Radiation by Stefan's Boltzmann Law Questions in English

(D) Given: $\lambda_1 = 0.26 \mu m$,$\lambda_2 = 0.13 \mu m$.
According to Wien's displacement law,$\lambda T = \text{constant}$.
Therefore,$\lambda_1 T_1 = \lambda_2 T_2$.
$\frac{T_1}{T_2} = \frac{\lambda_2}{\lambda_1} = \frac{0.13}{0.26} = \frac{1}{2}$.
According to Stefan-Boltzmann law,the emissive power $E \propto T^4$.
Thus,the ratio of emissive powers is $\frac{E_1}{E_2} = \left(\frac{T_1}{T_2}\right)^4$.
$\frac{E_1}{E_2} = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$.
So,the ratio is $1:16$.

(A) Given: $\lambda_m = 289.8 \ nm = 289.8 \times 10^{-9} \ m$.
Wien's constant $b = 2898 \ \mu m \ K = 2898 \times 10^{-6} \ m \ K$.
Stefan's constant $\sigma = 5.67 \times 10^{-8} \ W m^{-2} K^{-4}$.
From Wien's displacement law,$\lambda_m T = b$.
Therefore,the temperature of the star is $T = \frac{b}{\lambda_m} = \frac{2898 \times 10^{-6}}{289.8 \times 10^{-9}} = \frac{2898 \times 10^{-6}}{289.8 \times 10^{-9}} = 10^4 \ K$.
The radiation intensity (emissive power) $E$ is given by Stefan-Boltzmann law: $E = \sigma T^4$.
$E = (5.67 \times 10^{-8}) \times (10^4)^4$.
$E = 5.67 \times 10^{-8} \times 10^{16}$.
$E = 5.67 \times 10^8 \ W/m^2$.

(A) According to Stefan-Boltzmann Law,the power radiated by a body is given by $P = \sigma e A T^4$.
Since the spheres are of the same material,their emissivity $e$ is the same. The surface area $A$ of a sphere is $4 \pi r^2$.
Thus,the ratio of power radiated is $\frac{P_1}{P_2} = \frac{\sigma e (4 \pi r_1^2) T_1^4}{\sigma e (4 \pi r_2^2) T_2^4} = \left( \frac{r_1}{r_2} \right)^2 \left( \frac{T_1}{T_2} \right)^4$.
Given $r_1 = 5 \ m$,$r_2 = 2 \ m$,$T_1 = 200 \ K$,and $T_2 = 250 \ K$.
Substituting the values: $\frac{P_1}{P_2} = \left( \frac{5}{2} \right)^2 \left( \frac{200}{250} \right)^4$.
$\frac{P_1}{P_2} = \left( \frac{25}{4} \right) \left( \frac{4}{5} \right)^4 = \left( \frac{25}{4} \right) \left( \frac{256}{625} \right)$.
$\frac{P_1}{P_2} = \frac{25}{625} \times \frac{256}{4} = \frac{1}{25} \times 64 = \frac{64}{25}$.

(C) According to the Stefan-Boltzmann law,the thermal energy radiated per second $(Q)$ by a black body is given by $Q = \sigma A T^4$,where $\sigma$ is the Stefan-Boltzmann constant,$A$ is the surface area,and $T$ is the absolute temperature in Kelvin.
Given that the surface areas of both bodies $A$ and $B$ are equal $(A_A = A_B = A)$,the ratio of the energy radiated is:
$\frac{Q_A}{Q_B} = \frac{\sigma A T_A^4}{\sigma A T_B^4} = \left(\frac{T_A}{T_B}\right)^4$
Convert the temperatures from Celsius to Kelvin:
$T_A = 27^{\circ} C = 27 + 273 = 300 \ K$
$T_B = 177^{\circ} C = 177 + 273 = 450 \ K$
Substitute the values into the ratio:
$\frac{Q_A}{Q_B} = \left(\frac{300}{450}\right)^4 = \left(\frac{2}{3}\right)^4$
$\frac{Q_A}{Q_B} = \frac{16}{81}$
Thus,the ratio of the thermal energy radiated per second by $A$ to that by $B$ is $16: 81$.

(D) According to the Stefan-Boltzmann law,the total energy radiated per unit time (power) by a black body is proportional to the fourth power of its absolute temperature $(T)$:
$P \propto T^4$
Let the initial temperature of the Sun be $T_1 = T$ and the initial power received be $P_1$.
When the temperature is doubled,the new temperature is $T_2 = 2T$.
The new power received $P_2$ is given by:
$P_2 \propto (T_2)^4$
$P_2 \propto (2T)^4$
$P_2 \propto 16T^4$
Therefore,the ratio of the new power to the initial power is:
$\frac{P_2}{P_1} = \frac{16T^4}{T^4} = 16$
Thus,the rate of energy received on the Earth will increase by a factor of $16$.

(B) According to the Stefan-Boltzmann Law,the rate of heat energy radiated by a body at absolute temperature $T$ is given by $P = \sigma A e T^{4}$.
When a body at absolute temperature $T_{1}$ is placed in an enclosure at absolute temperature $T_{2}$,the net rate of heat exchange is given by $P_{net} = \sigma A e (T_{2}^{4} - T_{1}^{4})$.
Here,the absolute temperatures are $T_{1} = (t_{1} + 273) \ K$ and $T_{2} = (t_{2} + 273) \ K$.
Since the body is absorbing heat from the chamber,the rate of heat absorption is proportional to the difference of the fourth powers of their absolute temperatures.
Therefore,the rate of heat absorbed is proportional to $(t_{2} + 273)^{4} - (t_{1} + 273)^{4}$.

(D) According to Stefan-Boltzmann Law,the rate of loss of heat $E$ from a black body at temperature $T$ in an environment at temperature $T_0$ is given by $E \propto (T^4 - T_0^4)$.
Given temperatures are $T_1 = 327^{\circ} C = 600 \ K$,$T_2 = 427^{\circ} C = 700 \ K$,and $T_0 = 27^{\circ} C = 300 \ K$.
The ratio of the rates of loss of heat is $\frac{E_1}{E_2} = \frac{T_1^4 - T_0^4}{T_2^4 - T_0^4}$.
Substituting the values: $\frac{E_1}{E_2} = \frac{(600)^4 - (300)^4}{(700)^4 - (300)^4}$.
Factoring out $(100)^4$: $\frac{E_1}{E_2} = \frac{6^4 - 3^4}{7^4 - 3^4} = \frac{1296 - 81}{2401 - 81}$.
$\frac{E_1}{E_2} = \frac{1215}{2320}$.
Dividing both numerator and denominator by $5$,we get $\frac{243}{464}$.