Heat Transfer — Set 4

Correct Answer: B. Q/t = kA(ΔT/L)

• **Q/t = kA(ΔT/L)** = Fourier's law for one-dimensional steady conduction through a slab: heat flow rate = thermal conductivity × area × temperature difference / thickness. • **k, A, ΔT, L** — k is thermal conductivity (W m⁻¹ K⁻¹), A is cross-sectional area (m²), ΔT is temperature difference (K), L is thickness (m). • 💡 Wrong-option analysis: Q/t = σAT⁴: this is the Stefan-Boltzmann radiation law, not Fourier's conduction law; Q/t = hAΔT: this is Newton's law of cooling for convection; Q/t = mcΔT: this gives total heat stored/released, not the rate of heat flow.

Correct Answer: B. dT/dt ∝ -(T - Ts)

• **dT/dt ∝ −(T − Tₛ)** = Newton's law of cooling: the rate of temperature change is proportional to the negative of the excess temperature over the surroundings (Tₛ). • **Exponential decay** — This law leads to an exponential temperature decay: T(t) = Tₛ + (T₀ − Tₛ)e^(−kt), where k is a cooling constant. • 💡 Wrong-option analysis: P = σAT⁴: Stefan-Boltzmann law for radiation; v = c/n: speed of light in a medium (optics); Q = mcΔT: formula for heat stored, not rate of cooling.

Correct Answer: D. Convection

• **Convection** = When fluid is heated from below, it expands and becomes less dense; buoyancy forces drive warm fluid upward and cool fluid downward, making convection dominant. • **Natural convection** — This density-driven circulation is responsible for weather patterns, ocean currents, and the movement of molten rock in Earth's mantle. • 💡 Wrong-option analysis: No heat transfer occurs: heat transfer definitely occurs when heated from below; Radiation: radiation is not dominant for this scenario involving a fluid under gravity; Conduction: while conduction occurs, convection is far more effective in fluids heated from below.

Correct Answer: C. P ∝ T⁴

• **P ∝ T⁴** = The Stefan-Boltzmann law: 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** — A small rise in temperature causes a large increase in radiated power; doubling T increases P by a factor of 16. • 💡 Wrong-option analysis: P ∝ T³: not the Stefan-Boltzmann relationship; P ∝ T²: Wien's displacement law relates peak wavelength to T (λ_max ∝ 1/T), not T²; P ∝ T: linear dependence significantly underestimates radiation at high temperatures.

Correct Answer: C. Wool traps still air and reduces heat loss

• **Wool traps still air and reduces heat loss** = Wool fibres trap tiny pockets of still air — still air is a poor conductor and prevents convection, reducing heat loss from the body. • **k(still air) ≈ 0.026 W m⁻¹ K⁻¹** — The insulating effect comes from trapped still air, not the wool itself; many insulators (foam, feathers, fur) work the same way. • 💡 Wrong-option analysis: Wool increases radiation from the body: wool reduces heat loss, it does not increase radiation; Wool generates heat by chemical reaction: wool does not undergo exothermic reactions to produce heat; Wool increases convection strongly: wool blocks convection by trapping still air.

Correct Answer: B. They are poor conductors of heat

• **They are poor conductors of heat** = Wood and plastic have very low thermal conductivity (~0.1–0.3 W m⁻¹ K⁻¹), so very little heat reaches the hand from the hot pan. • **k(wood) ≈ 0.1–0.2 W m⁻¹ K⁻¹** — Compare this to k(steel) ≈ 50 W m⁻¹ K⁻¹; wood conducts ~250–500 times less heat than steel, making it safe to hold. • 💡 Wrong-option analysis: They are good conductors: good conductors would burn the hand; They are transparent: transparency has nothing to do with thermal insulation; They increase heat radiation: wood and plastic do not increase radiation — they simply insulate by poor conduction.

Correct Answer: C. higher emissivity

• **Higher emissivity** = A black or dull surface has emissivity closer to 1 (the ideal black body value), meaning it emits more thermal radiation per unit area at any temperature than a white or shiny surface. • **Kirchhoff's law** — Emissivity = absorptivity at thermal equilibrium; dull black surfaces absorb more radiation and also emit more. • 💡 Wrong-option analysis: Lower emissivity: that describes a white or shiny surface; Emissivity greater than 1 always: emissivity cannot exceed 1 for any real surface; Zero emissivity: zero emissivity describes a perfect reflector that emits no radiation.

Correct Answer: B. It reflects all incident radiation

• **It reflects all incident radiation** = A perfect reflector has reflectivity = 1 (absorptivity = 0, emissivity = 0) — it reflects every photon of incident radiation and emits none. • **ε = 0 (perfect reflector)** — By Kirchhoff's law, a perfect reflector is also a perfect non-emitter; a polished silver mirror approximates this. • 💡 Wrong-option analysis: It has emissivity equal to 1: emissivity 1 describes a perfect absorber/emitter (black body), not a reflector; It absorbs all incident radiation: that describes a perfect black body; It emits maximum radiation: maximum emission is a property of an ideal black body, not a reflector.

Correct Answer: B. 80% of black body

• **80% of black body** = Emissivity ε is defined as the ratio of a surface's emitted power to the black body's emitted power at the same temperature; ε = 0.8 means 80% of black body radiation. • **P = εσAT⁴** — For emissivity 0.8, P_surface = 0.8 × P_black body; the surface emits 80% of the maximum possible radiation. • 💡 Wrong-option analysis: Greater than black body: emissivity cannot exceed 1, so no surface can emit more than a black body; Equal to black body: only if ε = 1; 0% of black body: that would require ε = 0 (a perfect reflector).

Correct Answer: A. Feeling heat from a fire without touching it

• **Feeling heat from a fire without touching it** = When you feel warmth from a fire at a distance without physical contact, the heat reaches you by thermal radiation (infrared electromagnetic waves) through the air. • **Infrared radiation** — Fire emits significant infrared radiation that travels through air and is absorbed by skin, causing the sensation of warmth even without direct contact. • 💡 Wrong-option analysis: Heating a metal bar from one end: this is conduction — heat travels through the solid metal; Heating water in a pot by stirring: stirring promotes convection inside the water; Cooling by a fan only: a fan drives forced convection, removing heat by fluid motion.