Describe the fundamental concepts, principles and theories underlying the thermal properties of materials (heat transfer, kinetic theory of solids, liquids and gases, thermodynamics and thermal radiation)

Heat flows from higher temperature to lower temperature through three mechanisms.

2.1 Thermal Conduction

Conduction is heat transfer through direct molecular interaction without net movement of material. For a slab of area A and thickness dx with temperature difference (θ₁ - θ₂):

$\frac{dQ}{dt} = -kA\frac{d\theta}{dx}$

where k is the thermal conductivity (Wm⁻¹K⁻¹). Good conductors (copper: 380 Wm⁻¹K⁻¹) have high k; poor conductors (rubber: 0.2 Wm⁻¹K⁻¹) have low k.

Series and parallel conduction follow analogous electrical resistance formulas. For series:

$\frac{Q}{t} = \frac{A(\theta_1 - \theta_2)}{\frac{L_1}{k_1} + \frac{L_2}{k_2}}$

2.2 Convection

Convection involves fluid movement carrying heat. Newton's law of cooling states that the rate of heat loss is proportional to temperature excess:

$\frac{dQ}{dt} = k(\theta - \theta_s)$

The cooling equation is:

$\ln\left(\frac{\theta - \theta_s}{\theta_i - \theta_s}\right) = -\lambda t$

where θ is body temperature, θ_s is surrounding temperature, and λ is a constant.

2.3 Thermal Radiation

All bodies emit electromagnetic radiation. A blackbody is a perfect emitter/absorber with emissivity ε = 1.

Stefan-Boltzmann law (for net radiative loss):

$P_{net} = \varepsilon\sigma A(T^4 - T_s^4)$

where σ = 5.67 × 10⁻⁸ Wm⁻²K⁻⁴.

Wien's displacement law:

$\lambda_{max}T = 2.9 \times 10^{-3} \text{ m·K}$

Energy cannot be created or destroyed, only converted:

$dQ = dU + dW$

where dQ is heat added, dU is change in internal energy, and dW is work done by the system.

3.1 Specific Heat Capacity

The heat required to raise mass m by ΔT is:

$\Delta Q = mc\Delta T$

For gases, c_p (constant pressure) differs from c_v (constant volume). Mayer's equation relates them:

$C_p - C_v = R$

The heat capacity ratio is γ = C_p/C_v:

• Monoatomic: γ = 5/3 ≈ 1.67
• Diatomic: γ = 7/5 = 1.4

3.2 Thermodynamic Processes

Question: A copper rod 2 m long with radius 1 cm has one end maintained at 250°C. At steady state, heat flows at 2.1 Js⁻¹. What is the temperature of the other end? (k_copper = 380 Js⁻¹m⁻¹°C⁻¹)

Solution

Cross-sectional area: A = πr² = π(0.01 m)² = 3.14 × 10⁻⁴ m²

From the conduction equation:

$\frac{dQ}{dt} = kA\frac{(\theta_1 - \theta_2)}{L}$

Solving for θ₂:

$\theta_2 = \theta_1 - \frac{(dQ/dt)L}{kA}$

$\theta_2 = 250°C - \frac{2.1 \text{ Js}^{-1} \times 2 \text{ m}}{380 \text{ Js}^{-1}\text{m}^{-1}\text{°C}^{-1} \times 3.14 \times 10^{-4} \text{ m}^2}$

$\theta_2 = 250°C - 35.3°C = 214.7°C$

The temperature at the other end is approximately 215°C.

In Tanzanian households, thermal properties of materials are applied when cooking using sufuria (cooking pots). Aluminum pots conduct heat efficiently from the flame to food, while wooden handles use low thermal conductivity to protect hands from burns. Understanding heat transfer also helps in designing proper storage for perishable foods like fish and vegetables in hot climates—using insulated containers or placing ice blocks wrapped in cloth slows heat gain through conduction and convection, keeping food fresher for market sales.