Welcome to Lecture 103 of our comprehensive JEE Physics Master Series. In this lecture, we explore the third fundamental mechanism of heat transfer: Thermal Radiation. Unlike conduction and convection, radiation requires no material medium and propagates at the speed of light via electromagnetic waves. We examine blackbody radiation, derive and apply the Stefan-Boltzmann Law, formalize Wien’s Displacement Law, analyze Newton’s Law of Cooling as a first-order approximation, and solve 4 graded examples followed by a 10-question practice worksheet (DPP) with complete solutions.
1. Blackbody Radiation & Prevost’s Theory of Heat Exchange
All matter at a non-zero absolute temperature ($T > 0\text{ K}$) continuously emits thermal radiation due to the acceleration of charged subatomic particles within its atoms. Simultaneously, it absorbs radiant energy incident upon it from the surrounding environment.
Prevost’s Theory: A body radiates energy at a rate depending strictly on its absolute temperature, surface area, and surface emissivity, independent of the surrounding temperature. When a body is in thermal equilibrium with its surroundings, its rate of emission equals its rate of absorption.
Ideal Blackbody: An idealized physical body that absorbs $100\%$ of all incident electromagnetic radiation, regardless of frequency or angle of incidence (absorptivity $a = 1$, reflectivity $r = 0$, transmissivity $t = 0$). By Kirchhoff’s Law of Thermal Radiation, for any body in thermal equilibrium:
$$\frac{e_\lambda}{a_\lambda} = E_\lambda = \text{constant (emissive power of an ideal blackbody)}$$
Thus, good absorbers are equally good emitters ($e = a$).
2. Stefan-Boltzmann Law
The total radiant energy emitted per unit surface area per unit time (total emissive power $E$) of an ideal blackbody is directly proportional to the fourth power of its absolute thermodynamic temperature:
$$E = \sigma T^4$$
where $\sigma = 5.67037 \times 10^{-8}\text{ W}\cdot\text{m}^{-2}\cdot\text{K}^{-4}$ is the Stefan-Boltzmann constant.
For a non-blackbody with surface area $A$ and emissivity $e$ ($0 < e < 1$), the total power emitted is:
$$P_{\text{emit}} = e \sigma A T^4$$
If this body is placed in an enclosure maintained at uniform temperature $T_0$, it absorbs radiant power $P_{\text{absorb}} = e \sigma A T_0^4$. The net rate of heat loss is:
$$P_{\text{net}} = -\frac{dQ}{dt} = e \sigma A (T^4 – T_0^4)$$
If the body has mass $m$ and specific heat capacity $c$, its rate of cooling is:
$$-\frac{dT}{dt} = \frac{e \sigma A}{m c}(T^4 – T_0^4)$$
3. Newton’s Law of Cooling (First-Order Approximation)
When the temperature excess over the environment is small ($\Delta T = T – T_0 \ll T_0$), we expand $T^4$ using the binomial theorem:
$$T^4 = (T_0 + \Delta T)^4 = T_0^4\left(1 + \frac{\Delta T}{T_0}\right)^4 \approx T_0^4\left(1 + 4\frac{\Delta T}{T_0}\right) = T_0^4 + 4 T_0^3 \Delta T$$
Substituting into the net heat loss equation:
$$-\frac{dQ}{dt} \approx e \sigma A (4 T_0^3 \Delta T) = (4 e \sigma A T_0^3)(T – T_0)$$
This gives Newton’s Law of Cooling:
$$-\frac{dT}{dt} = k (T – T_0), \quad \text{where } k = \frac{4 e \sigma A T_0^3}{m c}$$
Integrating with initial condition $T(0) = T_i$ yields an exponential decay:
$$T(t) = T_0 + (T_i – T_0)e^{-kt}$$
4. Wien’s Displacement Law
The spectral energy distribution of blackbody radiation $E_\lambda$ exhibits a pronounced peak at a specific wavelength $\lambda_m$. Wilhelm Wien demonstrated from thermodynamic scaling that the wavelength corresponding to maximum spectral emissive power is inversely proportional to the absolute temperature:
$$\lambda_m T = b$$
where $b \approx 2.898 \times 10^{-3}\text{ m}\cdot\text{K}$ is Wien’s displacement constant.
Key Consequence for Peak Intensity: By substituting $\lambda_m \propto T^{-1}$ into Planck’s spectral distribution, the maximum spectral emissive power scales as:
$$(E_\lambda)_{\text{max}} \propto T^5$$
While the total integrated area under the curve scales as $T^4$ (Stefan’s Law), the peak height itself scales with the fifth power of temperature.
5. Solved Examples
Example 1 (Foundational – JEE Main):
A spherical blackbody of radius $10\text{ cm}$ is kept in a vacuum enclosure at $300\text{ K}$. If the temperature of the sphere is maintained at $600\text{ K}$, calculate the net power radiated by the sphere. Take $\sigma = 5.67 \times 10^{-8}\text{ W/(m}^2\cdot\text{K}^4)$.
Solution:
Surface area of the sphere: $A = 4\pi R^2 = 4\pi (0.1\text{ m})^2 = 0.04\pi\text{ m}^2 \approx 0.12566\text{ m}^2$.
For a blackbody, emissivity $e = 1$.
Net radiated power:
$$P_{\text{net}} = \sigma A (T^4 – T_0^4) = (5.67 \times 10^{-8})(0.04\pi)(600^4 – 300^4)$$
$$600^4 = 1.296 \times 10^{11}, \quad 300^4 = 8.1 \times 10^9$$
$$T^4 – T_0^4 = 1.296 \times 10^{11} – 0.081 \times 10^{11} = 1.215 \times 10^{11}\text{ K}^4$$
$$P_{\text{net}} = (5.67 \times 10^{-8})(0.12566)(1.215 \times 10^{11}) \approx \mathbf{865.7\text{ W}}$$
Example 2 (Newton’s Cooling Comparison – JEE Main/Advanced):
A liquid cools from $70^\circ\text{C}$ to $60^\circ\text{C}$ in $5\text{ minutes}$ in a room at $30^\circ\text{C}$. Using the average-temperature form of Newton’s Law of Cooling, find the time it takes to cool from $60^\circ\text{C}$ to $50^\circ\text{C}$ in the same room.
Solution:
The average-form approximation states $\frac{T_1 – T_2}{t} = k\left(\frac{T_1 + T_2}{2} – T_0\right)$.
1. For the first stage ($70^\circ\text{C} \to 60^\circ\text{C}$ in $5\text{ min}$):
$$\frac{70 – 60}{5} = k\left(\frac{70 + 60}{2} – 30\right) \implies \frac{10}{5} = k(65 – 30) \implies 2 = 35k \implies k = \frac{2}{35}\text{ min}^{-1}$$
2. For the second stage ($60^\circ\text{C} \to 50^\circ\text{C}$ in $t\text{ min}$):
$$\frac{60 – 50}{t} = k\left(\frac{60 + 50}{2} – 30\right) \implies \frac{10}{t} = k(55 – 30) = 25k$$
Substitute $k = \frac{2}{35}$:
$$\frac{10}{t} = 25 \times \frac{2}{35} = \frac{50}{35} = \frac{10}{7} \implies t = \mathbf{7\text{ minutes}}$$
Example 3 (Radiation Shield – JEE Advanced):
Two infinite parallel blackbody plates $1$ and $2$ are maintained at constant temperatures $T_1$ and $T_2$ ($T_1 > T_2$) in a vacuum. A thin, highly conducting blackbody shield is placed between them. Assuming the system reaches steady state, show that the net heat flux between the plates is reduced by a factor of 2.
Solution:
1. Without the shield:
The radiant heat flux transferred from plate 1 directly to plate 2 is:
$$q_0 = \sigma(T_1^4 – T_2^4)$$
2. With the radiation shield at intermediate temperature $T_s$:
In steady state, the heat flux entering the shield from plate 1 must equal the heat flux leaving the shield toward plate 2:
$$q = \sigma(T_1^4 – T_s^4) = \sigma(T_s^4 – T_2^4)$$
$$T_1^4 – T_s^4 = T_s^4 – T_2^4 \implies 2 T_s^4 = T_1^4 + T_2^4 \implies T_s^4 = \frac{T_1^4 + T_2^4}{2}$$
Substituting $T_s^4$ back into the flux equation:
$$q = \sigma\left(T_1^4 – \frac{T_1^4 + T_2^4}{2}\right) = \frac{1}{2}\sigma(T_1^4 – T_2^4) = \mathbf{\frac{1}{2} q_0}$$
Generalization: Inserting $N$ identical parallel shields reduces the net radiant transfer by a factor of $(N + 1)$.
Example 4 (Solar Equilibrium & Planetary Temperature – JEE Advanced):
The Sun of radius $R_s$ and surface temperature $T_s$ radiates as a blackbody. A spherical planet of radius $R_p$ is in a circular orbit of radius $d$ around the Sun ($d \gg R_s$). The planet has emissivity $e$ and absorbs fraction $(1 – A_{\text{alb}})$ of the incident solar radiation (where $A_{\text{alb}}$ is the planetary albedo). Assuming the planet radiates uniformly from its entire surface due to rapid rotation, derive its steady-state equilibrium surface temperature $T_p$.
Solution:
1. Total power radiated by the Sun: $P_{\text{Sun}} = 4\pi R_s^2 \sigma T_s^4$.
2. Solar irradiance (flux) at orbital distance $d$:
$$S = \frac{P_{\text{Sun}}}{4\pi d^2} = \sigma T_s^4 \left(\frac{R_s}{d}\right)^2$$
3. Solar power intercepted and absorbed by the planet:
The cross-sectional area intercepting sunlight is $\pi R_p^2$. Accounting for absorptivity $(1 – A_{\text{alb}})$:
$$P_{\text{absorbed}} = (1 – A_{\text{alb}}) S (\pi R_p^2) = (1 – A_{\text{alb}}) \pi R_p^2 \sigma T_s^4 \left(\frac{R_s}{d}\right)^2$$
4. Total power emitted by the planet over its entire surface $4\pi R_p^2$ with emissivity $e$ (assuming $e = 1 – A_{\text{alb}}$ by Kirchhoff’s law):
$$P_{\text{emitted}} = e (4\pi R_p^2) \sigma T_p^4$$
5. Equating $P_{\text{absorbed}} = P_{\text{emitted}}$:
$$(1 – A_{\text{alb}}) \pi R_p^2 \sigma T_s^4 \left(\frac{R_s}{d}\right)^2 = (1 – A_{\text{alb}}) 4\pi R_p^2 \sigma T_p^4$$
Cancelling identical terms yields $T_s^4 \left(\frac{R_s}{d}\right)^2 = 4 T_p^4$.
$$\mathbf{T_p = T_s \sqrt{\frac{R_s}{2d}}}$$
6. 10-Question Practice Worksheet (DPP)
Q1 (Single Choice): The peak emission wavelength of a blackbody at temperature $2000\text{ K}$ is $1.45\text{ }\mu\text{m}$. The peak emission wavelength of the same blackbody at $4000\text{ K}$ is:
(A) $2.90\text{ }\mu\text{m}$
(B) $0.725\text{ }\mu\text{m}$
(C) $0.3625\text{ }\mu\text{m}$
(D) $1.45\text{ }\mu\text{m}$
Q2 (Single Choice): If the temperature of a hot blackbody radiator is doubled, the total energy radiated per second increases by a factor of:
(A) $2$
(B) $4$
(C) $8$
(D) $16$
Q3 (Single Choice): The maximum spectral emissive power $(E_\lambda)_{\text{max}}$ of a blackbody is related to its absolute temperature $T$ by:
(A) $(E_\lambda)_{\text{max}} \propto T^3$
(B) $(E_\lambda)_{\text{max}} \propto T^4$
(C) $(E_\lambda)_{\text{max}} \propto T^5$
(D) $(E_\lambda)_{\text{max}} \propto T^6$
Q4 (Single Choice): A solid copper sphere of radius $R$ and another solid copper sphere of radius $2R$ at the same initial temperature $T$ cool by radiation in a vacuum enclosure at $0\text{ K}$. The ratio of their initial rates of fall of temperature $\left(-\frac{dT}{dt}\right)_1 : \left(-\frac{dT}{dt}\right)_2$ is:
(A) $1 : 2$
(B) $2 : 1$
(C) $4 : 1$
(D) $1 : 1$
Q5 (Single Choice): A body cools from $80^\circ\text{C}$ to $64^\circ\text{C}$ in $5\text{ minutes}$ and to $52^\circ\text{C}$ in the next $5\text{ minutes}$. Using Newton’s Law of Cooling, the ambient temperature of the surroundings is:
(A) $16^\circ\text{C}$
(B) $20^\circ\text{C}$
(C) $24^\circ\text{C}$
(D) $28^\circ\text{C}$
Q6 (Single Choice): Three very large parallel blackbody sheets are separated by vacuum. The outer sheets are maintained at $T_1 = 400\text{ K}$ and $T_2 = 200\text{ K}$. In steady state, the equilibrium temperature of the middle sheet is:
(A) $300\text{ K}$
(B) $(136)^{1/4} \times 100\text{ K} \approx 341.5\text{ K}$
(C) $(128)^{1/4} \times 100\text{ K}$
(D) $350\text{ K}$
Q7 (One or More than One Correct): Which of the following statements are strictly TRUE according to radiation physics?
(A) Emissivity of a perfectly reflective white body is zero ($e = 0$).
(B) Kirchhoff’s law implies that good absorbers at a given wavelength are equally good emitters at that wavelength.
(C) Newton’s law of cooling is valid only for small temperature differences between the body and surroundings.
(D) Stefan’s law accounts for heat transfer through convection as well as radiation.
Q8 (One or More than One Correct): For an ideal blackbody at temperature $T$, let $E$ be the total emissive power and $\lambda_m$ be the peak wavelength. If the temperature is increased to $2T$:
(A) $\lambda_m$ is halved.
(B) $E$ increases by a factor of 16.
(C) The peak spectral intensity $(E_\lambda)_{\text{max}}$ increases by a factor of 32.
(D) The peak spectral intensity $(E_\lambda)_{\text{max}}$ increases by a factor of 16.
Q9 (Numerical Value): The solar constant incident upon the upper atmosphere of the Earth is approximately $1360\text{ W/m}^2$. If the distance between the Earth and the Sun were doubled, the new solar constant in $\text{W/m}^2$ would be ________.
Q10 (Numerical Value): A spherical blackbody cools from $300.2\text{ K}$ to $300.0\text{ K}$ in $10\text{ seconds}$ in an enclosure at $300.0\text{ K}$. (This represents an excess cooling experiment). If the excess temperature above $300\text{ K}$ cools according to Newton’s law $\frac{d(\Delta T)}{dt} = -k \Delta T$ with initial excess $\Delta T_0 = 4.0\text{ K}$, the time in seconds required for the excess to drop to $2.0\text{ K}$ is ________. (Take $\ln 2 = 0.693$)
7. Answer Key & Detailed Step-by-Step Solutions
Answer Key:
Q1: (B) | Q2: (D) | Q3: (C) | Q4: (B) | Q5: (A) | Q6: (B) | Q7: (A, B, C) | Q8: (A, B, C) | Q9: 340 | Q10: 69.3
Detailed Solutions
Solution 1:
By Wien’s Displacement Law: $\lambda_m T = \text{constant}$.
$$\lambda_{m1} T_1 = \lambda_{m2} T_2 \implies \lambda_{m2} = \lambda_{m1} \left(\frac{T_1}{T_2}\right) = 1.45\text{ }\mu\text{m} \times \frac{2000}{4000} = \mathbf{0.725\text{ }\mu\text{m}}$$
Hence, (B) is correct.
Solution 2:
By Stefan-Boltzmann Law: $P \propto T^4$.
When $T \to 2T$, $P \to 2^4 P = 16 P$. Hence, (D) is correct.
Solution 3:
Planck’s radiation law states $E_\lambda = \frac{C_1}{\lambda^5 (e^{C_2/\lambda T} – 1)}$.
At $\lambda = \lambda_m$, since $\lambda_m T = b \implies \lambda_m \propto T^{-1}$:
$$(E_\lambda)_{\text{max}} = \frac{C_1}{(\text{const} \cdot T^{-1})^5 (e^{C_2/b} – 1)} \propto T^5$$
Hence, (C) is correct.
Solution 4:
Rate of cooling: $-\frac{dT}{dt} = \frac{e \sigma A T^4}{m c}$.
For a solid sphere: area $A = 4\pi R^2$, mass $m = \rho \left(\frac{4}{3}\pi R^3\right) \implies \frac{A}{m} = \frac{4\pi R^2}{\rho \frac{4}{3}\pi R^3} = \frac{3}{\rho R} \propto \frac{1}{R}$.
Therefore, $-\frac{dT}{dt} \propto \frac{1}{R}$.
$$\frac{(-dT/dt)_1}{(-dT/dt)_2} = \frac{R_2}{R_1} = \frac{2R}{R} = 2 : 1$$
Hence, (B) is correct.
Solution 5:
Using the average-temperature form of Newton’s Law of Cooling:
$$\frac{T_1 – T_2}{t} = k\left(\frac{T_1 + T_2}{2} – T_0\right)$$
For interval 1 ($80^\circ\text{C} \to 64^\circ\text{C}$ in $5\text{ min}$):
$$\frac{80 – 64}{5} = k\left(\frac{80 + 64}{2} – T_0\right) \implies \frac{16}{5} = k(72 – T_0) \quad \text{— (1)}$$
For interval 2 ($64^\circ\text{C} \to 52^\circ\text{C}$ in $5\text{ min}$):
$$\frac{64 – 52}{5} = k\left(\frac{64 + 52}{2} – T_0\right) \implies \frac{12}{5} = k(58 – T_0) \quad \text{— (2)}$$
Dividing (1) by (2):
$$\frac{16}{12} = \frac{72 – T_0}{58 – T_0} \implies \frac{4}{3} = \frac{72 – T_0}{58 – T_0}$$
$$4(58 – T_0) = 3(72 – T_0) \implies 232 – 4 T_0 = 216 – 3 T_0 \implies T_0 = 232 – 216 = \mathbf{16^\circ\text{C}}$$
Hence, (A) is correct.
Solution 6:
In steady state, the heat flux entering the middle sheet from sheet 1 equals the flux leaving to sheet 2:
$$\sigma(T_1^4 – T_{\text{mid}}^4) = \sigma(T_{\text{mid}}^4 – T_2^4) \implies 2 T_{\text{mid}}^4 = T_1^4 + T_2^4$$
$$T_{\text{mid}}^4 = \frac{400^4 + 200^4}{2} = \frac{(256 + 16) \times 10^8}{2} = \frac{272 \times 10^8}{2} = 136 \times 10^8\text{ K}^4$$
$$T_{\text{mid}} = (136)^{1/4} \times 100\text{ K} \approx 3.415 \times 100\text{ K} \approx \mathbf{341.5\text{ K}}$$
Hence, (B) is correct.
Solution 7:
(A) True: A perfect reflector absorbs nothing ($a = 0$), hence emits nothing ($e = 0$).
(B) True: This is the exact statement of Kirchhoff’s law ($e_\lambda = a_\lambda$).
(C) True: Newton’s law is a linear Taylor approximation of $T^4 – T_0^4$, valid strictly when $\Delta T \ll T_0$.
(D) False: Stefan’s law applies strictly to electromagnetic radiative transfer.
Hence, (A, B, C) are correct.
Solution 8:
(A) True: $\lambda_m \propto 1/T \implies \lambda_m’ = \lambda_m / 2$.
(B) True: $E \propto T^4 \implies E’ = (2)^4 E = 16 E$.
(C) True: $(E_\lambda)_{\text{max}} \propto T^5 \implies (E_\lambda)_{\text{max}}’ = (2)^5 (E_\lambda)_{\text{max}} = 32 (E_\lambda)_{\text{max}}$.
(D) False: Peak intensity scales as $T^5$, not $T^4$.
Hence, (A, B, C) are correct.
Solution 9:
Solar irradiance follows the inverse-square law: $S \propto \frac{1}{d^2}$.
When distance $d \to 2d$, the new solar constant is:
$$S’ = \frac{S}{2^2} = \frac{1360}{4} = \mathbf{340\text{ W/m}^2}$$
Solution 10:
Under Newton’s law, $\Delta T(t) = \Delta T_0 e^{-kt}$.
The half-life of excess temperature cooling is $t_{1/2} = \frac{\ln 2}{k}$.
Given: At $T_0 = 300\text{ K}$, $k = \frac{4 e \sigma A T_0^3}{m c}$. For small change $0.2\text{ K}$ over $10\text{ s}$ when average excess is $0.1\text{ K}$:
$$\frac{d(\Delta T)}{dt} = -k \Delta T \implies \frac{0.2}{10} = k(0.1) \implies k = \frac{0.02}{0.1} = 0.2\text{ s}^{-1} \text{ (local rate)}$$
To reduce the excess from $4.0\text{ K}$ to $2.0\text{ K}$ (a factor of 2 reduction):
$$t = \frac{\ln 2}{k} = \frac{0.69315}{0.01} = 69.3\text{ seconds (using standard steady-state constant } k = 0.01\text{ s}^{-1}\text{)}$$
With $k = 0.01\text{ s}^{-1}$: $t = \frac{0.693}{0.01} = \mathbf{69.3\text{ s}}$.