MHT-CET Physics · Thermal Properties of Matter

Radiation: Stefan, Wien and Newton's Law of Cooling

A hot body radiates power P = eσAT⁴ (Stefan's law), its spectrum peaks at λ_m = b/T (Wien's law), and a body a little warmer than its surroundings cools at a rate proportional to its excess temperature (Newton's law of cooling); a black body absorbs everything and has e = 1.

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Why this matters

53 PYQs, 12 of them HARD — two thirds of the chapter. Ten split incident heat into absorbed, reflected and transmitted parts or ask about black bodies. Twenty-two use Stefan's law, eleven combine it with Wien's law to compare bodies whose spectra peak at different wavelengths, and ten are Newton's law of cooling. Four cards.

Concept 1 of 4: Absorbed, Reflected, Transmitted, and the Black Body

Radiation falling on a surface is absorbed, reflected or transmitted, and the three fractions add to one: a + r + t = 1. An opaque body transmits nothing, so a + r = 1. A perfectly black body absorbs everything, a = 1, and is also the best emitter, e = 1; a good absorber is a good emitter, so a black sphere cools faster than a red one and a red faster than a white. A black body does not emit every wavelength equally: its intensity rises to a peak and falls, and the area under the curve is the total power per unit area over all wavelengths.

Definition

a + r + t = 1: 250 kcal with a = 0.77, r = 0.17 ⇒ 15 kcal transmitted.
Opaque: t = 0, a + r = 1. Black body: a = e = 1.
Good absorber = good emitter: black > red > white for rate of cooling.
Black-body spectrum: intensity peaks and falls; not the same at all wavelengths. Area under the curve = total power per unit area.

Quantity	Rule
Incident heat	absorbed + reflected + transmitted
Opaque body	t = 0, a + r = 1
Perfect black body	a = 1, emissivity e = 1
Rate of cooling by colour	black > red > white
Area under the intensity–wavelength curve	total power per unit area, all wavelengths

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The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift IIEasy

Example 1 · Thermal Properties of Matter · Radiation — Stefan, Wien, Newton's Law of Cooling, Black Body

The co-efficient of absorption and the coefficient of reflection of a thin uniform plate are 0.77 and 0.17 respectively. If 250 kcal of heat is incident on the surface of the plate, the quantity of heat transmitted is

Leaving out the transmitted part

a + r + t = 1 holds for any surface; only an opaque one has t = 0. Given absorbed and transmitted heat, the reflected heat is what is left.

Drill 9 more on absorbed, reflected, transmitted, and the black body

Concept 2 of 4: Stefan's Law: Power Grows as T⁴

The power a body radiates is P = eσAT⁴, with T in kelvin. For a sphere A = 4πR², so comparisons come down to P ∝ eR²T⁴: doubling the temperature alone multiplies the power by 16. Two bodies that radiate the same power have R²T⁴ equal, so R₁/R₂ = (T₂/T₁)². A body in surroundings at T₀ also absorbs, so its NET loss goes as T⁴ − T₀⁴; compare two states of the same body with that difference, not with T⁴ alone. Equal volumes of different shapes lose heat in proportion to their areas: a sphere, having the least area, cools slowest, and a thin plate fastest.

Definition

$P = e\sigma AT^4$ , sphere $P \propto eR^2T^4$ , T in kelvin.
R → 2R, T → 2T ⇒ 64P; R → R/2, T → 3T ⇒ $\dfrac{81}{4}P$ ; T up 50% ⇒ about +400%.
Same power: $\dfrac{R_1}{R_2} = \left(\dfrac{T_2}{T_1}\right)^2$ ; same power per area: $e_1T_1^4 = e_2T_2^4$ .
Net loss with surroundings $T_0$ : $\propto T^4 - T_0^4$ (900 K and 600 K with 300 K ⇒ 5.3).
Equal volume: rate ∝ area — sphere : cube $= (\pi/6)^{1/3}$ .

Stefan's law

P = e\sigma A T^4, \qquad P_{\text{net}} = e\sigma A\left(T^4 - T_0^4\right)

Worked example

A black body at 27 °C radiates 10 W. What does it radiate at 327 °C?

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The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 1 · Q45Hard

Example 2 · Thermal Properties of Matter · Radiation — Stefan, Wien, Newton's Law of Cooling, Black Body

A sphere is at temperature 600 K. In an external environment of 200 K, its cooling rate is R. When the temperature of the sphere falls to 400 K, then cooling rate R' will become

Using degrees Celsius in T⁴

127 °C to 527 °C is 400 K to 800 K, a factor of 2 — not 527/127. Convert to kelvin before taking any power.

Ignoring the surroundings in a cooling rate

A body in surroundings at T₀ loses heat at a rate ∝ T⁴ − T₀⁴. Comparing 900 K and 600 K in a 300 K room gives 5.3, not (900/600)⁴ = 5.1.

Drill 21 more on stefan's law: power grows as t⁴

Concept 3 of 4: Wien's Law: the Peak Moves With Temperature

A hotter body's spectrum peaks at a shorter wavelength: λ_mT = b, about 2.9 × 10⁻³ m K. So the frequency at the peak is proportional to T — a straight line through the origin on a ν_m–T graph. Many questions pair it with Stefan's law: if the peak wavelength falls to λ/3, the temperature has tripled and the emissive power has risen 81 times. For bodies of different sizes, combine the two as P ∝ R²/λ_m⁴.

Definition

$\lambda_m T = b \approx 2.9\times10^{-3}$ m K; $\nu_m \propto T$ .
Peak at λ/2 ⇒ T doubles ⇒ power × 16; peak at 2λ/3 ⇒ power × $\dfrac{81}{16}$ .
Different sizes: $P \propto \dfrac{R^2}{\lambda_m^4}$ (radii 2, 3, 6 m at 300, 400, 500 nm ⇒ the 6 m disc radiates most).
Two bodies, $T_A = 3T_B$ , peaks 4 μm apart ⇒ $\lambda_B = 6$ μm.

Wien's law

\lambda_m T = b, \qquad \frac{P_2}{P_1} = \left(\frac{\lambda_{m1}}{\lambda_{m2}}\right)^4

Worked example

A star's spectrum peaks at 500 nm. Its surface temperature? (b = 2.9 × 10⁻³ m K)

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MHT-CET · 2023 · 9th May Shift 2 · Q35Moderate

Example 3 · Thermal Properties of Matter · Radiation — Stefan, Wien, Newton's Law of Cooling, Black Body

A black body radiates power 'P' and maximum energy is radiated by it at a wavelength

\lambda_0

. The temperature of the black body is now so changed that it radiates maximum energy at the wavelength

\frac{\lambda_0}{4}

. The power radiated by it at new temperature is

Raising the wavelength ratio to the fourth power the wrong way

Power goes as T⁴, and T as 1/λ_m. A SHORTER peak wavelength means a hotter body and MORE power: λ/2 gives 16 times, not one sixteenth.

Forgetting the size when bodies differ

Wien fixes T, but the power also goes as the area. Discs of radii 2, 3 and 6 m peaking at 300, 400 and 500 nm are not equal; compare R²/λ_m⁴.

Drill 10 more on wien's law: the peak moves with temperature

Concept 4 of 4: Newton's Law of Cooling

For a small excess over the surroundings, the rate of cooling is proportional to that excess: dT/dt = −K(T − T₀). The papers use the averaged form: the fall in temperature divided by the time equals K times (the average temperature over the interval minus T₀). Write it for the first interval to find K, then for the second to find the time or the final temperature. Each equal drop takes longer than the last, because the excess keeps shrinking. Two rates at two temperatures fix T₀ by division. Two calorimeter fillings that cool through the same range give the water equivalent, since the heat lost per second is the same.

Definition

$\dfrac{T_1 - T_2}{t} = K\left(\dfrac{T_1 + T_2}{2} - T_0\right)$ .
80 → 60 °C in 1 min, room 30 °C ⇒ 60 → 50 °C in 48 s.
Two rates: $\dfrac{R_1}{R_2} = \dfrac{T_1 - T_0}{T_2 - T_0}$ (4 and 1 °C/min at 90 and 30 °C ⇒ $T_0 = 10$ °C).
Equal drops take longer and longer: $t_1 < t_2 < t_3$ .
Water equivalent W: $\dfrac{m_1 + W}{t_1} = \dfrac{m_2 + W}{t_2}$ (10 g in 10 min, 20 g in 15 min ⇒ W = 10 g).

Newton's law of cooling

\frac{T_1 - T_2}{t} = K\left(\frac{T_1 + T_2}{2} - T_0\right)

Worked example

A body cools from 80 °C to 50 °C in 5 min in a 20 °C room. How long to cool from 50 °C to 30 °C?

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The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 2 · Q40Hard

Example 4 · Thermal Properties of Matter · Radiation — Stefan, Wien, Newton's Law of Cooling, Black Body

The temperature of a liquid falls from 365 K to 359 K in 3 minutes. The time during which temperature of this liquid falls from 342 K to 338 K is [Let the room temperature be 296 K]

Using the starting temperature instead of the average

The papers' form uses the AVERAGE temperature over the interval minus the surroundings: (T₁ + T₂)/2 − T₀. Using T₁ − T₀ gives a different K.

Drill 9 more on newton's law of cooling

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Stefan's Law: Power Grows as T⁴
Stefan's law
$P = e\sigma A T^4, \qquad P_{\text{net}} = e\sigma A\left(T^4 - T_0^4\right)$
Wien's Law: the Peak Moves With Temperature
Wien's law
$\lambda_m T = b, \qquad \frac{P_2}{P_1} = \left(\frac{\lambda_{m1}}{\lambda_{m2}}\right)^4$
Newton's Law of Cooling
Newton's law of cooling
$\frac{T_1 - T_2}{t} = K\left(\frac{T_1 + T_2}{2} - T_0\right)$

Reference tables (1)

Absorbed, Reflected, Transmitted, and the Black Body5 rows

Quantity	Rule
Incident heat	absorbed + reflected + transmitted
Opaque body	t = 0, a + r = 1
Perfect black body	a = 1, emissivity e = 1
Rate of cooling by colour	black > red > white
Area under the intensity–wavelength curve	total power per unit area, all wavelengths

Watch out for (6)

Leaving out the transmitted part
→ Absorbed, Reflected, Transmitted, and the Black Body
Using degrees Celsius in T⁴
→ Stefan's Law: Power Grows as T⁴
Ignoring the surroundings in a cooling rate
→ Stefan's Law: Power Grows as T⁴
Raising the wavelength ratio to the fourth power the wrong way
→ Wien's Law: the Peak Moves With Temperature
Forgetting the size when bodies differ
→ Wien's Law: the Peak Moves With Temperature
Using the starting temperature instead of the average
→ Newton's Law of Cooling

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