Observations:
– A star with high luminosity appears brighter if it is closer, and dimmer if it is farther away.
– Two stars of different luminosities may appear the same brightness if one is much farther away.
p) Luminosity and Apparent Brightness
⇒ Luminosity (L)
Luminosity is the total power output of a radiating object (such as a star, planet, or light source). It is measured in watts (W) and represents the total energy emitted per second.
⇒ Apparent Brightness (b)
Apparent brightness is the power per unit area received by an observer from a luminous object. It decreases with distance from the source.
The relationship between luminosity and apparent brightness follows the inverse square law:
[math]b = \frac{L}{4πd^2}[/math]
Where:
– b = Apparent brightness (W/m²)
– L = Luminosity (W)
– d = Distance between the observer and the light source (m)
1. Inverse Square Law:
– If the distance doubles, the apparent brightness decreases by a factor of 4 ([math]b ∝ 1/d^2[/math]).
– If the distance triples, the apparent brightness decreases by a factor of 9.
2. Comparison of Stars:
– Two stars with the same luminosity will appear different in brightness if they are at different distances.
– A very luminous but far star can appear as bright as a less luminous but closer star.
3. Application to Astronomy:
– Used to estimate the distance of stars and galaxies if luminosity is known.
– Helps in identifying different types of stars based on their brightness and spectral characteristics.
Figure 10 Luminosity and apparent brightness
q) Black Body Radiation and Wien’s Displacement Law
⇒ Black Body Radiation
A black body is an idealized object that absorbs all incident radiation and emits radiation at all wavelengths based only on its temperature.
The emitted radiation follows a characteristic black body spectrum, where:
– Hotter objects emit more energy at shorter wavelengths (blue light).
– Cooler objects emit more energy at longer wavelengths (red light).
Figure 11 Black body radiation spectrum
⇒ Wien’s Displacement Law
The peak wavelength of radiation emitted by a black body is inversely proportional to its absolute temperature. This relationship is given by:
[math]\lambda_{\text{max}} \, T = 2.9 \times 10^{-3} \, \text{m·K}[/math]
Where:
– [math]\lambda_{\text{max}}[/math] = Wavelength at which maximum radiation is emitted (m)
– T = Absolute temperature of the black body (K)
⇒ Implications of Wien’s Law
1. Hotter objects emit shorter wavelengths
– The Sun (≈ 5800 K) has a peak wavelength in the visible spectrum (~500 nm, green-yellow light).
– A cooler red star (≈ 3000 K) has a peak wavelength in the infrared spectrum (~1000 nm).
– A hot blue star (≈ 10,000 K) emits most of its radiation in the ultraviolet (~300 nm).
2. Determining Temperature of Stars
– By measuring [math]\lambda_{\text{max}}[/math] from a star’s spectrum, its temperature can be calculated using Wien’s law.
– Hotter stars appear blue, while cooler stars appear red.
3. Temperature and Color of Common Objects
– Incandescent light bulb filament (~3000 K): Glows yellowish.
– Molten metal (~2000 K): Glows red.
– Flames & the Sun (~5800 K): Appear white/yellow.