Question bank — Heat transfer — conduction (Fourier's law k), convection, radiation (Stefan-Boltzmann σT⁴)

Sensation mein True, actually mein nahi — dono room temperature par hain. Metal ka thermal conductivity $k$ zyada hota hai isliye woh tumhare haath se heat bahar zyada tezi se conduct karta hai, jise tumhare nerves "thanda" padhte hain; tum heat-flow rate $H$ feel karte ho, temperature nahi.

Akele mein False, madad se True. Apne aap chhodne par, heat sirf hot→cold flow karti hai (Second law of thermodynamics); ek fridge ya heat pump ise doosri taraf move kar sakta hai, lekin sirf external work karke.

True. $H=kA\Delta T/\Delta x$ mein, thickness $\Delta x$ denominator mein hai, isliye double $\Delta x$ → aadha $H$. Physically, heat molecular collisions ki relay se cross karti hai; do guni moti wall matlab do guni chain of hand-offs aur do guni total resistance $R=\Delta x/(kA)$, isliye flow aadha ho jaata hai.

True — yeh thermal resistances parallel mein hain, aur parallel paths conductances ($1/R$) jodte hain, isliye total conductance double hoti hai aur $H$ same $\Delta T$ ke liye double hota hai.

False. "Black body" ka matlab hai woh saari incident radiation absorb karta hai aur best possible emitter hai; Sun almost ek black body hai phir bhi white-hot glows karta hai. Colour ek hot spectrum ke visible slice ke baare mein hai — dekho Black body radiation.

False. Kisi bhi $T>0$ K ke liye emission kabhi nahi rukti; thermal equilibrium par body $e\sigma A{T}^4$ emit karti hai aur identical $e\sigma A{T}_0^4$ absorb karti hai, isliye net exchange $P_{\text{net}}=e\sigma A(T^4-T_0^4)$ zero hoti hai jabki dono flows jaari rehti hain.

True — yeh Kirchhoff's law hai: ek given wavelength par emissivity $e$ absorptivity ke barabar hoti hai, isliye ek matte black surface dono efficiently soakta aur radiate karta hai, jabki shiny silver dono mein se koi bhi achha nahi karta.

False. Convection ke liye bulk flow of a fluid chahiye; ek rigid solid circulate nahi kar sakta, isliye ek metal bar heat sirf conduction se move karta hai. Note: us solid ke andar, heat abhi bhi microscopically travel karti hai — lattice vibrations (phonons) aur free-electron drift ke zariye — lekin woh hand-off conduction hai, convection nahi, kyunki koi matter bulk mein move nahi karta.

True. Agar ek layer mein zyada heat daakhil hoti aur kam nikalti, to energy pile up hoti aur temperatures change hoti rehti — jo "steady" ko contradict karti. Isliye $H$ har series layer se identical hai, jaise series resistors mein ek current.

Ek hi root error: $T$ absolute (kelvin) hona chahiye, Celsius nahi. Sahi value hai $5800^{\circ }\text{C}=6073$ K. Kyunki $T^4$ absolute temperature ko power tak uthata hai, $5800$ (ek Celsius number) plug karna "close enough" nahi hai — yeh bilkul alag quantity hai aur power ko underestimate karta hai.

Galat — minus hatana direction tod deta hai. Heat gradient ke neeche flow karti hai (lower $T$ ki taraf), isliye jahan $dT/dx<0$ hai wahan minus current $H$ ko positive banata hai; iske bina tum predict karte ki heat uphill chadh rahi hai, jo Second law of thermodynamics violate karta hai. Isse sirf tab chhodo jab deliberately magnitude $\Delta T=T_{\text{hot}}-T_{\text{cold}}$ se compute kar rahe ho.

Galat — woh conductances jodhta hai, jo sirf parallel paths ke liye apply hota hai. Series mein same current $H$ ko dono slabs cross karni padti hai, isliye unki resistances jodti hain: $R_{\text{tot}}=\Delta x_1/(k_{1}A)+\Delta x_2/(k_{2}A)$, phir $H=\Delta T/R_{\text{tot}}$. Yahan conductances jodhna ek moti insulator ko magically flow badhane deta, jo bakwaas hai.

Vacuum medium hatata hai, isliye yeh conduction aur convection ko khatam karta hai, dono ko matter chahiye — radiation ko yeh kuch nahi karta, jo waves hain aur vacuum mein freely cross karte hain. Radiation ek alag trick se roki jaati hai: silvered walls low emissivity $e$ ke saath. Dono features ko confuse karna miss karta hai ki thermos ko dono kyun chahiye.

Bahut zyada galat. Power $\propto T^4$, isliye factor hai $(600/300{)}^4=2^4=16$, $2$ nahi. Trap yeh hai ki fourth-power law ko linear ki tarah treat karna; Stefan–Boltzmann ka poora point yahi hai ki chhote temperature rises huge power jumps dete hain.

Zaroori nahi — yeh ek difference $T_s-T_{\text{fluid}}$ use karta hai, aur difference °C aur K mein numerically identical hota hai ($273$ offsets cancel ho jaate hain). Kelvin sirf tab zaruri hota hai jab tum ek absolute temperature ko power tak uthao, jaise $T^4$ mein.

Misleading — insulation depend karta hai resistance $R=\Delta x/(kA)$ par, sirf $k$ par nahi. Ek pane bahut patla hota hai ($\Delta x$ tiny), jo $R$ ko shrink karta hai aur $H=\Delta T/R$ ko bada banata hai. Asli insulation double glazing mein trapped air gap se aati hai, jahan $k_{\text{air}}$ bahut kam hota hai — dekho Thermal conductivity k.

Yeh Planck's law (blackbody spectrum) ko saari wavelengths par integrate karne se aata hai; us curve ke neeche ka total area $T^4$ ke saath scale karta hai. Yeh haath se choose nahi kiya gaya — yeh Black body radiation ki physics se nikalta hai.

Radiation electromagnetic waves hain jinhein koi medium nahi chahiye, isliye woh space ke vacuum mein cross karti hain; sound ek pressure wave hai jise travel karne ke liye matter chahiye, aur space mein essentially kuch nahi hai.

Still air ka $k\approx 0.025$ bahut chhota hai, isliye pure conduction bahut dheemi hoti hai. Convection physically hot air packets ko room mein buoyant circulation ke zariye transport karta hai, energy ko molecule-to-molecule hand-off se kahin zyada tezi se move karta hai.

Dono ek hi linear "flow = drive / resistance" structure follow karte hain, isliye Electrical resistance Ohm's law ke series/parallel combination rules directly carry over karte hain, jo tumhe composite walls ke baare mein bina kuch naye derivation ke reason karne deta hai.

Atmosphere Sun ki short-wavelength radiation ke liye transparent hai lekin Earth ki long-wavelength (infrared) re-emission ke liye partly opaque hai, outgoing energy trap karta hai aur equilibrium temperature badhata hai — yeh Greenhouse effect ke peeche ka mechanism hai.

Jab $T\gg T_0$ hota hai, $T_0^4$ term orders of magnitude chhoti hoti hai (jaise $300^4$ lagbhag 4800× chhota hai $2500^4$ se), isliye woh answer ko barely shift karti hai — ek legitimate 80/20 estimate, physics ka rule nahi.

Ek matte black surface ki high emissivity $e$ near 1 hoti hai, radiated power $P=e\sigma A{T}^4$ maximize karti hai; ek shiny low-$e$ surface poorly radiate karti aur kam heat shed karti.

Zero. $P=e\sigma A{T}^4$ mein $T=0$ hone par $P=0$ milta hai. Yeh ek limiting/degenerate case hai — charges ki koi thermal agitation nahi matlab koi thermal radiation nahi.

Haan — woh $e\sigma A{T}^4$ emit karti hai aur barabar amount absorb karti hai, isliye emission kabhi nahi rukti; sirf net flow $P_{\text{net}}=e\sigma A(T^4-T_0^4)=0$ equilibrium par khatam hoti hai.

Steady state mein ek single uniform rod ke liye nahi — nonzero $\Delta T$ nonzero $H$ force karta hai. Lekin agar $k\to 0$ (perfect insulator) ya $A\to 0$, to $H\to 0$; ek near-vacuum gap conduction ko completely khatam karke ise approximate karta hai.

$H=kA\Delta T/\Delta x\to \infty$. Infinitely patli wall zero resistance offer karti hai, isliye heat freely pass hoti hai — woh mathematical limit jo explain karta hai ki kyun patle, high-$k$ contacts (jaise ek metal chammach) itne effective lagte hain.

Zero — $P_{\text{net}}=0\cdot \sigma A(T^4-T_0^4)=0$. Woh na radiation emit karta hai na absorb, isliye woh sirf conduction ya convection se heat exchange kar sakta hai.

Haan — jab $T<T_0$ (body surroundings se thandi), $T^4-T_0^4<0$, isliye $P_{\text{net}}<0$: body balance mein radiant energy gain karti hai, jaise ek warm room mein ek thanda object warm ho raha ho.