🔥 RATE OF FLOW OF HEAT 🔥

Grade XI Physics | NEB Syllabus

Understanding Heat Transfer: Conduction, Convection, and Radiation

What You'll Learn Today:

Mechanisms of heat transfer: Conduction, Convection, Radiation
Thermal conductivity and its measurement
Black-body radiation and Stefan-Boltzmann law
Mathematical derivations of heat transfer equations
Real-world applications and problem-solving

1. Introduction: Mechanisms of Heat Transfer

What is Heat Transfer?

Heat transfer is the process by which thermal energy moves from a region of higher temperature to a region of lower temperature.

Fundamental Principle

"Heat always flows from a body at higher temperature to a body at lower temperature."

Three Modes of Heat Transfer

1. Conduction

Transfer through direct contact
Occurs in solids
No bulk motion of material
Example: Heat through metal rod

2. Convection

Transfer by fluid motion
Occurs in liquids and gases
Involves bulk motion of fluid
Example: Hot air rising

3. Radiation

Transfer by electromagnetic waves
No medium required
Can occur in vacuum
Example: Sun's heat reaching Earth

Figure 1: Three Modes of Heat Transfer

2. 12.1 Conduction: Thermal Conductivity

What is Conduction?

Conduction is the process of heat transfer through a material without any actual motion of the material particles. It occurs due to molecular collisions and vibrations.

Figure 2: Heat Conduction in a Metal Rod

Fourier's Law of Heat Conduction

Statement: The rate of heat flow through a conductor is directly proportional to the cross-sectional area, the temperature gradient, and inversely proportional to the length of the conductor.

Mathematical Expression:

Q/t = -kA(dT/dx)

Where:

Q/t = Rate of heat flow (J/s or W)
k = Thermal conductivity of the material (W/m⋅K)
A = Cross-sectional area (m²)
dT/dx = Temperature gradient (K/m)
Negative sign indicates heat flows from high to low temperature

3. Derivation: Rate of Heat Flow by Conduction

Consider a uniform rod of length l, cross-sectional area A, with temperatures T₁ and T₂ at its ends (T₁ > T₂).

Step 1: Assume steady state condition (temperature at any point remains constant with time)

Step 2: Temperature gradient (constant for uniform rod)

dT/dx = (T₂ - T₁)/l

Step 3: Apply Fourier's Law

Q/t = -kA(dT/dx)

Step 4: Substitute the temperature gradient

Q/t = -kA[(T₂ - T₁)/l]

Step 5: Simplify (since T₁ > T₂, T₂ - T₁ is negative)

Q/t = kA(T₁ - T₂)/l

Final Result:

H = kA(T₁ - T₂)/l

Where H = rate of heat flow (Watts)

Physical Meaning of Terms

k (Thermal Conductivity): Material property indicating heat conducting ability
A (Area): Larger area → More heat transfer
(T₁ - T₂): Larger temperature difference → More heat transfer
l (Length): Longer rod → Less heat transfer

Figure 3: Temperature Gradient in Conductor

4. Thermal Conductivity (k)

Definition

Thermal conductivity is a material property that measures the ability of a substance to conduct heat. It is numerically equal to the rate of heat flow per unit area per unit temperature gradient.

k = (Q/t) × (l)/(A × ΔT)

Unit: W/m⋅K (Watts per meter per Kelvin)

Physical Meaning

If a material has thermal conductivity k, then 1 W of heat flows through 1 m² area of the material of 1 m thickness for every 1 K temperature difference.

Thermal Conductivities of Common Materials

Material	Thermal Conductivity k (W/m⋅K)	Category
Silver	428	Excellent conductor
Copper	401	Excellent conductor
Aluminum	237	Good conductor
Iron	80	Good conductor
Water	0.6	Poor conductor
Glass	0.8	Poor conductor
Wood	0.12	Insulator
Air	0.026	Insulator

Important Notes:

Metals have high thermal conductivity due to free electrons
Non-metals generally have lower thermal conductivity
Gases have very low thermal conductivity
Thermal conductivity depends on temperature and pressure

5. Measurement of Thermal Conductivity

Searle's Method

Searle's method is a steady-state technique used to measure the thermal conductivity of good conductors like metals.

Figure 4: Searle's Apparatus

Working Principle

Step 1: Steam is passed through one end of the metal rod

Step 2: Water flows through the jacket at the other end

Step 3: In steady state, rate of heat flow through rod equals rate of heat absorbed by water

Heat flow through rod:

H_rod = kA(θ₁ - θ₂)/l

Heat absorbed by water:

H_water = ms(θ_out - θ_in)

Where m = mass flow rate of water, s = specific heat of water

In steady state:

kA(θ₁ - θ₂)/l = ms(θ_out - θ_in)

Solving for k:

k = [ms(θ_out - θ_in) × l] / [A(θ₁ - θ₂)]

6. 12.2 Convection

What is Convection?

Convection is the process of heat transfer by the actual movement of heated particles of a fluid (liquid or gas) from one place to another.

Types of Convection

Natural (Free) Convection

Movement due to density differences
Heated fluid becomes less dense
Rises due to buoyancy
Example: Hot air rising from radiator

Forced Convection

Movement caused by external force
Fans, pumps, or blowers
More efficient than natural convection
Example: Air conditioning system

Figure 5: Natural Convection in a Room

Newton's Law of Cooling

For convection, the rate of heat loss is proportional to the temperature difference between the body and its surroundings.

dQ/dt = hA(T - T₀)

Where:

h = convective heat transfer coefficient (W/m²⋅K)
A = surface area (m²)
T = temperature of body (K)
T₀ = temperature of surroundings (K)

7. 12.3 Radiation: Ideal Radiator

What is Thermal Radiation?

Thermal radiation is the process of heat transfer in the form of electromagnetic waves, primarily in the infrared region. It does not require a medium and can occur in vacuum.

Characteristics of Thermal Radiation

Travels at speed of light (3 × 10⁸ m/s)
Can travel through vacuum
Obey laws of reflection and refraction
Travel in straight lines
Carry energy and momentum

Ideal Radiator (Perfect Black Body)

Definition:

An ideal radiator (or perfect black body) is a body that absorbs all the radiant energy incident upon it and also emits maximum possible radiant energy at any given temperature.

Figure 6: Perfect Black Body

Real Surfaces vs. Ideal Radiator

Property	Ideal Radiator (Black Body)	Real Surfaces
Absorption	Absorbs 100% of incident radiation	Absorbs less than 100%
Emission	Emits maximum possible radiation	Emits less than maximum
Reflection	Reflects 0% of incident radiation	Reflects some radiation
Transmission	Transmits 0% of incident radiation	May transmit some radiation

8. 12.4 Black-Body Radiation

Black-Body Radiation Spectrum

When a black body is heated, it emits electromagnetic radiation of all wavelengths. The distribution of energy among different wavelengths depends on the temperature.

Figure 7: Black-Body Radiation Curves

Important Laws of Black-Body Radiation

1. Wien's Displacement Law

λₘT = b = 2.898 × 10⁻³ m⋅K

Where λₘ = wavelength at which intensity is maximum, T = absolute temperature

Implication: Hotter bodies emit radiation at shorter wavelengths

2. Stefan-Boltzmann Law

E = σT⁴

Where E = energy radiated per unit area per unit time, σ = Stefan-Boltzmann constant

Implication: Energy radiated increases very rapidly with temperature

9. 12.5 Stefan-Boltzmann Law

Statement of the Law

"The total energy radiated per unit surface area of a black body per unit time is directly proportional to the fourth power of its absolute temperature."

E = σT⁴

Where:

E = energy radiated per unit area per unit time (W/m²)
σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²⋅K⁴
T = absolute temperature (Kelvin)

Derivation (Conceptual)

Step 1: Experimental observation shows that E ∝ T⁴

Step 2: Introduce proportionality constant

E = σT⁴

Step 3: For a body of surface area A

Total energy radiated = E × A = σAT⁴

Step 4: For a real body with emissivity ε (0 < ε ≤ 1)

E_real = εσT⁴

Where ε = 1 for perfect black body, ε < 1 for real surfaces

Physical Significance

Temperature dependence is very strong (T⁴)
Doubling temperature increases radiation by factor of 16
Explains why hot objects glow brightly
Fundamental to understanding stellar physics

Figure 8: Energy vs Temperature (T⁴ Relationship)

10. Net Heat Transfer by Radiation

When Body is Surrounded by Another Body

When a hot body is placed in an environment at different temperature, there is both emission and absorption of radiation.

Consider: A body at temperature T₁ surrounded by environment at T₂

Step 1: Energy radiated by the body

E_emitted = εσAT₁⁴

Step 2: Energy absorbed from surroundings

E_absorbed = εσAT₂⁴

Step 3: Net energy loss

E_net = εσA(T₁⁴ - T₂⁴)

Step 4: Rate of heat loss

dQ/dt = εσA(T₁⁴ - T₂⁴)

Special Cases:

If T₁ = T₂: No net heat transfer
If T₂ = 0 (surrounded by vacuum at 0 K): dQ/dt = εσAT₁⁴
If body is perfect black body (ε = 1): dQ/dt = σA(T₁⁴ - T₂⁴)

Figure 9: Net Radiation Heat Transfer

11. Solved Numerical Problems

Problem 1: Conduction Through Metal Rod

Question: A copper rod of length 1 m and cross-sectional area 2 × 10⁻⁴ m² has one end at 100°C and the other end at 0°C. Calculate the rate of heat flow through the rod. (Thermal conductivity of copper = 401 W/m⋅K)

Given: l = 1 m, A = 2 × 10⁻⁴ m², T₁ = 100°C = 373 K, T₂ = 0°C = 273 K, k = 401 W/m⋅K

Solution:

Formula: H = kA(T₁ - T₂)/l

Calculation:

H = (401 × 2 × 10⁻⁴ × (373 - 273))/1

H = (401 × 2 × 10⁻⁴ × 100)/1

H = (401 × 2 × 10⁻²)/1

H = 8.02 W

Answer: The rate of heat flow is 8.02 W.

Solved Numerical Problems

Problem 2: Thermal Conductivity Measurement

Question: In Searle's experiment, a metal rod of length 60 cm and cross-sectional area 2 cm² is used. The temperature difference across the rod is 50°C. Water flows at the rate of 10 g/s and its temperature rises by 2°C. Calculate the thermal conductivity of the metal.

Given: l = 60 cm = 0.6 m, A = 2 cm² = 2 × 10⁻⁴ m², ΔT = 50°C, m = 10 g/s = 0.01 kg/s, Δθ = 2°C, s = 4200 J/kg⋅K

Solution:

Step 1: Heat absorbed by water per second

H_water = msΔθ = 0.01 × 4200 × 2 = 84 W

Step 2: Heat conducted through rod = Heat absorbed by water

H_rod = kAΔT/l = 84 W

Step 3: Solve for k

k = (H_rod × l)/(A × ΔT)

k = (84 × 0.6)/(2 × 10⁻⁴ × 50)

k = 50.4/(10⁻²) = 5040 W/m⋅K

Answer: The thermal conductivity is 5040 W/m⋅K.

Solved Numerical Problems

Problem 3: Stefan-Boltzmann Law

Question: A black body has a surface area of 0.01 m² and temperature of 2000 K. Calculate the energy radiated per second. (σ = 5.67 × 10⁻⁸ W/m²⋅K⁴)

Given: A = 0.01 m², T = 2000 K, σ = 5.67 × 10⁻⁸ W/m²⋅K⁴, ε = 1 (black body)

Solution:

Formula: P = εσAT⁴

Calculation:

P = 1 × 5.67 × 10⁻⁸ × 0.01 × (2000)⁴

P = 5.67 × 10⁻¹⁰ × 16 × 10¹²

P = 5.67 × 16 × 10²

P = 90.72 × 10² = 9072 W

Answer: The energy radiated per second is 9072 W.

Solved Numerical Problems

Problem 4: Wien's Displacement Law

Question: The Sun's surface temperature is approximately 5800 K. At what wavelength does it emit maximum energy? (Wien's constant = 2.898 × 10⁻³ m⋅K)

Given: T = 5800 K, b = 2.898 × 10⁻³ m⋅K

Solution:

Formula: λₘT = b

Calculation:

λₘ = b/T = (2.898 × 10⁻³)/5800

λₘ = 4.997 × 10⁻⁷ m = 499.7 nm

Answer: The Sun emits maximum energy at wavelength 500 nm (green light).

Solved Numerical Problems

Problem 5: Net Radiation Heat Transfer

Question: A copper sphere of radius 5 cm at temperature 500 K is placed in an environment at 300 K. If the emissivity of copper is 0.8, calculate the net rate of heat loss by radiation. (σ = 5.67 × 10⁻⁸ W/m²⋅K⁴)

Given: r = 5 cm = 0.05 m, T₁ = 500 K, T₂ = 300 K, ε = 0.8, σ = 5.67 × 10⁻⁸ W/m²⋅K⁴

Solution:

Step 1: Calculate surface area

A = 4πr² = 4π(0.05)² = 4π(0.0025) = 0.0314 m²

Step 2: Apply net radiation formula

dQ/dt = εσA(T₁⁴ - T₂⁴)

Step 3: Calculate T₁⁴ and T₂⁴

T₁⁴ = (500)⁴ = 6.25 × 10¹⁰

T₂⁴ = (300)⁴ = 8.1 × 10⁹

T₁⁴ - T₂⁴ = 6.25 × 10¹⁰ - 8.1 × 10⁹ = 5.44 × 10¹⁰

Step 4: Calculate heat loss

dQ/dt = 0.8 × 5.67 × 10⁻⁸ × 0.0314 × 5.44 × 10¹⁰

dQ/dt = 0.8 × 5.67 × 0.0314 × 5.44 × 10²

dQ/dt = 77.6 W

Answer: The net rate of heat loss is 77.6 W.

12. Multiple Choice Questions (MCQs)

Set 1: Questions 1-10

The unit of thermal conductivity is:
1. W/m
2. W/m²
3. W/m⋅K
4. J/kg⋅K
Answer: (c) W/m⋅K
Heat transfer by convection occurs in:
1. Solids only
2. Liquids and gases only
3. Solids, liquids and gases
4. Vacuum only
Answer: (b) Liquids and gases only
Which of the following has the highest thermal conductivity?
1. Copper
2. Aluminum
3. Iron
4. Silver
Answer: (d) Silver
According to Stefan-Boltzmann law, energy radiated is proportional to:
1. T
2. T²
3. T³
4. T⁴
Answer: (d) T⁴
A perfect black body:
1. Absorbs all incident radiation
2. Reflects all incident radiation
3. Transmits all incident radiation
4. Emits no radiation
Answer: (a) Absorbs all incident radiation

Multiple Choice Questions (MCQs)

Set 2: Questions 6-15

Wien's displacement law relates:
1. Temperature and energy
2. Wavelength and frequency
3. Maximum wavelength and temperature
4. Intensity and wavelength
Answer: (c) Maximum wavelength and temperature
Thermal radiation can travel through:
1. Solids only
2. Liquids only
3. Gases only
4. Vacuum
Answer: (d) Vacuum
The value of Stefan-Boltzmann constant is:
1. 1.38 × 10⁻²³ W/m²⋅K⁴
2. 5.67 × 10⁻⁸ W/m²⋅K⁴
3. 9 × 10⁹ W/m²⋅K⁴
4. 8.31 W/m²⋅K⁴
Answer: (b) 5.67 × 10⁻⁸ W/m²⋅K⁴
In Searle's method, the steady state condition means:
1. Temperature changes with time
2. Temperature at any point remains constant
3. No heat flow
4. All temperatures are equal
Answer: (b) Temperature at any point remains constant
Convection does not occur in:
1. Water
2. Air
3. Vacuum
4. Oil
Answer: (c) Vacuum

Multiple Choice Questions (MCQs)

Set 3: Questions 11-20

The rate of heat flow by conduction is directly proportional to:
1. Cross-sectional area
2. Temperature difference
3. Length of conductor
4. Both (a) and (b)
Answer: (d) Both (a) and (b)
Which has the lowest thermal conductivity?
1. Copper
2. Silver
3. Air
4. Aluminum
Answer: (c) Air
Forced convection is caused by:
1. Density differences
2. External forces
3. Temperature gradients
4. Pressure differences
Answer: (b) External forces
The emissivity of a perfect black body is:
1. 0
2. 0.5
3. 1
4. ∞
Answer: (c) 1
If the temperature of a black body is doubled, its energy radiation increases by:
1. 2 times
2. 4 times
3. 8 times
4. 16 times
Answer: (d) 16 times (2⁴ = 16)

13. Additional Practice Problems

Unsolved Numerical Questions

A steel rod of length 2 m and cross-sectional area 5 × 10⁻⁴ m² has one end at 200°C and the other at 50°C. Find the rate of heat flow if thermal conductivity of steel is 50 W/m⋅K.
In an experiment, water flows at 15 g/s through a calorimeter. Its temperature rises from 15°C to 17°C. If the heat supplied is 200 W, find the specific heat of water.
A tungsten filament of a bulb has an area of 0.5 cm² and temperature 2500 K. Calculate the energy radiated per second if emissivity is 0.3.
The peak wavelength of radiation from a star is 400 nm. Estimate its surface temperature using Wien's law.
A body at 600 K is placed in surroundings at 300 K. If its surface area is 0.02 m² and emissivity 0.7, find the net heat loss per second.
Two rods of same material and length but different diameters are connected in series. If diameter ratio is 1:2, find the ratio of temperature gradients.
A black body at 1000 K radiates energy. If its temperature is increased to 2000 K, by what factor does the total energy radiated increase?
Calculate the wavelength at which human body (temperature 310 K) emits maximum radiation.

14. Summary - Key Concepts

Important Formulas

Concept	Formula	Variables
Conduction Rate	H = kA(T₁ - T₂)/l	k = thermal conductivity
Stefan-Boltzmann Law	E = σT⁴	σ = 5.67 × 10⁻⁸ W/m²⋅K⁴
Wien's Law	λₘT = b	b = 2.898 × 10⁻³ m⋅K
Net Radiation	dQ/dt = εσA(T₁⁴ - T₂⁴)	ε = emissivity
Convection	dQ/dt = hA(T - T₀)	h = convective coefficient

Key Points to Remember

🔥 Conduction: H ∝ kA(T₁ - T₂)/l
🌀 Convection: Requires fluid motion
☀️ Radiation: E ∝ T⁴ (Stefan-Boltzmann)
📏 Wien's Law: λₘ ∝ 1/T
🎯 Black Body: ε = 1, absorbs and emits maximum
📊 Steady State: Temperature constant with time

Important Relationships

Doubling temperature increases radiation by factor of 16
Conduction rate increases with area and temperature difference
Conduction rate decreases with length
Hotter bodies emit shorter wavelength radiation

15. Real-World Applications

Heat Transfer in Daily Life and Technology

Conduction Applications

Cooking utensils: Metals for handles, insulators for grips
Thermos flask: Vacuum to prevent conduction
Heat sinks: High thermal conductivity for cooling
Building insulation: Low thermal conductivity materials

Convection Applications

Refrigerators: Natural convection circulation
Heating systems: Forced convection with fans
Weather systems: Atmospheric convection
Car radiators: Forced convection cooling

Radiation Applications

Solar panels: Absorb solar radiation
Greenhouse effect: Traps infrared radiation
Heat lamps: Emit infrared radiation
Remote sensing: Detects thermal radiation

Combined Applications

Steam engines: All three modes of heat transfer
Power plants: Complex heat transfer systems
Electronic cooling: Conduction + convection + radiation
Spacecraft design: Radiation is primary mode

Why It Matters

Understanding heat transfer is crucial for energy efficiency
Essential for designing heating and cooling systems
Fundamental to meteorology and climate science
Important for materials science and engineering design

Thank You! 🙏

Questions & Discussion

Important Constants to Remember:

Stefan-Boltzmann constant: σ = 5.67 × 10⁻⁸ W/m²⋅K⁴
Wien's displacement constant: b = 2.898 × 10⁻³ m⋅K
Specific heat of water: s = 4200 J/kg⋅K
Absolute zero: 0 K = -273.15°C

Problem-Solving Strategy:

Identify the mode of heat transfer (conduction/convection/radiation)
Write down the appropriate formula
Convert all temperatures to Kelvin
Check units and physical reasonableness
For radiation problems, consider net heat transfer

Mastering heat transfer concepts is fundamental to understanding thermodynamics! 🚀