Unit of Stefan’s constant is: (2002)
1. $\text{Watt-m}^2\text{-K}^4$
2. $\text{Watt-m}^2/\text{K}^4$
3. $\text{Watt}/\text{m}^2\text{-K}$
4. $\text{Watt}/\text{m}^2\text{K}^4$
View Answer
From Stefan's law, $E = \sigma T^4$. The unit of emissive power $E$ is $\text{J}/(\text{s m}^2)$ or $\text{W}/\text{m}^2$. Therefore, the unit of $\sigma$ is $\text{W}/(\text{m}^2 \text{K}^4)$.
A black body has wavelength $\lambda_m$ corresponding to maximum energy at $2000 \text{ K}$. Its wavelength corresponding to maximum energy at $3000 \text{ K}$ will be: (2001)
1. $\frac{3}{2}\lambda_m$
2. $\frac{2}{3}\lambda_m$
3. $\frac{16}{81}\lambda_m$
4. $\frac{81}{16}\lambda_m$
View Answer
According to Wien's displacement law, $\lambda_{m1} T_1 = \lambda_{m2} T_2$. $\lambda_m (2000) = \lambda_{m2} (3000) \Rightarrow \lambda_{m2} = \frac{2000}{3000} \lambda_m = \frac{2}{3} \lambda_m$.
A sphere maintained at temperature $600 \text{ K}$, has cooling rate $R$ in an external environment of $200 \text{ K}$ temperature. If its temperature, falls to $400 \text{ K}$ then its cooling rate will be: (1999)
1. $\frac{3}{16}R$
2. $\frac{16}{3}R$
3. $\frac{9}{27}R$
4. None
View Answer
Cooling rate $\propto (T^4 - T_0^4)$. $\frac{R'}{R} = \frac{400^4 - 200^4}{600^4 - 200^4} = \frac{2^4 - 1^4}{3^4 - 1^4} = \frac{15}{80} = \frac{3}{16}$. Since $\frac{3}{16} = \frac{9}{48}$, the correct cooling rate would be $\frac{3}{16}R$. The closest option is incorrectly printed as 9/27, the answer is 3/16 R
A black body is at a temperature of $500 \text{ K}$. It emits energy at a rate which is proportional to: (1997)
1. $(500)^3$
2. $(500)^4$
3. 500
4. $(500)^2$
View Answer
According to Stefan-Boltzmann law, the rate of emission of radiant energy by a black body is proportional to the fourth power of its absolute temperature, $E \propto T^4$. Here, $T = 500 \text{ K}$, so $E \propto (500)^4$.
The total radiant energy per unit area, normal to the direction of incidence, received at a distance $R$ from the center of a star of radius $r$, whose outer surface radiates as a black body at a temperature $T$ $K$ is given by: (2010 Pre)
(Where $\sigma$ is Stefan’s Constant)
1. $\frac{4\pi\sigma r^2 T^4}{R^2}$
2. $\frac{\sigma r^2 T^4}{R^2}$
3. $\frac{\sigma r^2 T^4}{4\pi r^2}$
4. $\frac{\sigma r^4 T^4}{r^2}$
View Answer
Radiant energy per unit area per unit time (Intensity) at distance $R$ is $I = \frac{P}{4\pi R^2} = \frac{\sigma (4\pi r^2) T^4}{4\pi R^2} = \frac{\sigma r^2 T^4}{R^2}$.