Heat Transfer Rate (Q/t)
0 W
In kW
0 kW

What is Heat Transfer?

Heat transfer is the movement of thermal energy from a hotter region to a colder region. There are three fundamental mechanisms: conduction, convection, and radiation. Understanding these is essential for thermal management in electronics, building design, and engineering systems.

1. Conduction

Conduction is heat transfer through a material without bulk movement of the material itself. Energy is transferred by vibrating atoms and free electrons passing energy to neighboring particles.

Q/t = kA(ΔT)/d
  • k = thermal conductivity (W/m·K) — higher k means better conductor
  • A = cross-sectional area (m²)
  • ΔT = temperature difference (K or °C)
  • d = thickness of material (m)

Good conductors: Copper (385 W/m·K), Aluminum (205 W/m·K)
Poor conductors (insulators): Air (0.024 W/m·K), Styrofoam (0.03 W/m·K)

2. Convection

Convection is heat transfer by the bulk movement of a fluid (liquid or gas). Hot fluid rises, cool fluid sinks, creating convection currents.

Q/t = hA(ΔT)
  • h = convection heat transfer coefficient (W/m²·K)
  • A = surface area in contact with fluid (m²)
  • ΔT = temperature difference between surface and fluid (K)

Typical h values: Natural convection in air: 5-25 W/m²·K | Forced air: 25-250 W/m²·K | Water: 100-15,000 W/m²·K

3. Radiation

Radiation is heat transfer through electromagnetic waves — no medium required. All objects above absolute zero emit thermal radiation.

Q/t = εσA(T⁴ - T₀⁴)
  • ε = emissivity (0 to 1) — black body = 1, shiny metal ≈ 0.1
  • σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²K⁴
  • T, T₀ = absolute temperatures in Kelvin
⚠️ Important: For radiation, temperatures MUST be in Kelvin! Convert: K = °C + 273.15

Worked Examples

✅ Example 1: Heat Conduction Through a Wall

A concrete wall is 20 cm thick with area 10 m². The inside temperature is 22°C and outside is 5°C. If k = 1.0 W/m·K, find the heat loss rate.

Given: k = 1.0, A = 10 m², ΔT = 22 - 5 = 17°C, d = 0.2 m
Formula: Q/t = kAΔT/d
Solution: Q/t = (1.0 × 10 × 17) / 0.2 = 170 / 0.2 = 850 W

✅ Example 2: Copper vs Steel Comparison

Two rods of equal dimensions (A = 1 cm², d = 30 cm) have a 100°C temperature difference. Compare heat flow through copper (k = 385) vs steel (k = 50).

Copper: Q/t = (385 × 0.0001 × 100) / 0.3 = 12.8 W
Steel: Q/t = (50 × 0.0001 × 100) / 0.3 = 1.67 W
Ratio: Copper conducts 12.8/1.67 = 7.7× more heat

✅ Example 3: Convection from a Heatsink

A CPU heatsink has surface area 0.02 m² and temperature 60°C. Room air is at 25°C with h = 50 W/m²·K. What is the heat dissipation?

Given: h = 50, A = 0.02, ΔT = 60 - 25 = 35°C
Formula: Q/t = hAΔT
Solution: Q/t = 50 × 0.02 × 35 = 35 W

✅ Example 4: Radiation from Hot Object

A black-body sphere of radius 5 cm is at 500 K in an environment at 300 K. Calculate the net radiated power.

Given: ε = 1, A = 4πr² = 4π(0.05)² = 0.0314 m²
Formula: Q/t = εσA(T⁴ - T₀⁴)
Solution: Q/t = 1 × 5.67×10⁻⁸ × 0.0314 × (500⁴ - 300⁴)
= 1.78×10⁻⁹ × (6.25×10¹⁰ - 8.1×10⁹) = 1.78×10⁻⁹ × 5.44×10¹⁰
= 96.8 W

✅ Example 5: Combined Heat Transfer

A 100W incandescent bulb (ε = 0.8) operates at 2500 K. Calculate the surface area needed to radiate this power to a 300 K room.

Given: P = 100 W, ε = 0.8, T = 2500 K, T₀ = 300 K
Formula: A = P / [εσ(T⁴ - T₀⁴)]
Solution: A = 100 / [0.8 × 5.67×10⁻⁸ × (2500⁴ - 300⁴)]
= 100 / [4.54×10⁻⁸ × 3.9×10¹³] = 100 / 1.77×10⁶
= 5.65 × 10⁻⁵ m² ≈ 0.57 cm²
💡 Maker's Tip: For electronics cooling, convection typically dominates. Increase heatsink fin area and use a fan (forced convection) to maximize h. At very high temperatures (>500°C), radiation becomes significant.