ARCHETYPE: Cooling & Radiation Numericals

Take two iron balls out of the same furnace at the same white heat — one the size of a marble, one the size of a football — and set them side by side on the anvil. Watch. The football glows on, sullen and orange, for minutes. The marble goes dark almost while you are looking at it. Yet the football is radiating far more power than the marble at every instant. How can the body that pours out more energy be the body that stays hot? That question is the entire archetype, and examiners across CBSE, ICSE, JEE and NEET have been asking it in one costume or another for forty years. The answer is that power and cooling rate are different quantities. Power follows Stefan's law, $P = e\sigma A(T^4 - T_s^4)$ , and grows with the surface area — so the big ball wins. But the rate at which the temperature falls is that power divided by the heat capacity $mc$ , and mass grows with the volume — so the big ball loses. Area over volume goes as $1/r$ , and the small body cools faster. Always. This article is the numerical workshop for that idea. We fix one geometry — a sphere of radius $r$ — and grind out every question a paper can build on it: the raw Stefan power, the net exchange with cooler walls, the rate of temperature fall in kelvin per second, the ratio problems where you must never substitute a single number, and the star whose colour and brightness fall out of Wien and Stefan together.

Key takeaways

The engine of every number here is the net Stefan power $P = e\sigma A(T^4 - T_s^4)$ , $\sigma = 5.67\times10^{-8}\ \mathrm{W\,m^{-2}K^{-4}}$ , $T$ in kelvin. Geometry sets $A$ , and $A$ sets the exponents: for a sphere $P \propto r^2T^4$ .
Divide that power by the heat capacity $mc$ and you get the rate of temperature fall, $\left|\dfrac{\Delta T}{\Delta t}\right| = \dfrac{e\sigma A (T^4 - T_s^4)}{mc}$ — an answer in kelvin per second.
For a solid sphere $\dfrac{A}{m} = \dfrac{3}{\rho r}$ , so the cooling rate goes as $\dfrac{1}{r}$ : halve the radius and the body cools twice as fast. The most-tested line in the chapter.
Ratio questions carry no calculator. Radii $1:2$ with temperatures $2:1$ give powers in the ratio $(1^2\cdot 2^4):(2^2\cdot 1^4) = 4:1$ — substitute numbers and you have already lost the marks.
Anchors: a black sphere of radius $5\ \mathrm{cm}$ has $A = 0.0314\ \mathrm{m^2}$ ; at $500\ \mathrm{K}$ it radiates $111\ \mathrm{W}$ ; against $300\ \mathrm{K}$ walls the net is $96.9\ \mathrm{W}$ ; if it is iron ( $4.12\ \mathrm{kg}$ , $c = 450$ ) it starts cooling at $0.052\ \mathrm{K/s}$ .
Key idea: radiated power follows the area and the cooling rate follows the area-to-mass ratio — so a big body radiates more watts while a small body loses temperature faster, and the whole archetype is knowing which of the two a question is asking for.

The core insight

A hot body loses energy to cooler surroundings at the net Stefan rate $P = e\sigma A(T^4 - T_s^4)$ , and its temperature therefore falls at $P/(mc)$ kelvin per second; because $A \propto r^2$ while $m \propto r^3$ , the power scales as $r^2 T^4$ but the cooling rate scales as $T^4/r$ — the same physics answering two different questions.

Where this lives today

Spacecraft thermal control. In orbit there is no air, so a satellite can shed waste heat only by radiation: radiator panels are sized straight from $P = e\sigma A(T^4 - T_s^4)$ with $T_s \approx 3\ \mathrm{K}$ , the temperature of deep space. Engineers pick surfaces by emissivity exactly as this article does — white coatings with high $e$ to dump heat, gold foil with $e$ near zero to keep it in.

Why small bodies lose heat fastest. The $1/r$ law is why a premature baby goes into an incubator and an adult does not: a smaller body carries more surface for every kilogram of warm flesh. It is why a hummingbird must eat almost continuously, and why a grinder's spark goes from white-hot to cold within a metre of flight.

Pyrometry and stars. You cannot put a thermometer in a blast furnace, so you read the radiation instead: a pyrometer runs Stefan and Wien backwards to get $T$ from emitted power or colour. Astronomers push it furthest — $L = 4\pi R^2\sigma T^4$ is how we know a red giant is enormous without ever resolving its disc.

The setup

Fix the picture once and reuse it for the rest of your life as a physics student. A body — take a sphere of radius $r$ , because that is what papers set — sits at absolute temperature $T$ inside a large room whose walls are at the lower temperature $T_s$ . Nothing touches it: no conduction, no draught. It is losing heat by radiation alone, and we want numbers. Two flows cross its surface at every instant, and confusing them is where most marks die. The sphere emits $e\sigma A T^4$ watts, set entirely by its own temperature. It simultaneously absorbs $e\sigma A T_s^4$ watts streaming in from the walls, set entirely by their temperature — and with the same $e$ , because Kirchhoff's law forces a good emitter to be an equally good absorber. What the sphere loses is the difference. That net power is the number every problem here is secretly asking for.

Figure

The net exchange — a 500 K black sphere pours out 111 W and drinks back 14 W from 300 K walls, so it loses 97 W and cools

Now the half of the setup that students skip. Knowing the power tells you how many joules per second leave the sphere — it does not tell you how fast the sphere gets colder. To convert watts into kelvin per second you must know what one kelvin costs: the body's heat capacity $mc$ , in joules per kelvin. A body with a huge $mc$ can bleed a hundred watts and barely notice. So the geometry enters twice, in opposite directions. It sets the area $A = 4\pi r^2$ , which controls how much power gets out — bigger is better. And it sets the mass $m = \rho \cdot \tfrac{4}{3}\pi r^3$ , which controls what that loss costs in temperature — bigger is worse. In an archetype the geometry is the physics, and here it is a tug-of-war between $r^2$ and $r^3$ . The winner is $1/r$ .

The master pattern

Every radiation numerical you will ever be set is one of four questions asked about that sphere. Learn which is being asked and the arithmetic is trivial; mistake one for another and no arithmetic saves you.

Question 1: how much power does it radiate? Stefan's law, applied to the body's own temperature and nothing else:

P_{\text{emit}} = e\,\sigma\,A\,T^4 \qquad (1)

where: $P_{\text{emit}}$ — power leaving the surface, in watts; $e$ — the emissivity, $0 \le e \le 1$ ( $e = 1$ for a black body); $\sigma = 5.67\times10^{-8}\ \mathrm{W\,m^{-2}K^{-4}}$ ; $A$ — the radiating surface area, $4\pi r^2$ for a sphere and $6a^2$ for a cube of side $a$ ; $T$ — the absolute temperature, in kelvin, to the fourth power. Physical reading: a body shouting into the dark, caring nothing for what is outside. Emission depends on $T$ alone.

Question 2: how much does it net lose? Put the sphere in a room at $T_s$ and it is being shouted at as well. Subtract what it drinks back:

P_{\text{net}} = e\,\sigma\,A\,(T^4 - T_s^4) \qquad (2)

where: $T_s$ — the surroundings' temperature, again in kelvin. Physical reading: this is the cooling power. Equation (1) says our $500\ \mathrm{K}$ sphere loses $111\ \mathrm{W}$ ; equation (2) says it really loses $96.9\ \mathrm{W}$ , because $14.4\ \mathrm{W}$ comes straight back from the walls. Whenever you read "cools", "loses heat", or "rate of fall of temperature", you want (2). When $T \gg T_s$ — a filament, a star — the $T_s^4$ term is negligible and (2) collapses back to (1).

Question 3: how fast does its temperature drop? Divide the cooling power by the heat capacity: watts over joules-per-kelvin is kelvin-per-second, which is exactly what "rate of cooling" means.

\left|\frac{\Delta T}{\Delta t}\right| = \frac{P_{\text{net}}}{mc} = \frac{e\,\sigma\,A\,(T^4 - T_s^4)}{m\,c} \qquad (3)

where: $m$ — the mass in kg; $c$ — the specific heat capacity in $\mathrm{J\,kg^{-1}K^{-1}}$ (iron: $450$ ); $mc$ — the heat capacity of the whole body, in joules per kelvin. Physical reading: the body is a bucket of thermal energy with a hole in it. Equation (2) is the size of the hole; $mc$ is the size of the bucket — and a large bucket with a large hole still drains slowly. Note that this is an initial rate: as $T$ falls, $T^4 - T_s^4$ collapses and the cooling slows.

Question 4: which of two bodies cools faster? Feed the sphere's geometry into equation (3): write $A = 4\pi r^2$ and $m = \rho\,\tfrac{4}{3}\pi r^3$ , and the $\pi$ 's and most of the $r$ 's cancel.

\frac{A}{m} = \frac{4\pi r^2}{\rho\,\tfrac{4}{3}\pi r^3} = \frac{3}{\rho\, r} \qquad\Longrightarrow\qquad \left|\frac{\Delta T}{\Delta t}\right| = \frac{3\,e\,\sigma\,(T^4 - T_s^4)}{\rho\, c\, r} \qquad (4)

where: $\rho$ — the density (iron: $7870\ \mathrm{kg\,m^{-3}}$ ); $r$ — the radius. Physical reading, and the sentence to carry into the exam hall: the cooling rate of a solid sphere is inversely proportional to its radius. Not $r^2$ , not $r^3$ — just $r$ . Halve the radius and the temperature falls twice as fast, even though the small sphere radiates four times less power: our $5\ \mathrm{cm}$ iron sphere starts cooling at $0.052\ \mathrm{K/s}$ , the $2.5\ \mathrm{cm}$ one at $0.105\ \mathrm{K/s}$ .

what the question asks	what you use	how it scales
power radiated, or net power lost	$e\sigma A T^4$ , or $e\sigma A(T^4 - T_s^4)$	$\propto r^2$ — the big body radiates more
mass / heat capacity	$m = \rho\,\tfrac{4}{3}\pi r^3$ , $mc$	$\propto r^3$ — the big body has far more to lose
rate of temperature fall	$\dfrac{e\sigma A (T^4 - T_s^4)}{mc} = \dfrac{3e\sigma(T^4-T_s^4)}{\rho c r}$	$\propto \dfrac{1}{r}$ — the small body cools faster
colour of the glow	Wien: $\lambda_{\max} = b/T$ , $b = 2.898\times10^{-3}\ \mathrm{m\,K}$	size-independent — colour is a thermometer

Derivation AFrom watts to kelvin per second — the initial rate of cooling

We derive equation (3), the master result of this archetype: the rate at which a hot body's temperature falls when it is losing heat by radiation alone. The goal is to get from a power — joules leaving per second — to a rate of temperature change, kelvin lost per second, which is what an examiner actually asks for. No calculus is needed to follow it: we ask how many joules it costs to cool the body by one kelvin, then how many joules per second it is losing, and divide. We finish on this article's locked anchor, a $5\ \mathrm{cm}$ iron sphere at $500\ \mathrm{K}$ in a $300\ \mathrm{K}$ room.

Step 1. Write down the two flows across the surface. The body at temperature $T$ emits $e\sigma A T^4$ watts by Stefan's law; the walls at $T_s$ send back radiation that the body absorbs at $e\sigma A T_s^4$ watts, with the same $e$ because emissivity equals absorptivity (Kirchhoff). Subtract:

P_{\text{net}} = e\sigma A (T^4 - T_s^4)

Why: the surroundings are not a passive void. Ignore the return flow and every cooling answer is too big — $111\ \mathrm{W}$ for our sphere instead of the true $96.9\ \mathrm{W}$ , an error that worsens as the body nears room temperature.

Step 2. Ask what it costs to cool the body. Removing heat $Q$ from a mass $m$ of specific heat capacity $c$ drops its temperature by $\Delta T$ , where $Q = mc\,\Delta T$ — the calorimetry relation you already own. Turn it around: the heat capacity of the whole body is $mc$ joules per kelvin. For our iron sphere $m = 4.12\ \mathrm{kg}$ and $c = 450\ \mathrm{J\,kg^{-1}K^{-1}}$ , so $mc = 1854\ \mathrm{J/K}$ — the price of one kelvin.

Step 3. Divide the loss rate by the cost per kelvin. If the body sheds $P_{\text{net}}$ joules every second, and each kelvin of cooling costs $mc$ joules, then the number of kelvin lost per second is simply

\left|\frac{\Delta T}{\Delta t}\right| = \frac{P_{\text{net}}}{mc} = \frac{e\sigma A (T^4 - T_s^4)}{mc}

Why: check the units. $\mathrm{W} \div (\mathrm{J/K}) = \mathrm{K\,s^{-1}}$ . If your answer does not carry those units you divided by the wrong thing — dividing by $m$ alone, forgetting $c$ , is the classic slip.

Step 4. Put the anchor numbers in. Area: $A = 4\pi (0.05)^2 = 0.0314\ \mathrm{m^2}$ . Fourth powers, computed once and written down: $500^4 = 6.25\times10^{10}$ and $300^4 = 8.10\times10^{9}$ , so $T^4 - T_s^4 = 5.44\times10^{10}\ \mathrm{K^4}$ . For a black sphere ( $e = 1$ ), $P_{\text{net}} = 5.67\times10^{-8} \times 0.0314 \times 5.44\times10^{10} = 96.9\ \mathrm{W}$ . Mass: $m = \rho\cdot\tfrac{4}{3}\pi r^3 = 7870 \times \tfrac{4}{3}\pi (0.05)^3 = 4.12\ \mathrm{kg}$ , so $mc = 1854\ \mathrm{J/K}$ . Therefore

\left|\frac{\Delta T}{\Delta t}\right| = \frac{96.9}{1854} = 0.052\ \mathrm{K\,s^{-1}}

Read that physically: about $19$ seconds for the first single kelvin — and it slows after that, because $T^4 - T_s^4$ shrinks as $T$ falls. A hot body cools fast at first and then dawdles; this initial rate is the fastest it will ever manage. Verified: $A = 0.0314\ \mathrm{m^2}$ , $P_{\text{emit}} = 111\ \mathrm{W}$ , absorbed $14.4\ \mathrm{W}$ , $P_{\text{net}} = 96.9\ \mathrm{W}$ , $m = 4.12\ \mathrm{kg}$ , $mc = 1854\ \mathrm{J/K}$ , rate $= 0.0523\ \mathrm{K/s}$ — all recomputed in verify_hs2-cooling-radiation-numericals.py

Go deeper · Derivation B · 2 minWhy the small body always cools faster — the 1/r law

Derivation BWhy the small body always cools faster — the 1/r law

We now derive the single most-examined result in radiative cooling: for two bodies of the same material and shape at the same temperature, the rate of temperature fall is inversely proportional to the linear size. This is what makes a marble cool faster than a cannonball, a baby lose heat faster than an adult, a thin wire quench faster than a thick bar. The strategy is to take equation (3) and refuse to substitute a single number — we let the geometry cancel instead, because that is what turns a calculation into a law.

Step 1. Write the two geometric quantities for a solid sphere of radius $r$ and density $\rho$ , and keep them symbolic. Surface area $A = 4\pi r^2$ . Mass $m = \rho V = \rho \cdot \tfrac{4}{3}\pi r^3$ . Why: plugging in $r = 0.05$ right now is exactly the trap. A scaling law is a statement about a whole family of spheres; the instant you substitute, you have thrown the family away and kept one member.

Step 2. Form the ratio equation (3) actually needs — the area-to-mass ratio, since $A$ and $m$ are the only size-dependent things in it:

\frac{A}{m} = \frac{4\pi r^2}{\rho \cdot \tfrac{4}{3}\pi r^3} = \frac{3}{\rho\, r}

Why: the $4\pi$ cancels, the $\tfrac{1}{3}$ inverts to a $3$ , and $r^2/r^3 = 1/r$ . Everything about the shape has vanished except the size. Sanity check against the anchor: $3/(7870 \times 0.05) = 7.62\times10^{-3}\ \mathrm{m^2/kg}$ , and directly $0.0314/4.12 = 7.62\times10^{-3}$ . They agree.

Step 3. Substitute into equation (3) to get the cooling rate in its most useful form:

\left|\frac{\Delta T}{\Delta t}\right| = \frac{e\sigma (T^4 - T_s^4)}{c}\cdot\frac{A}{m} = \frac{3\,e\,\sigma\,(T^4 - T_s^4)}{\rho\,c\,r}

Step 4. Read the exponents off. At fixed material ( $\rho$ , $c$ , $e$ ) and fixed temperatures, every symbol on the right is constant except $r$ — so the cooling rate $\propto 1/r$ , while the power radiated, $e\sigma AT^4 \propto r^2$ , goes the other way. For two spheres of the same iron at the same temperature with radii $r$ and $2r$ : area $\times 4$ , net power $\times 4$ , mass $\times 8$ — and therefore cooling rate $\times \tfrac12$ . The big one loses four times the watts and still cools half as fast.

Step 5. Check the algebra against real numbers. The $5\ \mathrm{cm}$ iron sphere: $P_{\text{net}} = 96.9\ \mathrm{W}$ , $m = 4.12\ \mathrm{kg}$ , rate $0.0523\ \mathrm{K/s}$ . The $2.5\ \mathrm{cm}$ sphere, same $500\ \mathrm{K}$ and same room: $A = 7.85\times10^{-3}\ \mathrm{m^2}$ (a quarter), $P_{\text{net}} = 24.2\ \mathrm{W}$ (a quarter), $m = 0.515\ \mathrm{kg}$ (an eighth) — so the rate is $24.2/(0.515\times450) = 0.105\ \mathrm{K/s}$ , twice as fast on a quarter of the power. The marble beats the cannonball not by radiating harder but by having almost nothing to cool. Verified: $A/m = 3/(\rho r) = 7.62\times10^{-3}\ \mathrm{m^2/kg}$ ; small sphere $A = 7.85\times10^{-3}\ \mathrm{m^2}$ , $P_{\text{net}} = 24.2\ \mathrm{W}$ , $m = 0.515\ \mathrm{kg}$ , rate $= 0.1045\ \mathrm{K/s} = 2 \times 0.0523\ \mathrm{K/s}$ — recomputed in verify_hs2-cooling-radiation-numericals.py

Figure

The punchline — double the radius and the power radiated goes up four times, but the mass goes up eight times, so the temperature falls half as fast: the small body cools quicker

Limiting cases

case	result	meaning
$T = T_s$	$P_{\text{net}} = 0$ , rate $= 0$	room temperature reached — still radiating hard, but absorbing just as hard
$T \gg T_s$	$P_{\text{net}} \to e\sigma A T^4$	drop the $T_s^4$ : a filament or a star does not care about a $300\ \mathrm{K}$ room
small excess, $T = T_s + \Delta T$	$P_{\text{net}} \approx 4e\sigma A T_s^3\,\Delta T$	net power $\propto$ the excess — this is Newton's law of cooling
$e \to 0$ (polished silver)	$P \to 0$ , rate $\to 0$	a mirror cannot cool by radiation; halve $e$ and you halve power and rate

The third row is worth a stare: put our iron sphere at only $310\ \mathrm{K}$ in the same $300\ \mathrm{K}$ room and the exact Stefan net power, $\sigma A(310^4 - 300^4) = 2.02\ \mathrm{W}$ , is matched to about $5\%$ by the linear $1.92\ \mathrm{W}$ . A fourth power only looks linear when you stand very close to it — which is why Newton's law is excellent for a cooling cup of tea and a catastrophe for a cooling furnace.

Common trapKelvin, fourth powers, and the net difference — the three ways this topic takes your marks

Kelvin, always. A sphere at $227\ \mathrm{^\circ C}$ radiates as $500^4 = 6.25\times10^{10}$ , not as $227^4 = 2.66\times10^{9}$ — wrong by a factor of $23.5$ , and that wrong answer sits in the options waiting for you. In conduction you may keep Celsius because only a difference appears; here $T$ itself is raised to a power and the $273$ refuses to cancel.

Fourth power, not "times four". Doubling the absolute temperature multiplies the power by $2^4 = 16$ : our black sphere radiates $111\ \mathrm{W}$ at $500\ \mathrm{K}$ and $1781\ \mathrm{W}$ at $1000\ \mathrm{K}$ .

Net means $T^4 - T_s^4$ — never $T^4$ alone, and never $(T - T_s)^4$ . A cooling body absorbs as well as emits: $T^4$ alone gives our sphere $111\ \mathrm{W}$ instead of the true $96.9\ \mathrm{W}$ , while $(T - T_s)^4 = 1.6\times10^{9}$ is a factor of $34$ away from $T^4 - T_s^4 = 5.44\times10^{10}$ . Fourth powers first; subtract second.

Worked examples

1 · Board level — the black sphere and its power (CBSE/ICSE pattern, 3 marks)

A solid black sphere of radius $5.0\ \mathrm{cm}$ is held at $500\ \mathrm{K}$ , with the surroundings negligibly cold. Take $\sigma = 5.67\times10^{-8}\ \mathrm{W\,m^{-2}K^{-4}}$ . Find (a) the power radiated, (b) the power at $1000\ \mathrm{K}$ , (c) the power at $500\ \mathrm{K}$ if the surface is painted so that $e = 0.60$ .

Step 1. Area before anything else — this is where the marks are won and lost. A sphere radiates from its whole curved surface: $A = 4\pi r^2 = 4\pi (0.050)^2 = \boxed{0.0314\ \mathrm{m^2}}$ . Note the radius in metres; leaving it as $5$ inflates the area ten-thousandfold.

Step 2. (a) Fourth power once, written down: $500^4 = 6.25\times10^{10}\ \mathrm{K^4}$ . A black body has $e = 1$ , so equation (1) gives

P = \sigma A T^4 = 5.67\times10^{-8} \times 0.0314 \times 6.25\times10^{10} = \boxed{111\ \mathrm{W}}

A five-centimetre ball at dull red heat pours out as much power as an old-style light bulb.

Step 3. (b) Do not start again. Doubling the absolute temperature multiplies the power by $2^4 = 16$ : $P = 16 \times 111 = \boxed{1781\ \mathrm{W}}$ (direct check: $5.67\times10^{-8} \times 0.0314 \times 10^{12} = 1781\ \mathrm{W}$ ✓). Sixteen times, not twice. Step 4. (c) Emissivity is a plain multiplier, touching nothing else: $P = 0.60 \times 111 = \boxed{66.8\ \mathrm{W}}$ .

2 · Board level — the iron sphere's initial rate of cooling (CBSE/NEET pattern, 5 marks)

A solid iron sphere of radius $5.0\ \mathrm{cm}$ is heated to $500\ \mathrm{K}$ and hung in a room whose walls are at $300\ \mathrm{K}$ . For iron, $\rho = 7870\ \mathrm{kg\,m^{-3}}$ , $c = 450\ \mathrm{J\,kg^{-1}K^{-1}}$ ; treat it as a black body. Find (a) the net rate of energy loss, (b) its heat capacity, (c) the initial rate of fall of temperature, (d) roughly how long the first kelvin takes.

Step 1. The word "room" changes the physics: the sphere now absorbs as well as emits, so this is equation (2), not (1). Both fourth powers first: $500^4 = 6.25\times10^{10}$ , $300^4 = 8.10\times10^{9}$ , so $T^4 - T_s^4 = 5.44\times10^{10}\ \mathrm{K^4}$ .

Step 2. (a) Net power, with $A = 0.0314\ \mathrm{m^2}$ from Example 1:

P_{\text{net}} = \sigma A (T^4 - T_s^4) = 5.67\times10^{-8} \times 0.0314 \times 5.44\times10^{10} = \boxed{96.9\ \mathrm{W}}

Against the $111\ \mathrm{W}$ of Example 1, the walls are handing $14.4\ \mathrm{W}$ straight back.

Step 3. (b) Mass, then heat capacity: $m = \rho \cdot \tfrac{4}{3}\pi r^3 = 7870 \times \tfrac{4}{3}\pi (0.050)^3 = \boxed{4.12\ \mathrm{kg}}$ , so $mc = 4.12 \times 450 = \boxed{1854\ \mathrm{J/K}}$ .

Step 4. (c) Divide the loss rate by the price of a kelvin, equation (3):

\left|\frac{\Delta T}{\Delta t}\right| = \frac{96.9}{1854} = \boxed{0.052\ \mathrm{K\,s^{-1}}}

Step 5. (d) One kelvin takes about $1/0.052 = \boxed{19\ \mathrm{s}}$ — say "about", because this is an initial rate: every later kelvin costs a little longer.

3 · Competitive level — the ratio problem, done without a calculator (JEE-Main/NEET, ~90 seconds)

Two black spheres have radii in the ratio $1:2$ and absolute temperatures in the ratio $2:1$ . Find (a) the ratio of the powers they radiate and (b) — same material, surroundings negligibly cold — the ratio of their initial rates of fall of temperature.

Step 1. Recognise the species: not one number with units appears in the question, so not one may appear in your working. Write proportionalities only, $P \propto r^2 T^4$ .

Step 2. (a) Take the ratio and let the constants die:

\frac{P_1}{P_2} = \left(\frac{r_1}{r_2}\right)^2 \left(\frac{T_1}{T_2}\right)^4 = \left(\frac{1}{2}\right)^2 (2)^4 = \frac{16}{4} = \boxed{4:1}

The small hot sphere out-radiates the big cold one four to one: a fourth power of temperature comfortably outguns a square of radius.

Step 3. (b) Now switch quantities — and here is where the paper tries to catch you. The rate is not the power; it is power per unit heat capacity, and by equation (4) it goes as $T^4/r$ at fixed material:

\frac{(\Delta T/\Delta t)_1}{(\Delta T/\Delta t)_2} = \frac{T_1^4 / r_1}{T_2^4 / r_2} = \frac{2^4/1}{1^4/2} = \frac{16}{0.5} = \boxed{32:1}

Step 4. Read the two answers together: sphere 1 radiates only $4$ times the power but has $8$ times less mass, so its temperature plummets $4 \times 8 = 32$ times faster. Power ratio and rate ratio are different numbers, and a paper offering both is testing precisely this. Timing note: ninety seconds, every one of them on paper. The instant you read "in the ratio", stop reaching for numbers — a scaling problem has none to reach for. Set up $P \propto r^2T^4$ and $\Delta T/\Delta t \propto T^4/r$ , substitute the two ratios, done.

4 · Competitive level — two spheres, a time estimate, and a star (JEE-Advanced stretch, ~4 minutes)

(a) Two solid iron spheres of radii $5.0\ \mathrm{cm}$ and $2.5\ \mathrm{cm}$ are heated to $500\ \mathrm{K}$ and placed in the same $300\ \mathrm{K}$ room. Which cools faster, and by what factor? (b) Estimate how long the larger takes to cool from $500\ \mathrm{K}$ to $490\ \mathrm{K}$ . (c) A star has surface temperature $6000\ \mathrm{K}$ and radius $7.0\times10^{8}\ \mathrm{m}$ . Find its peak wavelength and luminosity, and say how many times brighter a red giant of $100$ times the radius and half the temperature would be. Take $b = 2.898\times10^{-3}\ \mathrm{m\,K}$ .

Step 1. (a) No arithmetic needed. Same material, same temperatures, so equation (4) gives rate $\propto 1/r$ : halving the radius doubles the cooling rate. The $2.5\ \mathrm{cm}$ sphere cools $\boxed{2\times\ \text{faster}}$ while radiating only a quarter of the power. Small bodies are all surface and no substance.

Step 2. (b) The trap is to divide $10\ \mathrm{K}$ by the initial rate: $10/0.052 = 191\ \mathrm{s}$ underestimates, because the sphere slows as it cools. Get the rate at the far end too. At $490\ \mathrm{K}$ : $490^4 = 5.765\times10^{10}$ , so $P_{\text{net}} = 5.67\times10^{-8} \times 0.0314 \times (5.765 - 0.81)\times10^{10} = 88.3\ \mathrm{W}$ , a rate of $88.3/1854 = 0.0476\ \mathrm{K/s}$ .

Step 3. Average the two rates — legitimate because the drop is small and the rate near-linear across it: $\bar{R} = (0.0523 + 0.0476)/2 = 0.0499\ \mathrm{K/s}$ , so $t \approx \Delta T/\bar{R} = 10/0.0499 = \boxed{200\ \mathrm{s}}$ , about three and a half minutes. (The same average-rate tool you met in Newton's law of cooling — honest only over a small drop.)

Step 4. (c) Colour first, by Wien: $\lambda_{\max} = b/T = 2.898\times10^{-3}/6000 = \boxed{483\ \mathrm{nm}}$ — blue-green, mid-visible, which is why a star like this looks white rather than coloured. Brightness per square metre, by Stefan: $\sigma T^4 = 5.67\times10^{-8} \times 1.296\times10^{15} = \boxed{7.35\times10^{7}\ \mathrm{W\,m^{-2}}}$ .

Step 5. Luminosity is that flux times the whole surface, $A = 4\pi R^2 = 6.16\times10^{18}\ \mathrm{m^2}$ :

L = 4\pi R^2 \sigma T^4 = 6.16\times10^{18} \times 7.35\times10^{7} = \boxed{4.5\times10^{26}\ \mathrm{W}}

— reassuringly close to the Sun's real output, since we have essentially described the Sun. For the giant, refuse to recompute; $L \propto R^2T^4$ , so

\frac{L_{\text{giant}}}{L} = (100)^2 \times \left(\tfrac{1}{2}\right)^4 = \frac{10\,000}{16} = \boxed{625\times}

At this size ratio $R^2$ beats $T^4$ : a far cooler star that is far bigger is still overwhelmingly brighter, which is why red giants are visible across the galaxy. Timing note: four minutes, but only if you triage. Part (a) is a one-line scaling — fifteen seconds. Part (b) is the expensive one (two fourth powers and an average), so leave it till last. In part (c) never get the final ratio by recomputing $L$ : $(100)^2(1/2)^4$ is five seconds of mental arithmetic, a fresh calculation a minute and an invitation to a power-of-ten error.

Exam traps

Celsius in a fourth power. Fatal, not merely costly: $227\ \mathrm{^\circ C}$ must become $500\ \mathrm{K}$ , and using $227^4$ instead of $500^4$ is wrong by a factor of $23.5$ . Every board plants the un-converted answer among the options. Make the kelvin conversion your first written line. (CBSE, ICSE, JEE-Main, NEET — universal.)

Using $T^4$ when the question says "cools". Surroundings mean absorption: the cooling power is $e\sigma A(T^4 - T_s^4)$ , so our sphere loses $96.9\ \mathrm{W}$ , not $111\ \mathrm{W}$ . Drop the $T_s^4$ only when $T_s \ll T$ — and say so when you do. (JEE-Main, JEE-Adv, NEET.)

Confusing the power with the rate of cooling. The classic MCQ: two spheres of one material at one temperature, one twice the radius — which cools faster? The big one radiates $4\times$ the power and still cools half as fast, because it carries $8\times$ the mass. Watts $\Rightarrow$ $P$ ; "rate of cooling" or "which cools faster" $\Rightarrow$ $P/(mc)$ . (JEE-Main, NEET — almost every year.)

Substituting numbers into a ratio problem. A question that gives ratios has deliberately given no data, and one line of proportionality answers it. Train the reflex: "in the ratio" $\Rightarrow$ write $P \propto r^2T^4$ and stop. (JEE-Main, JEE-Adv, NEET.)

Getting the area wrong — or leaving $r$ in centimetres. Half the marks lost here go before any physics happens: using $\pi r^2$ (the cross-section) instead of $4\pi r^2$ (the surface), or forgetting metres, which inflates $A$ by $10^4$ . A sphere radiates from $4\pi r^2$ ; a cube of side $a$ from $6a^2$ . And divide by $mc$ , not by $m$ — check the units: $\mathrm{W}/(\mathrm{J\,K^{-1}}) = \mathrm{K\,s^{-1}}$ . (CBSE, ICSE, NEET — method marks.)

Applying Newton's law of cooling to a red-hot body. The linear law is only the small-excess approximation of Stefan's: fine for tea $10\ \mathrm{K}$ above the room, nonsense for a sphere $200\ \mathrm{K}$ above it. Large excess $\Rightarrow$ fourth powers. (JEE-Adv; a standard NEET assertion-reason item.)

Origins · 1 minMax Planck and how this was discovered

Max Planck

1858–1947 · Germany

Every formula on this page is a child of Max Planck, though nothing on this page needs quantum theory. Planck spent the 1890s trying to write down the exact shape of the black-body spectrum — how much energy a hot body radiates at each wavelength. In December 1900 he found it, but only by assuming energy is exchanged in discrete packets, $E = hf$ : a step he called an act of desperation and did not believe for years. It founded quantum physics. What matters for a school exam is what falls out of his curve: add it up over all wavelengths and you get exactly Stefan's $\sigma T^4$ ; find its peak and you get Wien's $\lambda_{\max}T = b$ . Stefan and Wien had measured the total and the peak; Planck supplied the curve they were shadows of. Nobel Prize, 1918.

History

1791

— Prévost's theory of exchanges: every body radiates continually, and cooling is merely emitting more than one absorbs.

1879

— Josef Stefan finds empirically that the total radiated power rises as the fourth power of the absolute temperature.

1884

— Ludwig Boltzmann derives the T⁴ law thermodynamically; it becomes the Stefan-Boltzmann law.

1893

— Wilhelm Wien states the displacement law, fixing the wavelength of peak emission at b/T.

1900

— Max Planck fits the exact radiation spectrum in October and derives it from energy quanta in December; its integral gives Stefan's σ and its peak gives Wien's b.

Going further

Texts. NCERT Physics Class 11, Ch. 11 (Thermal Properties of Matter) · HC Verma, Concepts of Physics Vol. 2, Ch. 28 (Heat Transfer) · Halliday, Resnick & Walker, Ch. 18.

Provenance. Every number in this article — areas, powers, masses, rates, ratios, the star, the linearisation — is recomputed from the stated constants in `verify_hs2-cooling-radiation-numericals.py`.

Prerequisites. Stefan-Boltzmann and Wien's Laws · Newton's Law of Cooling · Thermal Radiation and the Black Body · Specific Heat Capacity.

The seam upward. Where does $\sigma$ itself come from? Not from thermodynamics but from counting photon modes — the undergraduate deep-dive is Photon Gas and Planck Radiation Law, with Stefan-Boltzmann Law and Radiation Pressure for the momentum radiation also carries. Neither is needed for any school exam; both are where this chapter grows up.

Why does the small sphere cool faster if the big one radiates more power?

Because "cooling fast" is about temperature, not energy. The big sphere really does lose more joules every second — four times more if its radius is doubled. But it also has eight times the mass to cool, so those extra joules are spread over far more iron. Divide power by heat capacity and the big body comes out worse: rate $= P/(mc) \propto r^2/r^3 = 1/r$ . Energy loss follows the surface; temperature loss follows the surface per unit mass.

Is the cooling rate of 0.052 K/s constant? Can I use it to find a total cooling time?

No, and that is a trap. It is the initial rate, valid at $500\ \mathrm{K}$ . As the sphere cools, $T^4 - T_s^4$ collapses and the rate falls with it — at $490\ \mathrm{K}$ it is already $0.0476\ \mathrm{K/s}$ . Over a small drop you may average the initial and final rates, as Example 4b does ( $10\ \mathrm{K}$ in about $200\ \mathrm{s}$ ). Over a large drop, dividing the total fall by the initial rate badly underestimates the time.

Why must temperature be in kelvin here when conduction lets me use Celsius?

Because of what the temperature is doing in the formula. In conduction only the difference $\Delta T$ appears, and a difference of $20\ \mathrm{^\circ C}$ is a difference of $20\ \mathrm{K}$ — the $273$ cancels. In radiation, $T$ itself is raised to the fourth power, and $(T + 273)^4 \ne T^4 + 273^4$ . Nothing cancels. A body at $227\ \mathrm{^\circ C}$ is at $500\ \mathrm{K}$ , and $500^4$ is $23.5$ times $227^4$ — a wrong answer examiners deliberately list as an option.

Where does the 1/r law come from, in one line?

From area over volume. The cooling rate is $e\sigma A(T^4 - T_s^4)/(mc)$ , and the only size-dependent part is $A/m$ . For a sphere, $A/m = 4\pi r^2 / (\rho \cdot \tfrac{4}{3}\pi r^3) = 3/(\rho r)$ : the $r^2$ from the area and the $r^3$ from the volume leave one power of $r$ downstairs. Every compact shape behaves the same way — a cube gives $6a^2/(\rho a^3) = 6/(\rho a)$ , the identical $1/\text{size}$ form. Small means fast.

Does this replace Newton's law of cooling, or contradict it?

Neither — it contains it. Newton's linear law is what Stefan's fourth-power law becomes when the body is only slightly hotter than its surroundings: put $T = T_s + \Delta T$ with small $\Delta T$ and you get $P_{\text{net}} \approx 4e\sigma A T_s^3\,\Delta T$ , proportional to the excess, exactly as Newton said. Check the size of the approximation: at a $10\ \mathrm{K}$ excess the exact answer is $2.02\ \mathrm{W}$ and the linear one $1.92\ \mathrm{W}$ , about $5\%$ low. At a $200\ \mathrm{K}$ excess it is useless. Small excess, Newton; large excess, Stefan.

Does emissivity change the 1/r scaling?

No — it scales the answer without touching the exponents. Emissivity multiplies the power and therefore the rate: a sphere with $e = 0.60$ loses $58.1\ \mathrm{W}$ instead of $96.9\ \mathrm{W}$ and cools at $0.031\ \mathrm{K/s}$ instead of $0.052$ . But $e$ is a property of the surface finish, not the size, so it cancels in any ratio problem where both bodies share a finish, and the $1/r$ law survives untouched. Only when two bodies have different finishes must you carry $e$ through the ratio.