Heat Transfer — Set 2

Correct Answer: A. Argon

• **Argon** = Argon gas is used between double-glazed window panes because it has lower thermal conductivity than air and reduces both conduction and convection. • **k(Argon) < k(Air)** — Argon is denser and less conductive than air, making it a better thermal insulator; it also inhibits convection currents in the gap. • 💡 Wrong-option analysis: Hydrogen: highly flammable and dangerous for home use; Oxygen: supports combustion and is not used for this purpose; Chlorine: toxic and corrosive, not used in windows.

Correct Answer: A. difference between body temperature and surroundings

• **Difference between body temperature and surroundings** = Newton's law of cooling states: dT/dt ∝ −(T − Tₛ) — the rate of cooling is proportional to the temperature excess of the body over its surroundings. • **dT/dt ∝ −(T − Tₛ)** — This means a hotter object cools faster initially; as its temperature approaches the surroundings, the cooling rate decreases. • 💡 Wrong-option analysis: Mass of the body only: mass affects how much heat is stored but is not what Newton's law is proportional to; Volume of the body only: volume is not the stated proportionality; Square of temperature: Stefan-Boltzmann law involves T⁴ for radiation, not Newton's cooling law.

Correct Answer: B. conduction and convection

• **Conduction and convection** = The vacuum in a thermos flask eliminates almost all material medium, stopping both conduction (no particles to collide) and convection (no fluid to move). • **Silvered walls** — The silvered inner surfaces of the thermos also reduce radiation heat loss by reflecting it back, addressing the third mode. • 💡 Wrong-option analysis: Convection only: vacuum stops both conduction and convection, not just convection; Radiation only: radiation is reduced by silvering, not by vacuum; Conduction only: vacuum stops both conduction and convection.

Correct Answer: B. Air (still air)

• **Air (still air)** = Still air is an excellent thermal insulator with very low thermal conductivity (~0.026 W m⁻¹ K⁻¹); it also prevents convection when trapped in small spaces. • **k(still air) ≈ 0.026 W m⁻¹ K⁻¹** — This is why wool, foam, and double-glazed windows work well — they trap still air to reduce heat transfer. • 💡 Wrong-option analysis: Copper: an excellent conductor (k ≈ 400 W m⁻¹ K⁻¹), not an insulator; Iron: a good conductor (k ≈ 80 W m⁻¹ K⁻¹); Aluminium: also a good conductor (k ≈ 205 W m⁻¹ K⁻¹).

Correct Answer: C. Increasing area of cross-section

• **Increasing area of cross-section** = Fourier's law: Q/t = kAΔT/L — heat flow rate is directly proportional to the cross-sectional area A; larger area provides more pathways for heat. • **Q/t ∝ A** — Doubling the cross-sectional area doubles the heat flow rate for the same material and temperature difference. • 💡 Wrong-option analysis: Decreasing temperature difference: reduces the driving force for conduction, decreasing heat flow; Using vacuum instead of solid: vacuum has no material for conduction to occur; Increasing thickness: increases the path length, reducing the heat flow rate (Q/t ∝ 1/L).

Correct Answer: B. T⁴

• **T⁴** = The Stefan-Boltzmann law states P = σAT⁴ for an ideal black body — radiated power is proportional to the fourth power of absolute temperature. • **σ = 5.67×10⁻⁸ W m⁻² K⁻⁴** — This Stefan-Boltzmann constant links radiation power to temperature; doubling T increases radiated power 16-fold. • 💡 Wrong-option analysis: T³: not the correct power in the Stefan-Boltzmann law; T²: also incorrect — the law requires T⁴; T: linear dependence would imply proportional heat radiation, which is too weak.

Correct Answer: B. 1

• **1** = An ideal black body has emissivity ε = 1 — it emits the maximum possible thermal radiation at a given temperature, which defines the upper limit. • **ε = 1 (black body)** — Real surfaces have 0 < ε < 1; emissivity 1 is the theoretical ideal; a perfect reflector has ε = 0. • 💡 Wrong-option analysis: 0: emissivity 0 describes a perfect reflector that emits nothing; Depends on thickness only: emissivity is a surface property, not primarily determined by thickness; Greater than 1: emissivity cannot exceed 1 — that is physically impossible.

Correct Answer: B. emitters of heat radiation

• **Emitters of heat radiation** = Black surfaces have high emissivity (close to 1), making them superior emitters of thermal radiation — this is why a black-painted radiator emits more heat. • **Kirchhoff's law** — Good absorbers are equally good emitters; a radiator painted black both absorbs and emits thermal radiation more effectively than a shiny white one. • 💡 Wrong-option analysis: Electrical conductors: color has no direct effect on electrical conductivity; Reflectors: black surfaces are poor reflectors; Insulators: black paint does not make a material a thermal or electrical insulator.

Correct Answer: D. Metals conduct heat quickly to food

• **Metals conduct heat quickly to food** = Metal bottoms in cooking utensils have high thermal conductivity, rapidly transferring heat from the flame to the food uniformly. • **k(copper) ≈ 400, k(Al) ≈ 205 W m⁻¹ K⁻¹** — These high conductivities ensure fast, even heat distribution across the pan bottom. • 💡 Wrong-option analysis: Metals prevent convection: metals do not prevent convection; they conduct heat; Metals are poor conductors: this is the opposite — metals are excellent heat conductors; Metals stop radiation: metals do not stop radiation; a shiny metal surface may reflect it, but that is not why utensils have metal bottoms.

Correct Answer: D. Covering with a thick insulating layer

• **Covering with a thick insulating layer** = A thick insulating layer (low k) reduces heat loss by conduction, convection, and radiation simultaneously — the most comprehensive reduction. • **R = L/(kA)** — Thermal resistance R increases with thickness L and decreases with conductivity k; thick insulation maximises R and minimises heat loss. • 💡 Wrong-option analysis: Using a fan: a fan increases convection, raising heat loss; Painting it white and making it shiny: reduces radiation only, not conduction or convection; Increasing surface area: a larger area increases both convection and radiation heat loss.