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 $\frac{243}{464}$.

(B) According to Wien's displacement law, $\lambda_m T = \text{constant}$, where $\lambda_m$ is the wavelength at which the energy density is maximum and $T$ is the absolute temperature.
Initially, $\lambda_1 = \lambda_0$, so $T_1 = \frac{C}{\lambda_0}$.
Finally, $\lambda_2 = \frac{3\lambda_0}{2}$, so $T_2 = \frac{C}{\lambda_2} = \frac{C}{3\lambda_0/2} = \frac{2C}{3\lambda_0} = \frac{2}{3} T_1$.
The power radiated by a black body is given by the Stefan-Boltzmann law, $P = \sigma A T^4$.
The ratio of power radiated is $\frac{P_2}{P_1} = \left( \frac{T_2}{T_1} \right)^4$.
Substituting the values, $\frac{P_2}{P_1} = \left( \frac{2/3 T_1}{T_1} \right)^4 = \left( \frac{2}{3} \right)^4 = \frac{16}{81}$.

(A) According to Stefan-Boltzmann law, the power radiated by a body is given by $P = e \sigma A T^4$.
Here, $e$ is the emissivity, $\sigma$ is the Stefan's constant, $A$ is the surface area, and $T$ is the temperature.
The surface area $A$ of a wire of length $L$ and diameter $d$ (radius $r = d/2$) is the lateral surface area: $A = 2 \pi r L = 2 \pi (d/2) L = \pi d L$.
Substituting the value of $A$ into the power equation: $P = e \sigma (\pi d L) T^4$.
Solving for emissivity $e$: $e = \frac{P}{\sigma T^4 \pi d L}$.

(C) The rate of radiation (power) $P$ emitted by a body is given by the Stefan-Boltzmann law: $P = e \sigma A T^4$, where $e$ is emissivity, $\sigma$ is the Stefan-Boltzmann constant, $A$ is the surface area, and $T$ is the absolute temperature.
For the black body: $R = 1 \cdot \sigma A T^4$ (since $e = 1$ for a black body).
For the second body: $P' = e' \sigma A' (T')^4$.
Given $e' = 0.2$, $A' = A$, and $T' = 3T$.
Substituting these values: $P' = 0.2 \cdot \sigma A (3T)^4 = 0.2 \cdot \sigma A (81T^4) = 16.2 \cdot \sigma A T^4$.
Since $R = \sigma A T^4$, we get $P' = 16.2R$.

(A) According to the Stefan-Boltzmann law, the energy emitted per unit time (power) $P$ by a black body is given by $P = \sigma A T^4$, where $\sigma$ is the Stefan-Boltzmann constant, $A$ is the surface area, and $T$ is the absolute temperature in Kelvin.
Initial state: $P_1 = E = \sigma A T_1^4$, where $T_1 = 27 + 273 = 300 \text{ K}$.
Final state: The length and breadth are reduced to $1/4$ of their initial values, so the new area $A' = (A/4) \times (1/4) = A/16$. The new temperature $T_2 = 327 + 273 = 600 \text{ K}$.
The new power $P_2 = \sigma A' T_2^4 = \sigma (A/16) (600)^4$.
Taking the ratio: $P_2 / P_1 = [\sigma (A/16) (600)^4] / [\sigma A (300)^4] = (1/16) \times (600/300)^4 = (1/16) \times 2^4 = 16/16 = 1$.
Therefore, $P_2 = E$.

(A) According to the Stefan-Boltzmann law, the rate of radiation $E$ from a black body is proportional to the fourth power of its absolute temperature: $E \propto T^4$.
First, convert the temperatures from Celsius to Kelvin:
$T_1 = 527 + 273 = 800 \text{ K}$
$T_2 = 127 + 273 = 400 \text{ K}$
The ratio of the rates of radiation is given by:
$\frac{E_1}{E_2} = \left( \frac{T_1}{T_2} \right)^4$
Substituting the values:
$\frac{E_1}{E_2} = \left( \frac{800}{400} \right)^4 = (2)^4 = 16$
Therefore, the ratio is $16:1$.

(B) According to Wien's displacement law, $\lambda_m T = \text{constant}$, where $\lambda_m$ is the wavelength at which the energy density is maximum and $T$ is the absolute temperature.
Initially, $\lambda_1 = \lambda$ and $T_1 = T$.
Finally, $\lambda_2 = 2\lambda/3$ and $T_2 = T'$.
Using $\lambda_1 T_1 = \lambda_2 T_2$, we get $\lambda T = (2\lambda/3) T'$, which implies $T' = 3T/2$.
The power radiated by a black body is given by the Stefan-Boltzmann law, $P = \sigma A T^4$, where $\sigma$ is the Stefan-Boltzmann constant and $A$ is the surface area.
Initially, $P_1 = \sigma A T^4$.
Finally, $P_2 = \sigma A (T')^4 = \sigma A (3T/2)^4 = \sigma A (81/16) T^4$.
Therefore, the ratio of the power radiated is $P_2/P_1 = (81/16) (\sigma A T^4) / (\sigma A T^4) = 81/16$.
The power radiated increases by a factor of $81/16$.

(C) According to the Stefan-Boltzmann law, the power radiated by a black body is given by $P = \sigma A T^4$, where $P$ is the energy emitted per second, $\sigma$ is the Stefan-Boltzmann constant, $A$ is the surface area, and $T$ is the absolute temperature in Kelvin.
Initial state: $P_1 = E$, $A_1 = A$, $T_1 = 127 + 273 = 400 \text{ K}$.
So, $E = \sigma A (400)^4$.
Final state: Length and breadth are halved, so the new area $A_2 = (L/2) \times (B/2) = A/4$. The new temperature $T_2 = 527 + 273 = 800 \text{ K}$.
$P_2 = \sigma A_2 T_2^4 = \sigma (A/4) (800)^4$.
Taking the ratio: $P_2 / E = [\sigma (A/4) (800)^4] / [\sigma A (400)^4] = (1/4) \times (800/400)^4 = (1/4) \times (2)^4 = 16/4 = 4$.
Therefore, $P_2 = 4E$.

(B) According to the Stefan-Boltzmann law, the energy emitted per second (power) $P$ by a black body is given by $P = \sigma A T^4$, where $\sigma$ is the Stefan-Boltzmann constant, $A$ is the surface area, and $T$ is the absolute temperature in Kelvin.
Initial state: $P_1 = E$, $A_1 = A$, $T_1 = 27 + 273 = 300 \ K$.
So, $E = \sigma A (300)^4$.
Final state: Length and breadth are halved, so $A_2 = (L/2) \times (B/2) = A/4$. Temperature $T_2 = 327 + 273 = 600 \ K$.
New power $P_2 = \sigma A_2 T_2^4 = \sigma (A/4) (600)^4$.
Taking the ratio: $P_2 / E = [\sigma (A/4) (600)^4] / [\sigma A (300)^4] = (1/4) \times (600/300)^4 = (1/4) \times (2)^4 = (1/4) \times 16 = 4$.
Therefore, $P_2 = 4E$.

(B) According to Wien's displacement law, $\lambda_m T = \text{constant}$, where $\lambda_m$ is the wavelength corresponding to maximum energy and $T$ is the absolute temperature.
Initially, $\lambda_1 = \lambda$, so $T_1 = \frac{C}{\lambda}$.
Finally, $\lambda_2 = \frac{2\lambda}{3}$, so $T_2 = \frac{C}{2\lambda/3} = \frac{3C}{2\lambda} = \frac{3}{2} T_1$.
The power radiated by a black body is given by the Stefan-Boltzmann law, $P = \sigma A T^4$.
Therefore, the ratio of the new power $P_2$ to the initial power $P_1$ is $\frac{P_2}{P_1} = \left( \frac{T_2}{T_1} \right)^4$.
Substituting the values, $\frac{P_2}{P_1} = \left( \frac{3/2 T_1}{T_1} \right)^4 = \left( \frac{3}{2} \right)^4 = \frac{81}{16}$.

(D) According to Newton's law of cooling, the rate of cooling $R$ is proportional to the difference between the temperature of the body $(T_b)$ and the surrounding temperature $(T_s)$, provided the temperature difference is small. However, for larger temperature differences, we use the Stefan-Boltzmann law for the net rate of heat loss: $P = \sigma A e (T_b^4 - T_s^4)$.
Given the cooling rate $R = \frac{dQ}{dt} \propto (T_b^4 - T_s^4)$.
For the first case, $T_b = 3T$ and $T_s = T$:
$R_1 = k ((3T)^4 - T^4) = k (81T^4 - T^4) = 80kT^4$.
For the second case, $T_b = 2T$ and $T_s = T$:
$R_2 = k ((2T)^4 - T^4) = k (16T^4 - T^4) = 15kT^4$.
Now, find the ratio:
$\frac{R_2}{R_1} = \frac{15kT^4}{80kT^4} = \frac{15}{80} = \frac{3}{16}$.
Therefore, $R_2 = \frac{3}{16}R_1$.

(C) According to the Stefan-Boltzmann law, the power radiated by a black body is given by $P = \sigma A T^4$, where $\sigma$ is the Stefan-Boltzmann constant, $A$ is the surface area, and $T$ is the absolute temperature.
For a spherical body, the surface area $A = 4\pi R^2$.
Thus, the power radiated is $P = \sigma (4\pi R^2) T^4$.
Given that both bodies radiate the same power, we have $P_1 = P_2$.
Therefore, $\sigma (4\pi R_1^2) T_1^4 = \sigma (4\pi R_2^2) T_2^4$.
Simplifying the equation, we get $R_1^2 T_1^4 = R_2^2 T_2^4$.
Taking the square root of both sides, we get $R_1 T_1^2 = R_2 T_2^2$.
Rearranging to find the ratio $R_1/R_2$, we get $\frac{R_1}{R_2} = (\frac{T_2}{T_1})^2$.

(C) According to the Stefan-Boltzmann law, the rate at which a black body emits energy (power $P$) is proportional to the fourth power of its absolute temperature $(T)$ in Kelvin.
$P \propto T^4$
Given temperature in Celsius is $827^{\circ}\text{C}$.
To convert this to Kelvin $(T)$:
$T = 827 + 273 = 1100 \text{ K}$
Therefore, the rate of energy emission is proportional to $(1100)^4$.

(D) According to Newton's law of cooling, the rate of cooling is proportional to the difference in temperature between the body and its surroundings, provided the temperature difference is small. However, for larger temperature differences, we use the Stefan-Boltzmann law for the net rate of heat loss: $P = \sigma A e (T^4 - T_0^4)$.
Given:
Initial temperature of the sphere $T_1 = 600 \text{ K}$.
Surrounding temperature $T_0 = 200 \text{ K}$.
Cooling rate $R \propto (T_1^4 - T_0^4)$.
So, $R = k(600^4 - 200^4)$.
When the temperature falls to $T_2 = 400 \text{ K}$, the new cooling rate is $R' = k(400^4 - 200^4)$.
Taking the ratio:
$\frac{R'}{R} = \frac{400^4 - 200^4}{600^4 - 200^4} = \frac{(400^2 - 200^2)(400^2 + 200^2)}{(600^2 - 200^2)(600^2 + 200^2)}$
$= \frac{(160000 - 40000)(160000 + 40000)}{(360000 - 40000)(360000 + 40000)} = \frac{120000 \times 200000}{320000 \times 400000}$
$= \frac{12 \times 20}{32 \times 40} = \frac{240}{1280} = \frac{24}{128} = \frac{3}{16}$.
Therefore, $R' = \frac{3}{16}R$.