Rate of Heat Flow: Conduction, Convection, Radiation & Applications

TAKEAWAYS:

• Heat flows from hotter to colder regions via conduction (solids), convection (fluids), and radiation (electromagnetic waves).
• Thermal conductivity (k) measures how well a material conducts heat: metals (high k) vs. wool (low k).
• Fourier’s law (Q/t = kAΔT/d) calculates heat flow rate through a slab.
• Convection currents (e.g., in air or water) transfer heat via fluid motion.
• Radiation depends on surface properties (emissivity, color) and follows Stefan-Boltzmann’s law (P = εσAT⁴).
• Applications include insulation, refrigerators, solar cookers, and Earth’s energy balance.

1. Introduction to Heat Transfer

Heat is energy that moves from a hotter object to a colder object until both reach the same temperature. There are three main ways heat transfers:

1. Conduction (through solids)
2. Convection (through fluids like air/water)
3. Radiation (through empty space as electromagnetic waves)

2. Conduction: Heat Through Solids

Definition: Heat transfer through a solid material where particles vibrate and pass energy to neighbors.

How it works:

• In metals, free electrons carry heat quickly (high thermal conductivity, k).
• In non-metals (wood, plastic), atoms vibrate slowly (low k).
• Fourier’s Law describes conduction mathematically: $$\frac{Q}{t}=\frac{kA\Delta T}{d}$$
  • Q/t = heat flow rate (J/s or W)
  • k = thermal conductivity (W/m·K)
  • A = cross-sectional area (m²)
  • ΔT = temperature difference (K or °C)
  • d = thickness of material (m)

Factors Affecting Conduction:

Worked Example 1:

A copper rod (k = 400 W/m·K) of length 0.5 m and cross-sectional area 2 cm² connects two blocks at 100°C and 0°C. Calculate the heat flow rate. Solution:

1. Convert area: A = 2 cm² = 2 × 10⁻⁴ m²
2. Use Fourier’s law: $$\frac{Q}{t}=\frac{400\times 2\times 10^{-4}\times (100-0)}{0.5}=16\text{W}$$ Answer: Heat flows at 16 watts.

3. Convection: Heat Through Fluids

Definition: Heat transfer via fluid motion (liquids/gases). Occurs in natural convection (e.g., warm air rising) or forced convection (e.g., fans blowing air).

How it works:

1. Heating: Fluid near a hot surface expands, becomes less dense, and rises.
2. Cooling: Cooler, denser fluid sinks, replacing the rising fluid.
3. Cycle repeats, creating convection currents.

flowchart TD
    A["Hot Surface\n(Heats fluid)"] -->|"Fluid expands"| B["Rising\nHot Fluid"]
    B --> C["Cooler Fluid\nSinks"]
    C -->|"Replaces hot fluid"| A
    caption: **Convection Current in Air**

Applications of Convection:

Worked Example 2:

Why does a ceiling fan make you feel cooler in summer? Explanation:

• The fan forces convection by blowing air over your skin.
• Sweat evaporates faster due to forced air movement, increasing cooling.
• Without the fan, natural convection is slower, and you feel warmer.

4. Radiation: Heat Through Empty Space

Definition: Heat transfer via electromagnetic waves (infrared rays). No medium required—works even in a vacuum (e.g., Sun’s heat reaching Earth).

Key Concepts:

1. All objects emit radiation based on their temperature.
2. Good absorbers are good emitters (e.g., black surfaces).
3. Stefan-Boltzmann’s Law: $$P=ϵ\sigma A{T}^4$$
   • P = power radiated (W)
   • ε = emissivity (0 to 1; blackbody = 1)
   • σ = Stefan-Boltzmann constant (5.67 × 10⁻⁸ W/m²·K⁴)
   • A = surface area (m²)
   • T = temperature (K)

Factors Affecting Radiation:

Worked Example 3:

A human body (surface area 1.5 m², ε = 0.98, T = 37°C) radiates heat. Calculate the power lost. Solution:

1. Convert T to Kelvin: T = 37 + 273 = 310 K
2. Apply Stefan-Boltzmann’s law: $$P=0.98\times 5.67\times 10^{-8}\times 1.5\times (310{)}^4$$ $$P\approx 100\text{W}$$ Answer: The body loses ~100 watts of heat via radiation.

5. Comparing Heat Transfer Methods

6. Practical Applications

A. Insulation (Reducing Heat Loss/Gain)

• Materials: Wool, fiberglass, aerogels (low k).
• Examples:
  • Thermos flask: Double-walled glass with vacuum (no conduction/convection) + shiny surface (reduces radiation).
  • House walls: Cavity walls with insulating foam.

B. Refrigerators and ACs

• Principle: Uses convection (coolant circulates) + radiation (heat escapes through coils).
• Steps:
  1. Compressor heats refrigerant gas.
  2. Gas cools inside coils (releasing heat to outside air via convection).
  3. Cool air blows into the room.

C. Solar Cookers

• How it works: Black pot absorbs solar radiation, while a glass cover traps heat (greenhouse effect).
• Advantages:
  • No fuel needed.
  • Eco-friendly (reduces smoke).

D. Earth’s Energy Balance

• Incoming solar radiation (short wavelength) is absorbed by Earth.
• Outgoing radiation (long wavelength, infrared) is partially trapped by greenhouse gases (CO₂, methane), causing global warming.

7. NEB Exam-Style Questions

Short Answer Questions (SAQ)

1. Define thermal conductivity. Why is it higher in metals than in wood?

   • Thermal conductivity (k) is the ability of a material to conduct heat.
   • Metals have free electrons that transfer heat quickly, while wood has tightly bound atoms that vibrate slowly.

2. How does a refrigerator keep food cold? Explain the role of convection.

   • A refrigerator uses a coolant that absorbs heat from inside.
   • A fan forces convection to circulate cool air, while radiator coils release heat outside via convection.

3. Why do we feel cold when we step out of a swimming pool?

   • Water has high specific heat, so it removes heat from our body quickly.
   • Evaporation of water from the skin also cools us down.

Long Answer Questions (LAQ)

4. Derive Fourier’s law of heat conduction. A composite wall has two layers: brick (k = 0.6 W/m·K, d = 0.1 m) and wood (k = 0.1 W/m·K, d = 0.2 m). If the outside temperature is 40°C and inside is 20°C, calculate the heat flow per unit area. Solution:

   • For series conduction, total resistance R = R₁ + R₂: $$R=\frac{d_1}{k_1}+\frac{d_2}{k_2}=\frac{0.1}{0.6}+\frac{0.2}{0.1}=0.1667+2=2.1667\text{m²⋅K/W}$$
   • Heat flow per unit area (Q/tA): $$\frac{Q}{tA}=\frac{\Delta T}{R}=\frac{40-20}{2.1667}\approx 9.23\text{W/m²}$$ Answer: 9.23 W/m² flows through the wall.

5. Explain why a black surface absorbs more heat than a white surface. How is this used in solar panels?

   • Black surfaces have high absorptivity (α) and high emissivity (ε), meaning they absorb and emit radiation well.
   • White surfaces reflect most radiation (low α).
   • Solar panels use black coatings to absorb sunlight efficiently and convert it to electricity.

Numerical Problems

6. A steel rod (k = 50 W/m·K) of length 1 m and area 0.01 m² connects two reservoirs at 150°C and 50°C. Calculate: a) Heat flow rate. b) Temperature at the midpoint of the rod. Solution: a) Using Fourier’s law: $$\frac{Q}{t}=\frac{50\times 0.01\times (150-50)}{1}=50\text{W}$$ b) Temperature drops linearly in steady state: $$T(x)=T_{\text{hot}}-\left(\frac{T_{\text{hot}}-T_{\text{cold}}}{L}\right)x$$ At x = 0.5 m: $$T=150-\left(\frac{100}{1}\right)\times 0.5=100°C$$ Answers: a) 50 W b) 100°C

Exam Tip

1. Memorize formulas:
   • Fourier’s law: Q/t = kAΔT/d
   • Stefan-Boltzmann’s law: P = εσAT⁴
2. Unit consistency: Always convert units (e.g., cm² → m², °C → K).
3. Diagrams: Draw convection currents, insulation layers, or radiation setups in answers.
4. Real-world links: Relate questions to daily life (e.g., "Why do we wear woolen clothes in winter?").
5. Significant figures: Match the least precise given value (e.g., k = 400 has 1 sig fig → answer to 1 sig fig).
6. Common mistakes to avoid:
   • Forgetting to convert temperature to Kelvin in radiation problems.
   • Mixing up conduction (solids) and convection (fluids).
   • Ignoring surface area in radiation calculations.