Thermal resistance (R) in heat conduction is defined as:

$R = \frac{x}{KA}$

Where:

• x is the length of the conducting material
• K is the thermal conductivity of the material
• A is the cross-sectional area

This creates a thermal version of Ohm's law:

$\frac{Q}{t} = \frac{T_1 - T_2}{R}$

Analogy to electrical circuits:

• Temperature difference (T₁-T₂) is analogous to voltage difference (V)
• Heat current (Q/t) is analogous to electrical current (I)
• Thermal resistance (R) is analogous to electrical resistance (R)
• Materials with high thermal conductivity have low thermal resistance, similar to how good electrical conductors have low electrical resistance

Note: This analogy allows thermal conduction problems to be solved using techniques from electrical circuit analysis.

For two thermal conductors in parallel, the equivalent thermal resistance R is given by:

$\frac{1}{R} = \frac{1}{R₁} + \frac{1}{R₂}$

This follows the same mathematical form as parallel electrical resistors. The parallel arrangement always results in a lower equivalent thermal resistance than either individual resistance.

Example: If two rods with thermal resistances of 10 K/W and 15 K/W are connected in parallel, the equivalent thermal resistance would be: $\frac{1}{R} = \frac{1}{10} + \frac{1}{15} = 0.1 + 0.067 = 0.167$ $R = \frac{1}{0.167} = 6 \text{ K/W}$

A cavity-based blackbody works through multiple internal reflections that trap incoming radiation:

• When radiation enters through a small hole in an enclosed cavity, it undergoes multiple reflections inside • With each reflection, a portion of radiation is absorbed by the internal walls • The probability of radiation escaping back through the small entrance hole becomes extremely small • This results in nearly 100% absorption of incident radiation, regardless of the actual material properties of the cavity walls

The cavity design effectively creates an absorption coefficient approaching 1.0, making it the closest practical implementation of a theoretical perfect blackbody.

Note: This principle is used in laboratory blackbody standards for calibrating radiation measurement instruments.

A blackbody cavity's design relates to Kirchhoff's law through:

• Kirchhoff's law states: The ratio of emissive power (E) to absorptive power (a) equals the emissive power of a blackbody at the same temperature • The cavity design creates an absorptive power approaching 1.0 • By Kirchhoff's law, this maximizes emissive power for any given temperature • The cavity will therefore emit radiation with a spectral distribution following Planck's radiation law perfectly

Used as laboratory standards because:

1. They provide reproducible, theoretically predictable radiation patterns
2. The radiation emitted is independent of the cavity material properties
3. They exhibit precise temperature-radiation relationships following Stefan-Boltzmann law (E = σT⁴)
4. They create reference spectra following Wien's displacement law (λₘT = constant)

This makes cavity blackbodies ideal for calibrating spectrometers, pyrometers, and other radiation measuring instruments with high precision.

The solid angle (Δω) is critical in radiative heat transfer calculations because:

• Radiation is emitted in three-dimensional space with varying intensity in different directions • The solid angle measures a conical section of this 3D space (measured in steradians) • For most real surfaces, radiation intensity varies with direction (non-Lambertian surfaces)

Importance of directional analysis:

1. Radiation is not uniform in all directions for real surfaces
2. The intensity often follows a directional distribution (e.g., cosine law for diffuse emitters)
3. When calculating total emitted energy, integration over all solid angles is necessary: $U_{total} = \int_{A}\int_{Ω}E(\theta,\phi)dωdA$

Example application: When designing thermal shields for spacecraft, engineers must account for the directional distribution of thermal radiation to properly protect sensitive components from heat sources that emit radiation preferentially in certain directions.

If a star's dominant wavelength (λₘ) is half that of our Sun, its temperature must be twice as high.

Analysis:

• According to Wien's displacement law: λₘ·T = constant (b)
• For both stars, we know: λₘ₁·T₁ = λₘ₂·T₂ = b
• Given: λₘ₂ = ½λₘ₁
• Solving for T₂: T₂ = T₁·(λₘ₁/λₘ₂) = T₁·(λₘ₁/(½λₘ₁)) = 2T₁

Therefore:

• The star's surface temperature is exactly twice that of our Sun
• If the Sun's surface temperature is approximately 5800K, the other star's temperature would be about 11,600K
• This hotter star would appear more blue-white compared to our yellow Sun
• The star would emit significantly more total energy (16× more) according to Stefan-Boltzmann law (T⁴)

Note: Stellar color is a reliable indicator of surface temperature following Wien's law.

Starting with heat transfer equation: Q = KA(θ₁-θ₂)t/x

For ice melting experiment:

1. Heat transferred equals latent heat absorbed: Q = mL (where m = mass of ice melted, L = latent heat of fusion)

2. Substituting: mL = KA(θ₁-θ₂)t/x

3. Solving for K: K = (mL·x)/(A·(θ₁-θ₂)·t)

Where:

• m = mass of ice melted (kg)
• L = latent heat of fusion (J/kg)
• x = thickness of conducting slab (m)
• A = surface area (m²)
• (θ₁-θ₂) = temperature difference (°C)
• t = time (s)

This derivation shows how to experimentally determine thermal conductivity by measuring ice melted over time.

Step 1: Calculate total surface area A = 2(60×60 + 60×30 + 60×30) cm² = 1.44 m²

Step 2: Calculate rate of heat flow ΔQ/Δt = KA(θ₁-θ₂)/x = (0.04 W/m·°C)(1.44 m²)(40°C)/(0.015 m) = 154 W

Step 3: Calculate melting rate Rate = (Heat flow)/(Latent heat) = 154 W/(3.36×10⁵ J/kg) = 0.46 g/s

This represents the steady-state rate at which ice melts due to heat conduction through the container walls.

Reducing ambient pressure would primarily affect heat transfer through:

1. Conduction (primary mechanism in the experiment):

   • Minimal direct effect on conduction through solid stone
   • Heat transfer equation Q = KA(θ₁-θ₂)t/x remains unchanged
   • No significant change to melting rate through this mechanism

2. Convection (secondary effect):

   • Reduced at lower pressures due to fewer gas molecules
   • Would decrease heat transfer from ambient air to the sides of the slab
   • Might slightly decrease the overall melting rate

3. Radiation (minimal contribution):

   • Unaffected by pressure changes
   • Continues at the same rate

Quantitatively, if assuming a 10% contribution from convection to the total heat transfer, reducing pressure to near-vacuum would decrease the ice melting rate by approximately 10%, from 4.8g/hour to about 4.3g/hour (exact values would depend on the specific pressure reduction and system geometry).

Kirchhoff's Law of Thermal Radiation

Mathematical Formulation

Kirchhoff's Law states that the ratio of emissive power (E) to absorptive power (a) is constant for all bodies at a given temperature and equals the emissive power of a blackbody:

$\frac{E(\text{body})}{a(\text{body})} = E(\text{blackbody})$

Physical Interpretation

• Fundamental principle: Good absorbers are good emitters; poor absorbers are poor emitters
• Thermodynamic basis: Derived from principles of thermodynamic equilibrium
• Blackbody reference: A blackbody (a=1) represents the theoretical maximum emitter

Material Implications

1. High emissivity materials (dark, matte surfaces):

   • High absorptive power (~0.9)
   • Correspondingly high emissive power
   • Example: Black-painted surfaces, carbon black

2. Low emissivity materials (reflective surfaces):

   • Low absorptive power (~0.1)
   • Correspondingly low emissive power
   • Example: Polished metals, silver surfaces

3. Theoretical limits:

   • Perfect reflector (a=0): Cannot emit radiation (E=0)
   • Perfect blackbody (a=1): Maximum possible emission at any temperature

Practical Applications

• Thermal insulation design (reflective barriers)
• Radiator and heat sink engineering
• Solar collector optimization
• Infrared temperature measurement calibration
• Spacecraft thermal control systems