Radiation by Stefan's Boltzmann Law Questions in English

(C) Stefan's law is given by $E = \sigma T^4$,where $E$ is the power radiated per unit area.
Rearranging for $\sigma$,we get $\sigma = \frac{E}{T^4}$.
The unit of $E$ (power per unit area) is $\frac{\text{Watt}}{\text{m}^2} = W\,m^{-2}$.
The unit of temperature $T$ is Kelvin $(K)$.
Therefore,the unit of $\sigma$ is $\frac{W\,m^{-2}}{K^4} = W\,m^{-2}\,K^{-4}$.
Thus,the correct option is $C$.

(B) According to the Stefan-Boltzmann law,the power radiated by a black body is given by the formula:
$P = \sigma A T^4$
Where $P$ is the power (energy per unit time,$J\,s^{-1}$ or $W$),$\sigma$ is the Stefan-Boltzmann constant,$A$ is the surface area $(m^2)$,and $T$ is the absolute temperature $(K)$.
Rearranging the formula to solve for $\sigma$:
$\sigma = \frac{P}{A T^4}$
Substituting the units:
Unit of $\sigma = \frac{W}{m^2 \cdot K^4} = W\,m^{-2}\,K^{-4}$
Since $W = J\,s^{-1}$,the unit can also be expressed as $J\,s^{-1}\,m^{-2}\,K^{-4}$ or $J\,m^{-2}\,s^{-1}\,K^{-4}$.
Thus,option $(b)$ is the correct unit.

(C) According to Stefan-Boltzmann law,the total energy $E$ radiated per unit surface area of a black body per unit time is directly proportional to the fourth power of its absolute temperature $T$.
Mathematically,this is expressed as $E = \sigma T^4$,where $\sigma$ is the Stefan-Boltzmann constant.
Therefore,the amount of radiation emitted is proportional to the fourth power of the temperature on the ideal gas scale.

(C) The rate of heat loss $P$ by radiation is given by Stefan-Boltzmann Law: $P = \sigma e A (T^4 - T_0^4)$.
Given:
Surface area $A = 200 \; cm^2 = 200 \times 10^{-4} \; m^2 = 0.02 \; m^2$.
Emissivity $e = 0.4$.
Temperature of the ball $T = 527^{\circ}C = 527 + 273 = 800 \; K$.
Temperature of the surroundings $T_0 = 27^{\circ}C = 27 + 273 = 300 \; K$.
Stefan's constant $\sigma = 5.67 \times 10^{-8} \; J/(m^2 \cdot s \cdot K^4)$.
Substituting the values:
$P = 5.67 \times 10^{-8} \times 0.4 \times 0.02 \times (800^4 - 300^4)$.
$P = 5.67 \times 10^{-8} \times 0.008 \times (4096 \times 10^8 - 81 \times 10^8)$.
$P = 5.67 \times 0.008 \times (4015)$.
$P = 0.04536 \times 4015 \approx 182.12 \; J/s$.
Thus,the rate of loss of heat is approximately $182 \; J/s$.

(A) According to Stefan-Boltzmann Law,the rate of radiation $E$ is proportional to the fourth power of the absolute temperature $T$,i.e.,$E \propto T^4$.
Given:
Initial temperature $T_1 = 0^{\circ}C = 0 + 273 = 273 \ K$.
Final temperature $T_2 = 273^{\circ}C = 273 + 273 = 546 \ K$.
Initial rate of radiation = $E$.
Using the ratio formula: $\frac{E_1}{E_2} = \left( \frac{T_1}{T_2} \right)^4$.
Substituting the values: $\frac{E}{E_2} = \left( \frac{273}{546} \right)^4 = \left( \frac{1}{2} \right)^4 = \frac{1}{16}$.
Therefore,$E_2 = 16 \ E$.

(A) According to Stefan-Boltzmann Law,the energy radiated per unit area per unit time (emissive power) is proportional to the fourth power of the absolute temperature: $E \propto T^4$.
Let the initial energy be $E_1 = E$ at temperature $T_1 = T$.
Let the final energy be $E_2$ at temperature $T_2 = \frac{T}{2}$.
Using the proportionality,we have $\frac{E_2}{E_1} = \left( \frac{T_2}{T_1} \right)^4$.
Substituting the values: $\frac{E_2}{E} = \left( \frac{T/2}{T} \right)^4 = \left( \frac{1}{2} \right)^4$.
$\frac{E_2}{E} = \frac{1}{16}$.
Therefore,$E_2 = \frac{E}{16}$.

(D) According to Stefan-Boltzmann Law,the energy radiated per unit time $E$ is proportional to the fourth power of the absolute temperature $T$,i.e.,$E \propto T^4$.
Given,initial temperature $T_1 = 400^{\circ}C = 400 + 273 = 673\;K$.
Let the final temperature be $T_2$.
We are given that the energy radiated becomes twice,so $E_2 = 2E_1$.
Using the ratio: $\frac{E_2}{E_1} = \left( \frac{T_2}{T_1} \right)^4$.
Substituting the values: $2 = \left( \frac{T_2}{673} \right)^4$.
Taking the fourth root on both sides: $T_2 = 673 \times 2^{1/4}$.
Since $2^{1/4} \approx 1.189$,$T_2 \approx 673 \times 1.189 \approx 800\;K$.

(B) The unit of Stefan's constant $\sigma$ in the $MKS$ system is $\frac{J}{m^2 \cdot s \cdot K^4}$.
We know that $1 \ J = 10^7 \ erg$ and $1 \ m^2 = 10^4 \ cm^2$.
Substituting these values into the unit expression:
$\sigma_{MKS} = \frac{1 \ J}{1 \ m^2 \cdot 1 \ s \cdot 1 \ K^4} = \frac{10^7 \ erg}{10^4 \ cm^2 \cdot 1 \ s \cdot 1 \ K^4}$.
Simplifying the expression:
$\sigma_{MKS} = 10^3 \frac{erg}{cm^2 \cdot s \cdot K^4}$.
Thus,the multiplying factor in the $CGS$ system is $1000$.

(B) According to the Stefan-Boltzmann Law,the rate of energy radiation $P$ from a black body is proportional to the fourth power of its absolute temperature $T$ (in Kelvin).
$P \propto T^4$
Initial temperature $T_1 = 7^oC = 7 + 273 = 280 \ K$.
Final temperature $T_2 = 287^oC = 287 + 273 = 560 \ K$.
The ratio of the rates of energy radiation is:
$\frac{P_2}{P_1} = (\frac{T_2}{T_1})^4$
Substituting the values:
$\frac{P_2}{P_1} = (\frac{560}{280})^4 = (2)^4 = 16$.
Therefore,the rate of energy radiation increases by a factor of $16$.

(B) According to Stefan-Boltzmann Law,the rate of energy radiation $E$ is proportional to the fourth power of the absolute temperature $T$,i.e.,$E \propto T^4$.
Given initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \text{ K}$.
Final temperature $T_2 = 151^{\circ}C = 151 + 273 = 424 \text{ K}$.
The ratio of the rates of radiation is given by $\frac{E_2}{E_1} = \left( \frac{T_2}{T_1} \right)^4$.
Substituting the values: $\frac{E_2}{Q} = \left( \frac{424}{300} \right)^4$.
Calculating the ratio: $\frac{424}{300} \approx 1.413$.
Then,$1.413^4 \approx 3.99 \approx 4$.
Therefore,the new rate of radiation is approximately $4Q \text{ kW m}^{-2}$.

(B) According to Stefan-Boltzmann Law,the rate of emission of radiation $E$ is proportional to the fourth power of the absolute temperature $T$,i.e.,$E \propto T^4$.
Given temperatures in Celsius: $T_A = 727^{\circ}C$ and $T_B = 127^{\circ}C$.
Convert these to Kelvin: $T_A = 727 + 273 = 1000 \ K$ and $T_B = 127 + 273 = 400 \ K$.
The ratio of the rate of emission is $\frac{E_A}{E_B} = \left( \frac{T_A}{T_B} \right)^4$.
Substituting the values: $\frac{E_A}{E_B} = \left( \frac{1000}{400} \right)^4 = \left( \frac{10}{4} \right)^4 = \left( \frac{5}{2} \right)^4$.
Calculating the power: $\frac{5^4}{2^4} = \frac{625}{16}$.

(C) According to Stefan-Boltzmann Law,the power radiated by a black body is given by $P = A \sigma T^4$.
Given: Area $A = 1 \text{ m}^2$,Power $P = 1 \text{ J/s}$,and Stefan's constant $\sigma = 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4$.
Substituting the values: $1 = 1 \times 5.67 \times 10^{-8} \times T^4$.
$T^4 = \frac{1}{5.67 \times 10^{-8}} \approx 0.176 \times 10^8 = 1.76 \times 10^7$.
$T = (1.76 \times 10^7)^{1/4} \approx 64.7 \text{ K} \approx 65 \text{ K}$.
Thus,the correct option is $C$.

(C) According to Stefan-Boltzmann law, the power radiated is given by $P = \varepsilon \sigma A T^4$.
Given:
Area $A = 10^{-4} \ m^2$,
Emissivity $\varepsilon = 0.80$,
Stefan's constant $\sigma = 5.67 \times 10^{-8} \ W/(m^2 \cdot K^4)$,
Heat radiated $Q = 1.58 \times 10^5 \ \text{calories} = 1.58 \times 10^5 \times 4.2 \ J = 6.636 \times 10^5 \ J$,
Time $t = 1 \ \text{hour} = 3600 \ s$.
Power $P = Q/t = (6.636 \times 10^5) / 3600 \approx 184.33 \ W$.
Using $P = \varepsilon \sigma A T^4$:
$184.33 = 0.80 \times 5.67 \times 10^{-8} \times 10^{-4} \times T^4$.
$184.33 = 4.536 \times 10^{-12} \times T^4$.
$T^4 = 184.33 / (4.536 \times 10^{-12}) \approx 4.06 \times 10^{13}$.
$T \approx (406 \times 10^{11})^{1/4} \approx 2520 \ K \approx 2500 \ K$.

(C) According to the Stefan-Boltzmann Law,the total energy radiated per unit time by a black body is given by $E = A \sigma T^4$,where $A$ is the surface area,$\sigma$ is the Stefan-Boltzmann constant,and $T$ is the absolute temperature.
For a sphere,the surface area $A = 4 \pi r^2$.
Therefore,the energy radiated $E \propto r^2 T^4$.
The ratio of energy radiated by spheres $P$ and $Q$ is given by:
$\frac{E_P}{E_Q} = \left( \frac{r_P}{r_Q} \right)^2 \left( \frac{T_P}{T_Q} \right)^4$.
Given: $r_P = 8 \; cm$,$r_Q = 2 \; cm$.
Temperatures in Kelvin: $T_P = 127 + 273 = 400 \; K$ and $T_Q = 527 + 273 = 800 \; K$.
Substituting the values:
$\frac{E_P}{E_Q} = \left( \frac{8}{2} \right)^2 \left( \frac{400}{800} \right)^4 = (4)^2 \left( \frac{1}{2} \right)^4 = 16 \times \frac{1}{16} = 1$.

(C) According to Stefan-Boltzmann Law,the power radiated by a body is given by $P = A \varepsilon \sigma T^4$.
Since $A$,$\varepsilon$,and $\sigma$ are constants,we have $P \propto T^4$.
Therefore,$\frac{P_2}{P_1} = \left( \frac{T_2}{T_1} \right)^4$.
Given $P_1 = 5 \ W$,$T_1 = 127 + 273 = 400 \ K$,and $T_2 = 927 + 273 = 1200 \ K$.
Substituting the values: $\frac{P_2}{5} = \left( \frac{1200}{400} \right)^4$.
$\frac{P_2}{5} = (3)^4 = 81$.
$P_2 = 5 \times 81 = 405 \ W$.

(B) The rate of radiated energy (power) is given by Stefan-Boltzmann law: $P = A \varepsilon \sigma T^4$.
Here,the area $A$ of the square plate is $(10 \text{ cm})^2 = (0.1 \text{ m})^2 = 0.01 \text{ m}^2$.
The emissivity $\varepsilon = 1$,and Stefan's constant $\sigma = 5.67 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}$.
Given $P = 1134 \text{ W}$.
Substituting the values: $1134 = 0.01 \times 1 \times 5.67 \times 10^{-8} \times T^4$.
$1134 = 5.67 \times 10^{-10} \times T^4$.
$T^4 = \frac{1134}{5.67 \times 10^{-10}} = 200 \times 10^{10} = 2 \times 10^{12}$.
$T = (2 \times 10^{12})^{1/4} = (2)^{1/4} \times 10^3 \approx 1.189 \times 1000 = 1189 \text{ K}$.

(D) According to Stefan-Boltzmann Law,the rate of heat radiation $H$ is proportional to the fourth power of the absolute temperature $T$ $(H \propto T^4)$.
First,convert the temperatures from Celsius to Kelvin:
$T_A = 727 + 273 = 1000 \ K$
$T_B = 327 + 273 = 600 \ K$
The ratio of the rates of heat radiated is given by:
$\frac{H_A}{H_B} = \left( \frac{T_A}{T_B} \right)^4$
$\frac{H_A}{H_B} = \left( \frac{1000}{600} \right)^4 = \left( \frac{10}{6} \right)^4 = \left( \frac{5}{3} \right)^4$
$\frac{H_A}{H_B} = \frac{5^4}{3^4} = \frac{625}{81}$
Therefore,the ratio is $625:81$.

(D) According to Stefan-Boltzmann law,the energy emitted per second $(P)$ by a black body is given by $P = A\sigma T^4$,where $A$ is the surface area,$\sigma$ is the Stefan-Boltzmann constant,and $T$ is the absolute temperature in Kelvin.
Since $A$ and $\sigma$ are constants,we have $P \propto T^4$.
Given $T_1 = 27^{\circ}C = 27 + 273 = 300\;K$ and $P_1 = 10\;J/s$.
Given $T_2 = 327^{\circ}C = 327 + 273 = 600\;K$.
Using the ratio: $\frac{P_2}{P_1} = \left( \frac{T_2}{T_1} \right)^4$.
Substituting the values: $P_2 = P_1 \left( \frac{600}{300} \right)^4 = 10 \times (2)^4 = 10 \times 16 = 160\;J/s$.
Therefore,the energy emitted per second is $160\;J$.

(C) According to the Stefan-Boltzmann law,the radiant energy emitted by a black body is proportional to the fourth power of its absolute temperature,i.e.,$E \propto T^4$.
Let $E_1$ be the initial radiant energy incident on the earth and $T_1$ be the initial temperature of the sun.
Given $E_1 = 20 \, kcal/(m^2 \cdot min)$.
Let $E_2$ be the new radiant energy when the temperature of the sun becomes $T_2 = 2T_1$.
Using the ratio formula:
$\frac{E_2}{E_1} = \left( \frac{T_2}{T_1} \right)^4$
Substituting the values:
$\frac{E_2}{20} = \left( \frac{2T_1}{T_1} \right)^4$
$\frac{E_2}{20} = (2)^4 = 16$
Therefore,$E_2 = 16 \times 20 = 320 \, kcal/(m^2 \cdot min)$.

(D) The power radiated by a black body is given by the Stefan-Boltzmann law: $P = A\sigma T^4$,where $A = 4\pi r^2$ is the surface area.
Therefore,$P \propto r^2 T^4$.
Given the initial state: $P_1 = 440\;W$,$r_1 = 24\;cm$,$T_1 = 500\;K$.
For the final state: $r_2 = r_1 / 2 = 12\;cm$ and $T_2 = 2T_1 = 1000\;K$.
Using the proportionality ratio:
$\frac{P_2}{P_1} = \left( \frac{r_2}{r_1} \right)^2 \left( \frac{T_2}{T_1} \right)^4$
$\frac{P_2}{440} = \left( \frac{1}{2} \right)^2 \left( \frac{2}{1} \right)^4$
$\frac{P_2}{440} = \frac{1}{4} \times 16 = 4$
$P_2 = 440 \times 4 = 1760\;W$.

(D) According to the Stefan-Boltzmann Law,the power $P$ radiated by a black body is proportional to the fourth power of its absolute temperature $T$,i.e.,$P \propto T^4$.
Since the rate of energy received on Earth is directly proportional to the power radiated by the Sun,we have $E \propto T^4$.
If the temperature is doubled $(T' = 2T)$,the new rate of energy $E'$ will be $E' \propto (2T)^4 = 16T^4$.
Therefore,the rate of energy received increases by a factor of $16$.

(D) According to the Stefan-Boltzmann Law,the total energy $E$ radiated per unit time by a black body is proportional to the fourth power of its absolute temperature: $E \propto T^4$.
Given temperatures are $T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$ and $T_2 = 927^{\circ}C = 927 + 273 = 1200 \ K$.
The ratio of the energy emitted is given by:
$\frac{E_1}{E_2} = \left( \frac{T_1}{T_2} \right)^4$
Substituting the values:
$\frac{E_1}{E_2} = \left( \frac{300}{1200} \right)^4 = \left( \frac{1}{4} \right)^4 = \frac{1}{256}$.
Therefore,the ratio is $1:256$.

(A) According to Stefan-Boltzmann Law,the power radiated by a black body is proportional to the fourth power of its absolute temperature: $E \propto T^4$.
Given temperatures in Celsius:
$T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$
$T_2 = 327^{\circ}C = 327 + 273 = 600 \ K$
The ratio of radiation emitted is given by:
$\frac{E_2}{E_1} = \left( \frac{T_2}{T_1} \right)^4$
Substituting the values:
$\frac{E_2}{E_1} = \left( \frac{600}{300} \right)^4 = (2)^4 = 16$
Thus,the radiation emitted increases by a factor of $16$.

(D) According to Stefan-Boltzmann law,the rate of energy emission $P$ is given by $P = A \sigma T^4$,where $A$ is the area,$\sigma$ is the Stefan-Boltzmann constant,and $T$ is the absolute temperature in Kelvin.
Initial state: $A_1 = 8 \text{ cm} \times 4 \text{ cm} = 32 \text{ cm}^2$,$T_1 = 127 + 273 = 400 \text{ K}$. Rate $P_1 = E$.
Final state: Length and breadth are halved,so $A_2 = (8/2) \text{ cm} \times (4/2) \text{ cm} = 4 \text{ cm} \times 2 \text{ cm} = 8 \text{ cm}^2$. Note that $A_2 = A_1 / 4$. $T_2 = 327 + 273 = 600 \text{ K}$.
Using the ratio: $\frac{P_2}{P_1} = \frac{A_2}{A_1} \times \left( \frac{T_2}{T_1} \right)^4$.
$\frac{P_2}{E} = \frac{1}{4} \times \left( \frac{600}{400} \right)^4 = \frac{1}{4} \times \left( \frac{3}{2} \right)^4 = \frac{1}{4} \times \frac{81}{16} = \frac{81}{64}$.
Therefore,$P_2 = \frac{81}{64}E$.

(D) According to Stefan-Boltzmann Law,the power radiated $P$ by a black body is directly proportional to the fourth power of its absolute temperature $T$.
$P \propto T^4$
Given that at temperature $T_1 = T$,the power radiated is $P_1 = Q$.
We need to find the power radiated $P_2$ at temperature $T_2 = 3T$.
Using the ratio formula:
$\frac{P_2}{P_1} = \left( \frac{T_2}{T_1} \right)^4$
Substituting the values:
$\frac{P_2}{Q} = \left( \frac{3T}{T} \right)^4$
$\frac{P_2}{Q} = (3)^4$
$\frac{P_2}{Q} = 81$
$P_2 = 81Q$
Therefore,the power radiated at temperature $3T$ is $81Q$.

(A) According to the Stefan-Boltzmann law,the power $P$ radiated by a black body of surface area $A$ and temperature $T$ is given by $P = A \sigma T^4$.
Since the bodies are spherical,the surface area $A = 4 \pi r^2$.
Thus,$P = 4 \pi r^2 \sigma T^4$.
Given that both bodies radiate the same power,we have $P_1 = P_2$.
$4 \pi r_1^2 \sigma T_1^4 = 4 \pi r_2^2 \sigma T_2^4$.
Simplifying this,we get $r_1^2 T_1^4 = r_2^2 T_2^4$.
Rearranging for the ratio $\frac{r_1}{r_2}$,we get $(\frac{r_1}{r_2})^2 = (\frac{T_2}{T_1})^4$.
Taking the square root on both sides,we get $\frac{r_1}{r_2} = (\frac{T_2}{T_1})^2$.

(A) According to the Stefan-Boltzmann Law,the energy radiated by a black body is proportional to the fourth power of its absolute temperature: $Q \propto T^4$.
Given initial temperature $T_1 = 327^{\circ}C = 327 + 273 = 600 \ K$.
Given final temperature $T_2 = 927^{\circ}C = 927 + 273 = 1200 \ K$.
Initial energy $Q_1 = 2 \ kJ$.
Using the ratio: $\frac{Q_2}{Q_1} = \left( \frac{T_2}{T_1} \right)^4$.
$\frac{Q_2}{2} = \left( \frac{1200}{600} \right)^4$.
$\frac{Q_2}{2} = (2)^4 = 16$.
$Q_2 = 16 \times 2 = 32 \ kJ$.

(B) According to the Stefan-Boltzmann law,the total radiant energy $Q$ emitted by a black body is proportional to the fourth power of its absolute temperature $T$,i.e.,$Q \propto T^4$.
Given the initial temperature $T_1 = 727^{\circ}C = 727 + 273 = 1000 \ K$.
We want to find the new temperature $T_2$ such that the radiant energy doubles,i.e.,$Q_2 = 2Q_1$.
Using the ratio: $\frac{Q_2}{Q_1} = \left( \frac{T_2}{T_1} \right)^4$.
Substituting the values: $2 = \left( \frac{T_2}{1000} \right)^4$.
Taking the fourth root on both sides: $\frac{T_2}{1000} = (2)^{1/4}$.
Since $(2)^{1/4} \approx 1.1892$,we have $T_2 = 1000 \times 1.1892 = 1189.2 \ K$.
Rounding to the nearest integer,$T_2 \approx 1190 \ K$.

(A) According to the Stefan-Boltzmann law,the energy radiated per second $(P)$ by a sphere of radius $r$ and temperature $T$ is given by $P = \sigma A T^4$,where $A = 4 \pi r^2$ is the surface area and $\sigma$ is the Stefan-Boltzmann constant.
Thus,$P = \sigma (4 \pi r^2) T^4$.
The ratio of the energy radiated by the two spheres is $\frac{P_1}{P_2} = \frac{\sigma (4 \pi r_1^2) T_1^4}{\sigma (4 \pi r_2^2) T_2^4} = \left( \frac{r_1}{r_2} \right)^2 \left( \frac{T_1}{T_2} \right)^4$.
Given $r_1 = 1 \; m$,$r_2 = 4 \; m$,$T_1 = 4000 \; K$,and $T_2 = 2000 \; K$.
Substituting these values: $\frac{P_1}{P_2} = \left( \frac{1}{4} \right)^2 \left( \frac{4000}{2000} \right)^4 = \left( \frac{1}{16} \right) \times (2)^4 = \frac{16}{16} = 1$.
Therefore,the ratio is $1:1$.

(A) According to Wien's displacement law,$\lambda_m T = \text{constant}$.
Given initial wavelength $\lambda_{m1} = \lambda_o$ and final wavelength $\lambda_{m2} = \frac{3\lambda_o}{4}$.
Using the relation $\lambda_{m1} T_1 = \lambda_{m2} T_2$,we get $T_2 = \frac{\lambda_{m1}}{\lambda_{m2}} T_1 = \frac{\lambda_o}{3\lambda_o/4} T_1 = \frac{4}{3} T_1$.
According to the Stefan-Boltzmann law,the power radiated $P$ is proportional to $T^4$,i.e.,$P \propto T^4$.
Therefore,the ratio of the powers is $\frac{P_2}{P_1} = \left( \frac{T_2}{T_1} \right)^4$.
Substituting the value of $T_2$,we get $\frac{P_2}{P_1} = \left( \frac{4/3 T_1}{T_1} \right)^4 = \left( \frac{4}{3} \right)^4 = \frac{256}{81}$.

(D) According to Stefan-Boltzmann law,the total energy $E$ radiated per unit surface area of a black body per unit time is directly proportional to the fourth power of its absolute temperature $T$.
Mathematically,$E \propto T^4$.
Given the temperature $T = 300 K$,the rate of energy emission is proportional to $(300)^4$.

(D) According to Stefan-Boltzmann Law,the power radiated by a black body is proportional to the fourth power of its absolute temperature: $Q \propto T^4$.
Let the initial temperature be $T_1 = T$ and the initial heat radiation be $Q_1 \propto T^4$.
The temperature is increased by $50\%$,so the new temperature $T_2 = T + 0.5T = 1.5T = \frac{3}{2}T$.
The new heat radiation $Q_2 \propto T_2^4 = (1.5T)^4 = (\frac{3}{2})^4 T^4 = \frac{81}{16} T^4$.
Thus,$Q_2 = \frac{81}{16} Q_1 = 5.0625 Q_1$.
The percentage increase in the quantity of emitted heat radiation is given by:
$\text{Percentage increase} = \frac{Q_2 - Q_1}{Q_1} \times 100\%$
$= \frac{5.0625 Q_1 - Q_1}{Q_1} \times 100\%$
$= 4.0625 \times 100\% = 406.25\%$.
Note: Based on standard textbook approximations for this specific problem type where $T_2 = 1.5T$,the result is $406.25\%$. Given the options provided,$400\%$ is the closest intended answer.

(D) According to Stefan-Boltzmann law,the net rate of heat loss by radiation from a body at temperature $T$ to the surroundings at temperature $T_0$ is given by $P = A\varepsilon \sigma (T^4 - T_0^4)$.
Since the balls are identical,$A$ and $\varepsilon$ are the same for both.
Therefore,the net heat loss $Q$ is proportional to $(T^4 - T_0^4)$.
Given $T_1 = 200^{\circ}C = (200 + 273) K = 473 K$.
Given $T_2 = 400^{\circ}C = (400 + 273) K = 673 K$.
Given $T_0 = 27^{\circ}C = (27 + 273) K = 300 K$.
The ratio of net heat loss is $\frac{Q_1}{Q_2} = \frac{T_1^4 - T_0^4}{T_2^4 - T_0^4}$.
Substituting the values,we get $\frac{Q_1}{Q_2} = \frac{473^4 - 300^4}{673^4 - 300^4}$.

(C) According to the Stefan-Boltzmann Law,the heat radiation energy emitted per second $(Q)$ by a body is given by $Q = A \varepsilon \sigma T^4$.
Since both spheres are made of the same material,their emissivity $(\varepsilon)$ is the same. Given that they are at the same temperature $(T)$,we have $Q \propto A$.
The surface area $(A)$ of a sphere is $4 \pi r^2$,so $Q \propto r^2$.
Therefore,the ratio of heat radiation energy emitted per second is $\frac{Q_1}{Q_2} = \frac{r_1^2}{r_2^2}$.
Given the ratio of radii $\frac{r_1}{r_2} = \frac{1}{2}$,we get $\frac{Q_1}{Q_2} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
Thus,the ratio is $1:4$.

(A) According to Stefan-Boltzmann Law,the rate of heat radiation $E$ is proportional to the fourth power of the absolute temperature $T$,i.e.,$E \propto T^4$.
Given:
$T_1 = 127^{\circ}C = 127 + 273 = 400 \ K$
$E_1 = 1 \ cal/cm^2 \cdot s$
$T_2 = 527^{\circ}C = 527 + 273 = 800 \ K$
Using the ratio formula:
$\frac{E_2}{E_1} = \left( \frac{T_2}{T_1} \right)^4$
$\frac{E_2}{1} = \left( \frac{800}{400} \right)^4$
$\frac{E_2}{1} = (2)^4 = 16$
Therefore,$E_2 = 16 \ cal/cm^2 \cdot s$.

(C) According to Stefan-Boltzmann Law,the power radiated by a black body is given by $P = A\sigma T^4$,where $A$ is the surface area,$\sigma$ is the Stefan-Boltzmann constant,and $T$ is the absolute temperature in Kelvin.
Given:
$P_1 = 20\,W$
$T_1 = 227^{\circ}C = 227 + 273 = 500\,K$
$T_2 = 727^{\circ}C = 727 + 273 = 1000\,K$
Since $P \propto T^4$,we have:
$\frac{P_2}{P_1} = \left( \frac{T_2}{T_1} \right)^4$
Substituting the values:
$\frac{P_2}{20} = \left( \frac{1000}{500} \right)^4$
$\frac{P_2}{20} = (2)^4$
$\frac{P_2}{20} = 16$
$P_2 = 16 \times 20 = 320\,W$
Therefore,the radiating power will be $320\,W$.

(C) According to the Stefan-Boltzmann law,the power radiated by a sphere is given by $P = A \varepsilon \sigma T^4$,where $A = 4\pi r^2$ is the surface area,$\varepsilon$ is the emissivity,and $\sigma$ is the Stefan-Boltzmann constant.
Since both spheres are made of the same material,$\varepsilon$ is the same for both.
Thus,$P \propto r^2 T^4$.
For the first sphere $(P_1)$: $P_1 \propto (1)^2 \times (4000)^4 = 1 \times (4000)^4$.
For the second sphere $(P_2)$: $P_2 \propto (4)^2 \times (2000)^4 = 16 \times (2000)^4$.
Calculating the ratio: $\frac{P_1}{P_2} = \frac{1^2}{4^2} \times \left( \frac{4000}{2000} \right)^4 = \frac{1}{16} \times (2)^4 = \frac{1}{16} \times 16 = 1$.
Therefore,$P_1 = P_2$,which means the energy radiated per second is equal in both cases.

(A) According to Stefan-Boltzmann Law,the power radiated by a black body is proportional to the fourth power of its absolute temperature: $P \propto T^4$.
Let $P_1 = W$ be the initial power at temperature $T_1 = T$.
Let $P_2$ be the final power at temperature $T_2 = T/3$.
Using the ratio method:
$\frac{P_2}{P_1} = \left( \frac{T_2}{T_1} \right)^4$
Substituting the values:
$\frac{P_2}{W} = \left( \frac{T/3}{T} \right)^4$
$\frac{P_2}{W} = \left( \frac{1}{3} \right)^4$
$\frac{P_2}{W} = \frac{1}{81}$
Therefore,$P_2 = \frac{W}{81}$ Watts.

(C) According to the Stefan-Boltzmann Law,the power $P$ radiated by a star is given by $P = \sigma A T^4$,where $A = 4\pi r^2$ is the surface area and $T$ is the surface temperature.
Thus,$P \propto r^2 T^4$.
For star $A$: $P_A \propto r^2 T^4$.
For star $B$: $P_B \propto (4r)^2 (T/2)^4 = 16r^2 \times (T^4 / 16) = r^2 T^4$.
Therefore,the ratio $P_A : P_B = (r^2 T^4) : (r^2 T^4) = 1:1$.

(B) According to the Stefan-Boltzmann law,the total power (energy emitted per unit time) radiated by a black body is given by $P = \sigma A T^4$,where $A = 4\pi r^2$ is the surface area.
Therefore,$P \propto r^2 T^4$.
Let the initial radius be $r_1$ and initial temperature be $T_1$. Let the final radius be $r_2 = 100r_1$ and final temperature be $T_2 = T_1 / 2$.
The ratio of the final energy emitted $(P_2)$ to the initial energy emitted $(P_1)$ is:
$\frac{P_2}{P_1} = \left( \frac{r_2}{r_1} \right)^2 \times \left( \frac{T_2}{T_1} \right)^4$
Substituting the values:
$\frac{P_2}{P_1} = (100)^2 \times \left( \frac{1}{2} \right)^4$
$\frac{P_2}{P_1} = 10000 \times \frac{1}{16} = 625$.
Thus,the total energy emitted increases by a factor of $625$.

(D) According to the Stefan-Boltzmann law,the total power radiated by a spherical body of radius $R$ and temperature $T$ is given by $P = \sigma (4\pi R^2) T^4$.
Since the radiant energy $Q$ received on Earth is proportional to the power radiated by the sun,we have $Q \propto R^2 T^4$.
Let the initial energy be $Q_1 = k R^2 T^4$ and the final energy be $Q_2 = k (2R)^2 (2T)^4$.
Taking the ratio,we get $\frac{Q_2}{Q_1} = \left( \frac{2R}{R} \right)^2 \times \left( \frac{2T}{T} \right)^4$.
$\frac{Q_2}{Q_1} = (2)^2 \times (2)^4 = 4 \times 16 = 64$.
Therefore,the ratio of the radiant energy received on the earth is $64$.

(C) According to Stefan-Boltzmann Law,the energy radiated per unit time (power) $P$ is proportional to the fourth power of the absolute temperature $T$,i.e.,$P \propto T^4$.
Given $P_1 = 2.7 \times 10^{-3} \text{ J/s}$ at $T_1 = 127^{\circ}C = 127 + 273 = 400 \text{ K}$.
Given $P_2 = 4.32 \times 10^{6} \text{ J/s}$.
Using the ratio: $\frac{P_2}{P_1} = \left( \frac{T_2}{T_1} \right)^4$.
$\frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{1/4} = \left( \frac{4.32 \times 10^{6}}{2.7 \times 10^{-3}} \right)^{1/4}$.
$\frac{T_2}{T_1} = \left( 1.6 \times 10^{9} \right)^{1/4}$ is incorrect based on the provided values. Let's re-calculate: $\frac{4.32}{2.7} = 1.6$. $1.6 \times 10^9 = 1600 \times 10^6$. $(1600 \times 10^6)^{1/4} = (16 \times 10^8)^{1/4} = 2 \times 10^2 = 200$.
Thus,$T_2 = 200 \times T_1 = 200 \times 400 \text{ K} = 80000 \text{ K}$.

(A) According to the Stefan-Boltzmann Law,the energy radiated per unit area per unit time $(E)$ is proportional to the fourth power of the absolute temperature $(T)$: $E \propto T^4$.
Given:
$E_1 = 1 \times 10^5 \ J/s \cdot m^2$
$T_1 = 227^\circ C = 227 + 273 = 500 \ K$
$E_2 = 1 \times 10^9 \ J/s \cdot m^2$
Using the ratio formula:
$\frac{E_2}{E_1} = \left( \frac{T_2}{T_1} \right)^4$
Substituting the values:
$\frac{1 \times 10^9}{1 \times 10^5} = \left( \frac{T_2}{500} \right)^4$
$10^4 = \left( \frac{T_2}{500} \right)^4$
Taking the fourth root on both sides:
$10 = \frac{T_2}{500}$
$T_2 = 10 \times 500 = 5000 \ K$

(B) According to Stefan-Boltzmann Law,the energy emitted per second $P$ is proportional to the fourth power of the absolute temperature,i.e.,$P \propto T^4$.
First,convert the temperatures from Celsius to Kelvin:
$T_1 = -73 + 273 = 200 \ K$
$T_2 = 327 + 273 = 600 \ K$
Now,find the ratio of energy emitted per second:
$\frac{P_1}{P_2} = \left( \frac{T_1}{T_2} \right)^4$
$\frac{P_1}{P_2} = \left( \frac{200}{600} \right)^4$
$\frac{P_1}{P_2} = \left( \frac{1}{3} \right)^4 = \frac{1}{81}$
Thus,the ratio of energy emitted per second is $1:81$.

(A) According to the Stefan-Boltzmann Law,the power radiated by a black body is proportional to the fourth power of its absolute temperature: $Q \propto T^4$.
Let the initial temperature be $T_1 = T$ and the initial radiation be $Q_1 \propto T^4$.
If the temperature is increased by $10\%$,the new temperature is $T_2 = T + 0.1T = 1.1T$.
The new radiation emitted is $Q_2 \propto (1.1T)^4$.
Taking the ratio: $\frac{Q_2}{Q_1} = \left( \frac{1.1T}{T} \right)^4 = (1.1)^4 = 1.4641$.
Thus,$Q_2 = 1.4641 Q_1$.
The percentage increase in radiation is given by: $\frac{Q_2 - Q_1}{Q_1} \times 100 = (1.4641 - 1) \times 100 = 46.41\%$.
Rounding to the nearest integer provided in the options,the answer is $46\%$.

(A) According to Stefan-Boltzmann law,the power radiated per unit area (emissive power) $E$ of a black body is given by $E = \sigma T^4$,where $\sigma$ is the Stefan-Boltzmann constant and $T$ is the absolute temperature.
Given: $E = 6.3 \times 10^7 \ W m^{-2}$ and $\sigma = 5.7 \times 10^{-8} \ W m^{-2} K^{-4}$.
Rearranging the formula to solve for $T$: $T^4 = \frac{E}{\sigma}$.
Substituting the values: $T^4 = \frac{6.3 \times 10^7}{5.7 \times 10^{-8}} = \frac{6.3}{5.7} \times 10^{15} \approx 1.105 \times 10^{15} = 1105 \times 10^{12}$.
Taking the fourth root: $T = (1105 \times 10^{12})^{1/4} \approx 5.77 \times 10^3 \ K$.
Rounding to the nearest given option,$T \approx 5.8 \times 10^3 \ K$.

(A) According to Stefan-Boltzmann law,the rate of cooling $H$ is proportional to the difference of the fourth powers of the temperatures of the body and the surroundings: $H \propto (T^4 - T_0^4)$.
Given $T_1 = 600\,K$,$T_0 = 200\,K$,and the initial cooling rate is $H$.
So,$H = k(600^4 - 200^4)$.
When the temperature reduces to $T_2 = 400\,K$,the new cooling rate $H'$ is $H' = k(400^4 - 200^4)$.
Taking the ratio:
$\frac{H'}{H} = \frac{400^4 - 200^4}{600^4 - 200^4} = \frac{(400^2 - 200^2)(400^2 + 200^2)}{(600^2 - 200^2)(600^2 + 200^2)}$
$= \frac{(160000 - 40000)(160000 + 40000)}{(360000 - 40000)(360000 + 40000)} = \frac{120000 \times 200000}{320000 \times 400000} = \frac{12 \times 20}{32 \times 40} = \frac{240}{1280} = \frac{3}{16}$.
Therefore,$H' = \frac{3}{16}H$.

(A) Stefan's constant,denoted by $\sigma$,is a physical constant that appears in the Stefan-Boltzmann law.
The Stefan-Boltzmann law states that the total power radiated per unit surface area of a black body is proportional to the fourth power of its thermodynamic temperature: $P = \sigma A T^4$.
The value of Stefan's constant is approximately $5.67 \times 10^{-8} \text{ W}/\text{m}^2\text{K}^4$.
Therefore,the correct option is $A$.