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    Heat Transfer

    Flashcards for topic Heat Transfer

    Intermediate39 cardsphysics

    Preview Cards

    Card 1

    Front

    How does the human body perceive thermal comfort, and why might two materials at identical temperatures create different sensory experiences?

    Back

    The human perception of thermal comfort depends on several factors:

    • Thermal receptors in skin respond primarily to rate of heat transfer, not absolute temperature
    • Materials in contact with skin create different experiences based on:
      • Thermal conductivity (rate of heat transfer)
      • Specific heat capacity (energy needed to change temperature)
      • Density (amount of material in contact with skin)
    • The product of these properties (thermal effusivity) determines initial tactile sensation
    • Higher thermal effusivity materials feel colder when below body temperature and hotter when above body temperature
    • Skin temperature changes of just 1-2°C from normal can trigger discomfort sensations

    Example: Standing on a marble floor (high conductivity) versus carpet (low conductivity) at identical 18°C temperatures creates different comfort levels because marble rapidly conducts heat from feet, triggering cold receptors, while carpet allows a thin layer of warm air to remain near the skin.

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    Card 2

    Front

    How does Searle's apparatus measure the thermal conductivity of a solid material?

    Back

    Searle's apparatus measures thermal conductivity through:

    1. A cylindrical rod of the test material is arranged with one end in a steam chamber (heat source) and the other end wrapped with a copper tube carrying water (heat sink)
    2. Two thermometers (T₁ and T₂) measure the temperature gradient in the rod at known distance x apart
    3. Two more thermometers (T₃ and T₄) measure the temperature of outgoing and incoming water
    4. The thermal conductivity K is calculated using: K = (x·m·s·(θ₃-θ₄))/(A·(θ₁-θ₂)·t)

    Where:

    • x = distance between temperature measuring points
    • m = mass of water collected in time t
    • s = specific heat capacity of water
    • A = cross-sectional area of the rod
    • (θ₁-θ₂) = temperature difference between points in the rod
    • (θ₃-θ₄) = temperature difference between outgoing and incoming water

    This method works by establishing a steady-state heat flow through the rod.

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    Card 3

    Front

    How is the solid angle (Δω) used in calculating radiative heat transfer, and why is directional analysis important when quantifying thermal radiation?

    Back

    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: Utotal=∫A∫ΩE(θ,ϕ)dωdAU_{total} = \int_{A}\int_{Ω}E(\theta,\phi)dωdAUtotal​=∫A​∫Ω​E(θ,ϕ)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.

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    Card 4

    Front

    What does Wien's displacement law state about the relationship between temperature and the peak wavelength of thermal radiation?

    Back

    Wien's displacement law states that the wavelength of maximum intensity (λₘ) in thermal radiation is inversely proportional to the absolute temperature (T) of the blackbody:

    λmT=bλₘT = bλm​T=b

    Where:

    • b is Wien's constant (0.288 cm·K for a blackbody)
    • λₘ is the wavelength of peak emission
    • T is the absolute temperature in Kelvin

    This means:

    • Higher temperature bodies have peak emissions at shorter wavelengths
    • Lower temperature bodies have peak emissions at longer wavelengths
    • The product of peak wavelength and temperature remains constant

    Example: The sun's surface (≈6000K) has peak emission in visible light (≈500nm), while human body temperature (310K) emits primarily in infrared (≈9.3μm).

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    Card 5

    Front

    If the dominant wavelength of radiation from a star is measured to be half that of our Sun, what can you determine about the star's surface temperature relative to our Sun?

    Back

    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.

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    Card 6

    Front

    Why is emissivity numerically equal to absorptive power for any surface, and how does this relate to the Stefan-Boltzmann law?

    Back

    Emissivity equals absorptive power because:

    1. From Kirchhoff's Law: E/a = Eₒ (where Eₒ is blackbody emissive power)
    2. For a non-blackbody: E = e·Eₒ (where e is emissivity)
    3. Substituting: (e·Eₒ)/a = Eₒ
    4. Therefore: e = a

    In the Stefan-Boltzmann law, this relationship appears as:

    • For a blackbody: u = σAT⁴ (energy emitted per unit time)
    • For a non-blackbody: u = eσAT⁴ (where e is emissivity)

    Practical implications:

    • A surface that absorbs 70% of incident radiation (a=0.7) will emit 70% as efficiently as a blackbody (e=0.7)
    • When calculating radiative heat exchange, e dictates both emission and absorption rates
    • For thermal equilibrium in an enclosure, a body with low absorptivity must have correspondingly low emissivity

    Example: Reflective emergency blankets have low absorptivity and low emissivity, helping maintain body temperature by minimizing both heat absorption and emission.

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    Card 7

    Front

    Explain the Seebeck effect and how it's applied in a thermocouple.

    Back

    The Seebeck effect is the generation of an electromotive force (voltage) when two dissimilar metals form junctions that are maintained at different temperatures.

    In a thermocouple:

    • Two different metals (A and B) form two junctions (J₁ and J₂)
    • A galvanometer connects between the junctions through one metal
    • When junction temperatures are equal → no deflection
    • When temperatures differ → galvanometer deflects
    • The magnitude of deflection correlates with the temperature difference

    This allows thermocouples to convert temperature differences directly into measurable electrical signals.

    Card 8

    Front

    How does a bolometer's sensitivity change when placed in a vacuum, and why?

    Back

    A bolometer's sensitivity increases when placed in a vacuum (enclosed in an evacuated glass bulb) because:

    1. Elimination of convective heat loss - heat cannot be carried away by air molecules
    2. Reduction of conductive heat loss - fewer gas molecules to transport heat energy
    3. Improved thermal isolation - heat remains concentrated in the sensing elements
    4. Lower thermal equilibrium time - faster response to radiation changes
    5. Reduced noise from ambient temperature fluctuations

    This vacuum environment ensures that temperature changes in the resistive elements are primarily due to the radiation being measured rather than being diminished by environmental heat transfer mechanisms.

    Card 9

    Front

    Explain the physical principle behind a thermopile's operation and the specific function of each component in the radiation detection process.

    Back

    A thermopile operates on the Seebeck effect principle:

    1. Physical Principle: When junctions of dissimilar metals are at different temperatures, an electromotive force (voltage) is generated proportional to the temperature difference.

    2. Component Functions:

      • Bismuth and Antimony metals: Chosen for their high thermoelectric coefficient
      • Hot junctions: Absorb incoming radiation, converting it to thermal energy
      • Blackened surface: Maximizes radiation absorption efficiency
      • Cold junctions: Maintained at reference temperature to create temperature differential
      • Metallic cover: Shields cold junctions from radiation to maintain temperature gradient
      • Metallic cone (often included): Concentrates radiation onto the hot junctions
      • Galvanometer: Measures the current generated, indicating radiation intensity
    3. Process sequence:

      • Radiation → absorption at hot junctions → temperature increase → voltage generation → galvanometer deflection

    The sensitivity is proportional to the number of thermocouple junctions in series.

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    Card 10

    Front

    Explain Kirchhoff's Law of thermal radiation, including its mathematical formulation, physical interpretation, and practical implications for different materials.

    Back

    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:

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