IIT JEE Physics (1978-2016: 39 Years) Topic-wise Complete Solutions

Stefan's Law

Problems from IIT JEE

Problem (IIT JEE 2005): A body with area $A$ and emissivity $e=0.6$ is kept inside a spherical black body. Total heat radiated by the body at temperature $T$ is,

1. $0.6\sigma eAT^4$
2. $0.8\sigma eAT^4$
3. $1.0\sigma eAT^4$
4. $0.4\sigma eAT^4$

Solution: By Stefan-Boltzmann law, energy radiated per unit time by a body of surface area $A$, emissivity $e$, and absolute temperature $T$ is $e\sigma A T^4$.

Problem (IIT JEE 1994): Two bodies A and B have thermal emissivities of $0.01$ and $0.81$, respectively. The outer surface area of the two bodies are the same. The two bodies emit total radiant power at the same rate. The wavelength $\lambda_B$ corresponding to maximum spectral radiance in the radiation from B is shifted from the wavelength corresponding to maximum spectral radiance in the radiation from A by $1.00\;\mathrm{\mu m}$. If the temperature of A is 5802 K,

1. the temperature of B is 1934 K.
2. $\lambda_B=1.5\;\mathrm{\mu m}$.
3. the temperature of B is 11604 K.
4. the temperature of B is 2901 K.

Solution: Let $e_A=0.01$, $e_B=0.81$, and $T_A=5802 \mathrm{K}$. Stefan's law gives radiant power of two bodies as, \begin{align} {\mathrm{d}Q_A}/{\mathrm{d}t}=\sigma A e_A {T_A}^{\!4},\nonumber\\ {\mathrm{d}Q_B}/{\mathrm{d}t}=\sigma A e_B {T_B}^{\!4}.\nonumber \end{align} Equate ${\mathrm{d}Q_A}/{\mathrm{d}t}={\mathrm{d}Q_B}/{\mathrm{d}t}$ to get, \begin{align} T_B=\left({e_A}/{e_B}\right)^{\!1/4}T_A=\left({0.01}/{0.81}\right)^{\! 1/4}\times 5802=1934\;\mathrm{K}.\nonumber \end{align} Wien's displacement law, $\lambda_m T=b$, gives, \begin{align} \lambda_B=(T_A/T_B)\lambda_A=(5802/1934)\lambda_A=3\lambda_A. \end{align} Also, since $\lambda_B > \lambda_A$ and $|\lambda_B-\lambda_A|=1\;\mathrm{\mu m}$, we get, \begin{align} \lambda_B-\lambda_A=1\;\mathrm{\mu m}. \end{align} Solve above equations to get $\lambda_A=0.5\;\mathrm{\mu m}$ and $\lambda_B=1.5\;\mathrm{\mu m}$. Thus, answer is A and B.

Problem (IIT JEE 2010): Two spherical bodies $A$ (radius 6 cm) and $B$ (radius 18 cm) are at temperatures $T_1$ and $T_2$, respectively. The maximum intensity in the emission spectrum of $A$ is at 500 nm and in that of $B$ is at 1500 nm. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by $A$ to that of $B$?

Solution: Stefan-Boltzmann law gives the energy radiated per unit time by a spherical black body of area $A=4\pi r^2$ and temperature $T$ as, \begin{align} E=\sigma A T^4=4\pi\sigma r^2 T^4. \end{align} The Wien's displacement law relates temperature of the black body to the wavelength at maximum intensity by \begin{align} \lambda_m T=b. \end{align} Eliminate $T$ from above equations to get, \begin{align} E=4\pi\sigma b^4 (r^2/\lambda^4),\nonumber \end{align} which gives, \begin{align} E_1/E_2=(r_1/r_2)^2\, (\lambda_2/\lambda_1)^4=(6/18)^2\, (1500/500)^4=9.\nonumber \end{align}