Radiation by Stefan's Boltzmann Law Questions in English

(A) According to Stefan-Boltzmann Law,the rate of energy emission (power) $P$ from a black body is directly proportional to the fourth power of its absolute temperature $T$.
Mathematically,$P \propto T^{4}$.
Given the temperature $T = 500 \; K$,the rate of energy emission is proportional to $(500)^{4}$.

(A) According to the Stefan-Boltzmann law,the power radiated by a black body is given by $P = \sigma A T^4$.
Since the energy emitted $Q$ in time $t$ is $Q = P \cdot t = \sigma A t T^4$,the ratio of energy emitted by two spheres is:
$\frac{Q_1}{Q_2} = \frac{A_1}{A_2} \left( \frac{T_1}{T_2} \right)^4$.
Given $r_1 = 4 \, m$,$r_2 = 1 \, m$,$T_1 = 2000 \, K$,and $T_2 = 4000 \, K$.
Since $A = 4 \pi r^2$,the ratio of areas is $\frac{A_1}{A_2} = \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{4}{1} \right)^2 = 16$.
The ratio of temperatures is $\frac{T_1}{T_2} = \frac{2000}{4000} = \frac{1}{2}$.
Substituting these values:
$\frac{Q_1}{Q_2} = 16 \times \left( \frac{1}{2} \right)^4 = 16 \times \frac{1}{16} = 1$.
Thus,the ratio is $1:1$.

(C) According to the 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.
Since the spheres are of the same material and at the same temperature,$\sigma$ and $T$ are constant.
The surface area of a sphere is $A = 4 \pi r^2$,so $P \propto r^2$.
Therefore,the ratio of the emissive power is $\frac{P_1}{P_2} = \frac{r_1^2}{r_2^2}$.
Given $\frac{r_1}{r_2} = \frac{1}{2}$,we have $\frac{P_1}{P_2} = (\frac{1}{2})^2 = \frac{1}{4}$.

(A) According to the Stefan-Boltzmann law,the total power $P$ radiated by a sphere of radius $r$ and temperature $T$ is given by $P = A \sigma T^4$,where $A = 4 \pi r^2$ is the surface area and $\sigma$ is the Stefan-Boltzmann constant.
Since the material is the same,the emissivity is the same. Given that the total power radiated $P$ is the same for both spheres,we have $P_1 = P_2$.
Substituting the formula: $4 \pi r_1^2 \sigma T_1^4 = 4 \pi r_2^2 \sigma T_2^4$.
Simplifying the equation: $r_1^2 T_1^4 = r_2^2 T_2^4$.
Rearranging to find the ratio of radii: $(\frac{r_1}{r_2})^2 = (\frac{T_2}{T_1})^4$.
Taking the square root on both sides: $\frac{r_1}{r_2} = (\frac{T_2}{T_1})^2$.

(A) According to Stefan-Boltzmann law,the rate of energy radiated $(E)$ by a black body is directly proportional to the fourth power of its absolute temperature ($T$ in Kelvin).
The formula is given by $E \propto T^4$.
Given the temperature in Celsius is $727^\circ C$,we must convert it to Kelvin:
$T = 727 + 273 = 1000 \ K$.
Substituting this value into the proportionality relation:
$E \propto (1000)^4$.
Therefore,the rate of energy emission is proportional to $(1000)^4$.

(C) The total power $P$ radiated by the sun,acting as a black body with radius $r$ and absolute temperature $T = (t + 273) \ K$,is given by the Stefan-Boltzmann law: $P = \sigma A T^4 = \sigma (4\pi r^2) (t + 273)^4$.
At a distance $R$ from the center of the sun,this power is distributed over a spherical surface area of $4\pi R^2$.
The power received per unit area (intensity $S$) by a surface normal to the incident rays is given by: $S = \frac{P}{4\pi R^2}$.
Substituting the value of $P$: $S = \frac{\sigma (4\pi r^2) (t + 273)^4}{4\pi R^2} = \frac{r^2 \sigma (t + 273)^4}{R^2}$.

(B) 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:
Initial temperature $T_1 = 227^o C = (227 + 273) K = 500 K$.
Initial rate of radiation $E_1 = 7 \; cal/cm^2 s$.
Final temperature $T_2 = 727^o C = (727 + 273) K = 1000 K$.
Using the ratio formula:
$\frac{E_2}{E_1} = \left( \frac{T_2}{T_1} \right)^4$
Substituting the values:
$\frac{E_2}{7} = \left( \frac{1000}{500} \right)^4$
$\frac{E_2}{7} = (2)^4$
$\frac{E_2}{7} = 16$
$E_2 = 7 \times 16 = 112 \; cal/cm^2 s$.
Therefore,the rate of heat radiated at $727^o C$ is $112 \; cal/cm^2 s$.

(A) According to the Stefan-Boltzmann law,the total power $P$ radiated by a star of radius $r$ acting as a black body at temperature $T$ is given by:
$P = \sigma A T^4 = \sigma (4\pi r^2) T^4$
At a distance $R$ from the center of the star,this energy is spread over a spherical surface area of $4\pi R^2$.
The radiant energy per unit area (intensity $S$) received at distance $R$ is:
$S = \frac{P}{4\pi R^2}$
Substituting the value of $P$:
$S = \frac{\sigma (4\pi r^2) T^4}{4\pi R^2} = \sigma \frac{r^2}{R^2} T^4$

(C) According to Stefan's law,the rate of energy emission (power) $Q$ from a black body is given by $Q = \sigma A T^4$.
Here,$A$ is the surface area of the star,which is $4\pi R^2$.
Substituting the value of $A$ into the equation,we get $Q = \sigma (4\pi R^2) T^4$.
Rearranging the formula to solve for temperature $T$,we have $T^4 = \frac{Q}{4\pi \sigma R^2}$.
Taking the fourth root on both sides,we get $T = \left( \frac{Q}{4\pi \sigma R^2} \right)^{1/4}$.

(C) According to the $Stefan-Boltzmann$ law,the rate of energy radiated by a black body is given by:
$E = \sigma A T^4 = \sigma (4 \pi R^2) T^4$
Given:
$E_1 = 450 \ W$,$T_1 = 500 \ K$,$R_1 = 12 \ cm$
$R_2 = \frac{R_1}{2}$,$T_2 = 2T_1$
Since $E \propto R^2 T^4$,we have:
$\frac{E_2}{E_1} = \left( \frac{R_2}{R_1} \right)^2 \left( \frac{T_2}{T_1} \right)^4$
$\frac{E_2}{E_1} = \left( \frac{1}{2} \right)^2 \times (2)^4$
$\frac{E_2}{E_1} = \frac{1}{4} \times 16 = 4$
$E_2 = 4 \times E_1 = 4 \times 450 \ W = 1800 \ W$

(D) According to $Wien's$ displacement law,$\lambda_{\max} T = \text{constant}$.
Let the initial temperature be $T$ and the final temperature be $T'$.
Given $\lambda_{\max, 1} = \lambda_0$ and $\lambda_{\max, 2} = \frac{3}{4}\lambda_0$.
Using the law: $\lambda_0 T = \left(\frac{3}{4}\lambda_0\right) T' \Rightarrow T' = \frac{4}{3}T$.
According to the $Stefan-Boltzmann$ law,the power radiated by a black body is $P = A \sigma T^4$,which implies $P \propto T^4$.
Therefore,$\frac{P_2}{P_1} = \left(\frac{T'}{T}\right)^4$.
Substituting the values: $n = \left(\frac{4/3 T}{T}\right)^4 = \left(\frac{4}{3}\right)^4 = \frac{256}{81}$.

(A) The rate of heat loss per unit area due to radiation,also known as the emissive power $e$,is given by the Stefan-Boltzmann law:
$e = \varepsilon \sigma (T^4 - T_0^4)$
Given:
Emissivity $\varepsilon = 0.6$
Stefan-Boltzmann constant $\sigma = \frac{17}{3} \times 10^{-8} \, W/m^2 K^4$
Surface temperature $T = 127^\circ C = 127 + 273 = 400 \, K$
Surrounding temperature $T_0 = 27^\circ C = 27 + 273 = 300 \, K$
Substituting the values:
$e = 0.6 \times \left( \frac{17}{3} \times 10^{-8} \right) \times [ (400)^4 - (300)^4 ]$
$e = 0.2 \times 17 \times 10^{-8} \times [ 256 \times 10^8 - 81 \times 10^8 ]$
$e = 3.4 \times 10^{-8} \times (175 \times 10^8)$
$e = 3.4 \times 175 = 595 \, J/m^2 \cdot s$

(D) According to Stefan-Boltzmann Law,the energy radiated per unit time by a black body is proportional to the fourth power of its absolute temperature: $E \propto T^4$.
Given temperatures in Celsius: $T_1 = 27^oC$ and $T_2 = 127^oC$.
Convert these to Kelvin: $T_1 = 27 + 273 = 300 \ K$ and $T_2 = 127 + 273 = 400 \ K$.
The ratio of the energies emitted is given by: $\frac{E_1}{E_2} = \left( \frac{T_1}{T_2} \right)^4$.
Substituting the values: $\frac{E_1}{E_2} = \left( \frac{300}{400} \right)^4 = \left( \frac{3}{4} \right)^4 = \frac{81}{256}$.
Thus,the ratio is $81:256$.

(B) The power $P$ received by a small area at a distance $d$ from a spherical source at temperature $T$ is proportional to the Stefan-Boltzmann law and the inverse square law of radiation.
According to the Stefan-Boltzmann law,the total power radiated by the sphere is $P_{source} \propto T^4$.
The intensity of radiation at a distance $d$ follows the inverse square law,$I \propto \frac{1}{d^2}$.
Thus,the power received by the foil is $P \propto \frac{T^4}{d^2}$.
Let the initial state be $P_1 = k \frac{T^4}{d^2} = P$.
In the final state,the temperature becomes $T' = 2T$ and the distance becomes $d' = 2d$.
The new power $P'$ is given by:
$P' = k \frac{(2T)^4}{(2d)^2} = k \frac{16 T^4}{4 d^2} = 4 \left( k \frac{T^4}{d^2} \right) = 4P$.

(C) Let the radius of sphere $P$ be $r$ and the radius of sphere $Q$ be $3r$.
The rate of heat loss by radiation is given by Stefan-Boltzmann Law: $\frac{dQ}{dt} = \sigma e A T^4$.
Also,the rate of heat loss is related to the rate of fall of temperature by the equation: $\frac{dQ}{dt} = -ms \frac{dT}{dt}$,where $m$ is mass,$s$ is specific heat capacity,and $\frac{dT}{dt}$ is the rate of fall of temperature.
Equating the two: $ms \frac{dT}{dt} = \sigma e A T^4$.
Since $m = \rho V = \rho (\frac{4}{3} \pi r^3)$,we have $\rho (\frac{4}{3} \pi r^3) s \frac{dT}{dt} = \sigma e (4 \pi r^2) T^4$.
Thus,the rate of fall of temperature is $\frac{dT}{dt} = \frac{\sigma e (4 \pi r^2) T^4}{\rho (\frac{4}{3} \pi r^3) s} = \frac{3 \sigma e T^4}{\rho r s}$.
This shows that $\frac{dT}{dt} \propto \frac{1}{r}$.
Therefore,$\frac{(\frac{dT}{dt})_P}{(\frac{dT}{dt})_Q} = \frac{r_Q}{r_P} = \frac{3r}{r} = 3$.
Given $(\frac{dT}{dt})_P = x (\frac{dT}{dt})_Q$,we find $x = 3$.

(B) According to the Stefan-Boltzmann law,the radiant power $E$ emitted by a body is given by $E = \varepsilon \sigma A T^4$,where $\varepsilon$ is emissivity,$\sigma$ is the Stefan-Boltzmann constant,$A$ is the surface area,and $T$ is the absolute temperature.
Given that the radiant power $E$ and surface area $A$ are the same for both bodies $P$ and $Q$,we have:
$E_P = E_Q$
$\varepsilon_P \sigma A_P \theta_P^4 = \varepsilon_Q \sigma A_Q \theta_Q^4$
Since $A_P = A_Q$ and $\sigma$ is constant,the equation simplifies to:
$\varepsilon_P \theta_P^4 = \varepsilon_Q \theta_Q^4$
Rearranging for $\theta_Q$:
$\theta_Q^4 = \frac{\varepsilon_P}{\varepsilon_Q} \theta_P^4$
$\theta_Q = \left( \frac{\varepsilon_P}{\varepsilon_Q} \right)^{1/4} \theta_P$
Thus,the correct option is $B$.

(A) According to the Stefan-Boltzmann law,the rate of emission of radiation $E$ from a black body is proportional to the fourth power of its absolute temperature $T$,given by $E = \sigma A T^4$.
Step $1$: Convert temperatures from Celsius to Kelvin.
$T_1 = 273^{\circ} C = (273 + 273) K = 546 K$.
$T_2 = 0^{\circ} C = (0 + 273) K = 273 K$.
Step $2$: Set up the ratio of emission rates.
$\frac{E_2}{E_1} = \left( \frac{T_2}{T_1} \right)^4$.
Step $3$: Substitute the values.
$\frac{E_2}{E} = \left( \frac{273}{546} \right)^4 = \left( \frac{1}{2} \right)^4 = \frac{1}{16}$.
Therefore,$E_2 = \frac{E}{16}$.

(C) The total power radiated by the Sun is given by the Stefan-Boltzmann law: $P_{sun} = \sigma T^4 \times (4\pi R^2)$.
The intensity $I$ of the radiation at a distance $r$ from the Sun is the power per unit area: $I = \frac{P_{sun}}{4\pi r^2} = \frac{\sigma T^4 \times 4\pi R^2}{4\pi r^2} = \frac{\sigma T^4 R^2}{r^2}$.
The Earth intercepts this radiation over its cross-sectional area,which is $\pi r_0^2$.
Therefore,the total radiant power incident on Earth is $P_{earth} = I \times \pi r_0^2 = \frac{\sigma T^4 R^2}{r^2} \times \pi r_0^2 = \frac{\pi r_0^2 R^2 \sigma T^4}{r^2}$.

(C) For the middle plate to be in a steady state,the net power received by the plate must equal the net power emitted by the plate.
Let the temperature of the middle plate be $\theta$.
The power received by the middle plate from the first plate (at $2T$) and the third plate (at $3T$) is given by the Stefan-Boltzmann law for black bodies: $P_{\text{received}} = \sigma A (2T)^4 + \sigma A (3T)^4$.
The middle plate emits radiation from both its sides,so the total power emitted is: $P_{\text{emitted}} = 2 \times \sigma A \theta^4$.
Equating the power received and emitted:
$\sigma A (2T)^4 + \sigma A (3T)^4 = 2 \sigma A \theta^4$
$16 T^4 + 81 T^4 = 2 \theta^4$
$97 T^4 = 2 \theta^4$
$\theta^4 = \frac{97}{2} T^4$
$\theta = \left( \frac{97}{2} \right)^{1/4} T$

(B) The total power radiated by the sun is given by the Stefan-Boltzmann law: $P = \sigma A_{sun} T^4 = \sigma (4\pi R^2) T^4$.
At a distance $r$ from the sun,this power is spread over a spherical surface area of $4\pi r^2$.
The intensity $I$ of radiation at distance $r$ is $I = P / (4\pi r^2) = (\sigma 4\pi R^2 T^4) / (4\pi r^2) = \sigma R^2 T^4 / r^2$.
The Earth intercepts this radiation with its cross-sectional area,which is $\pi r_0^2$.
Therefore,the total radiant power incident on Earth is $P_{Earth} = I \times \pi r_0^2 = (\sigma R^2 T^4 / r^2) \times \pi r_0^2 = \pi r_0^2 R^2 \sigma T^4 / r^2$.

(A) According to the Stefan-Boltzmann law,the power radiated by a body is given by $P = e A \sigma T^4$,where $e$ is the emissivity,$A$ is the surface area,$\sigma$ is the Stefan-Boltzmann constant,and $T$ is the absolute temperature.
Since both lamps radiate the same power $(P_A = P_B)$ and have the same dimensions (same surface area $A$),we have:
$e_A A \sigma T_A^4 = e_B A \sigma T_B^4$
Canceling the common terms $A$ and $\sigma$,we get:
$e_A T_A^4 = e_B T_B^4$
Rearranging the terms to find the ratio $\frac{T_A}{T_B}$:
$\frac{T_A^4}{T_B^4} = \frac{e_B}{e_A}$
Taking the fourth root on both sides:
$\frac{T_A}{T_B} = (\frac{e_B}{e_A})^{1/4}$

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

(B) According to the Stefan-Boltzmann law,the total power $P$ radiated by a black body of surface area $A$ at temperature $T$ is given by $P = A \sigma T^4$.
The power received by a small detector at distance $d$ is proportional to the intensity $I$,which follows the inverse square law: $I = \frac{P}{4 \pi d^2} = \frac{A \sigma T^4}{4 \pi d^2}$.
Given that the power received by the detector remains unchanged when the temperature changes from $T_1$ to $T_2$ and the distance changes from $d_1$ to $d_2$,we have:
$\frac{A \sigma T_1^4}{4 \pi d_1^2} = \frac{A \sigma T_2^4}{4 \pi d_2^2}$
Canceling the common constants $A, \sigma,$ and $4 \pi$,we get:
$\frac{T_1^4}{d_1^2} = \frac{T_2^4}{d_2^2}$
Rearranging the terms to find the ratio $d_2/d_1$:
$(\frac{d_2}{d_1})^2 = (\frac{T_2}{T_1})^4$
Taking the square root of both sides:
$\frac{d_2}{d_1} = (\frac{T_2}{T_1})^2$

(C) According to Stefan-Boltzmann's Law,the power radiated per unit area $(E)$ is given by:
$E = \sigma T^4$
Where $E$ is in $watt/m^2$ and $T$ is in $Kelvin$ $(K)$.
Rearranging for $\sigma$:
$\sigma = \frac{E}{T^4}$
Substituting the units:
Unit of $\sigma = \frac{watt/m^2}{K^4} = \frac{watt}{m^2 \times K^4}$.

(C) The rate of heat emission by radiation is given by the Stefan-Boltzmann law: $\Delta E = e A \sigma (T^4 - T_0^4)$.
Since the material,temperature,and surroundings are the same,the ratio of radiation emitted depends only on the surface area $A$: $\frac{(\Delta E)_{\text{sphere}}}{(\Delta E)_{\text{cube}}} = \frac{A_{\text{sphere}}}{A_{\text{cube}}} = \frac{4 \pi R^2}{6 a^2}$.
Given that the volumes are equal: $\frac{4}{3} \pi R^3 = a^3$,which implies $a = R \left( \frac{4\pi}{3} \right)^{1/3}$.
Substituting $a$ into the area ratio: $\frac{4 \pi R^2}{6 [R (4\pi/3)^{1/3}]^2} = \frac{4 \pi}{6 (4\pi/3)^{2/3}} = \frac{4 \pi}{6} \cdot \frac{3^{2/3}}{(4\pi)^{2/3}} = \frac{(4\pi)^{1/3}}{6 \cdot 3^{-2/3}} = \frac{(4\pi)^{1/3}}{6^{1/3} \cdot 6^{2/3} \cdot 3^{-2/3}} = \left( \frac{4\pi}{6 \cdot (4/3)^{2/3}} \right) = \left( \frac{\pi}{6} \right)^{1/3}$.
Thus,the ratio is $\left( \frac{\pi}{6} \right)^{1/3} : 1$.

(C) According to the Stefan-Boltzmann law,the radiated energy $E$ is given by $E = \sigma A T^4$,where $A$ is the surface area and $T$ is the absolute temperature.
For a spherical body of diameter $d$,the surface area $A = 4\pi r^2 = 4\pi (d/2)^2 = \pi d^2$.
Initial energy $E_1 = \sigma (\pi d^2) T^4 = E$.
New temperature $T_2 = 2T$ and new diameter $d_2 = d/4$.
New surface area $A_2 = \pi (d_2)^2 = \pi (d/4)^2 = \pi d^2 / 16 = A_1 / 16$.
New radiated energy $E_2 = \sigma A_2 T_2^4 = \sigma (A_1 / 16) (2T)^4$.
$E_2 = \sigma (A_1 / 16) (16 T^4) = \sigma A_1 T^4 = E$.

(D) 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:
Initial temperature $T_{1} = 227 + 273 = 500\,K$.
Final temperature $T_{2} = 727 + 273 = 1000\,K$.
Initial rate of radiation $E_{1} = 7\,cal\,cm^{-2}\,s^{-1}$.
Using the ratio formula:
$E_{2} = E_{1} \left(\frac{T_{2}}{T_{1}}\right)^{4}$
$E_{2} = 7 \times \left(\frac{1000}{500}\right)^{4}$
$E_{2} = 7 \times (2)^{4}$
$E_{2} = 7 \times 16 = 112\,cal\,cm^{-2}\,s^{-1}$.
Therefore,the rate of heat radiated at $727\,^{\circ}C$ is $112\,cal\,cm^{-2}\,s^{-1}$.

(A) According to the Stefan-Boltzmann law,the rate of cooling $R$ is proportional to $(T^4 - T_0^4)$,where $T$ is the temperature of the body and $T_0$ is the temperature of the surroundings.
For the first case: $R_1 = R \propto (600^4 - 200^4) = (6^4 - 2^4) \times 10^8 = (1296 - 16) \times 10^8 = 1280 \times 10^8$.
For the second case: $R_2 \propto (400^4 - 200^4) = (4^4 - 2^4) \times 10^8 = (256 - 16) \times 10^8 = 240 \times 10^8$.
Taking the ratio: $\frac{R_2}{R_1} = \frac{240 \times 10^8}{1280 \times 10^8} = \frac{24}{128} = \frac{3}{16}$.
Therefore,$R_2 = \frac{3}{16} R$.

(A) According to the Stefan-Boltzmann law,the total energy radiated per unit area per unit time by a black body is proportional to the fourth power of its absolute temperature $(E \propto T^4)$.
The area under the intensity-wavelength curve represents the total emissive power of the black body at that temperature.
Therefore,the ratio of the areas $A_1$ and $A_2$ corresponding to temperatures $T_1$ and $T_2$ is given by $\frac{A_2}{A_1} = \frac{T_2^4}{T_1^4}$.
Given temperatures are $T_1 = 27\,^oC = (27 + 273)\,K = 300\,K$ and $T_2 = 327\,^oC = (327 + 273)\,K = 600\,K$.
Substituting these values into the ratio:
$\frac{A_2}{A_1} = \left( \frac{600}{300} \right)^4 = (2)^4 = 16$.
Thus,the ratio $\frac{A_2}{A_1}$ is $16 : 1$.

(C) The rate of energy radiation is given by the Stefan-Boltzmann law: $P = e \sigma A T^4$.
For a black body,emissivity $e = 1$.
Let $R_1$ be the radius of the tungsten body and $R$ be the radius of the black body.
The power radiated by the tungsten body is $P_T = e_T \sigma (4 \pi R_1^2) T^4$,where $e_T = 0.3$.
The power radiated by a black body of the same radius $R_1$ is $P_B = 1 \cdot \sigma (4 \pi R_1^2) T^4$.
Given that the tungsten body radiates $30 \%$ of the energy of a black body of the same radius,we have $P_T = 0.3 P_B$.
We want to find the radius $R_{new}$ of a black body that radiates at the same rate as the tungsten body at the same temperature: $P_{new} = 1 \cdot \sigma (4 \pi R_{new}^2) T^4 = P_T$.
Since $P_T = 0.3 \cdot \sigma (4 \pi R_1^2) T^4$,we equate the two:
$\sigma (4 \pi R_{new}^2) T^4 = 0.3 \sigma (4 \pi R_1^2) T^4$.
$R_{new}^2 = 0.3 R_1^2$.
$R_{new} = \sqrt{0.3} \cdot R_1$.
Given diameter $d = 2.3 \ cm$,so $R_1 = 1.15 \ cm$.
$R_{new} = \sqrt{0.3} \cdot 1.15 \approx 0.5477 \cdot 1.15 \approx 0.6299 \ cm \approx 0.63 \ cm$.

(D) According to the Stefan-Boltzmann law,the power radiated by a black body is given by $E = \sigma A T^4$.
Since the surface area $A$ and the Stefan-Boltzmann constant $\sigma$ are constant,the power radiated is directly proportional to the fourth power of the absolute temperature: $E \propto T^4$.
Let the initial temperature be $T_1 = T$ and the final temperature be $T_2 = T/2$.
The ratio of the final power $E_2$ to the initial power $E_1$ is given by:
$E_2 / E_1 = (T_2 / T_1)^4 = ( (T/2) / T )^4 = (1/2)^4 = 1/16$.
Therefore,the amount of radiation emitted reduces to $1/16$ of its initial value.

(D) According to the Stefan-Boltzmann Law,the rate of heat radiation $E$ from a black body is directly proportional to the fourth power of its absolute temperature $T$,i.e.,$E \propto T^{4}$.
Given:
Initial temperature $T_{1} = 227 + 273 = 500\, K$.
Initial rate of radiation $E_{1} = 7\, cal\, cm^{-2} \,s^{-1}$.
Final temperature $T_{2} = 727 + 273 = 1000\, K$.
Using the ratio formula:
$E_{2} = E_{1} \left(\frac{T_{2}}{T_{1}}\right)^{4}$
Substituting the values:
$E_{2} = 7 \times \left(\frac{1000}{500}\right)^{4}$
$E_{2} = 7 \times (2)^{4}$
$E_{2} = 7 \times 16 = 112\, cal\, cm^{-2} \,s^{-1}$.
Thus,the rate of heat radiated is $112$ units.

(B) The rate of heat dissipation for a black body is given by the Stefan-Boltzmann law: $Q = \sigma A T^4$.
For another body with emissivity $e = 0.25$ and temperature $T' = 2T$,the rate of heat dissipation $Q'$ is given by: $Q' = e \sigma A (T')^4$.
Substituting the given values: $Q' = 0.25 \times \sigma A (2T)^4$.
$Q' = 0.25 \times \sigma A \times 16 T^4$.
$Q' = 0.25 \times 16 \times (\sigma A T^4)$.
Since $Q = \sigma A T^4$,we have $Q' = 4 \times Q = 4Q$.

(A) According to the 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: $T_1 = 7\,^oC$ and $T_2 = 287\,^oC$.
Convert these to Kelvin: $T_1 = 7 + 273 = 280\,K$ and $T_2 = 287 + 273 = 560\,K$.
The ratio of radiation emitted is $\frac{E_1}{E_2} = \left(\frac{T_1}{T_2}\right)^4$.
Substituting the values: $\frac{E_1}{E_2} = \left(\frac{280}{560}\right)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$.
Thus,the ratio is $1:16$.

(D) According to Stefan-Boltzmann law,the total energy radiated per unit area per unit time (emissive power $E$) by a black body is directly proportional to the fourth power of its absolute temperature $(T)$.
$E \propto T^4$
Since the area under the intensity-wavelength curve represents the total emissive power,we have $A \propto T^4$.
Given temperatures are $T_1 = 27\,^oC = (27 + 273)\,K = 300\,K$ and $T_2 = 327\,^oC = (327 + 273)\,K = 600\,K$.
Therefore,the ratio of the areas is:
$\frac{A_2}{A_1} = \frac{T_2^4}{T_1^4} = \left(\frac{600}{300}\right)^4 = (2)^4 = 16$.
Thus,the value of $\frac{A_2}{A_1}$ is $16 : 1$.

(C) The energy radiated per unit area per unit time is given by the Stefan-Boltzmann law: $I = \frac{P}{A} = \sigma T^4$.
Given,$I = 5.67 \times 10^4 \, W/m^2$ and Stefan-Boltzmann constant $\sigma = 5.67 \times 10^{-8} \, W/m^2 \cdot K^4$.
Substituting the values: $5.67 \times 10^4 = 5.67 \times 10^{-8} \times T^4$.
$T^4 = \frac{5.67 \times 10^4}{5.67 \times 10^{-8}} = 10^{12}$.
Taking the fourth root: $T = (10^{12})^{1/4} = 10^3 \, K = 1000 \, K$.
To convert to Celsius: $T(^{\circ}C) = T(K) - 273 = 1000 - 273 = 727 \, ^{\circ}C$.

(C) The rate of heat loss by the sphere is given by the Stefan-Boltzmann law: $dQ/dt = \sigma A (T^4 - T_0^4)$,where $A = 4\pi R^2$.
Also,the heat lost by the sphere is $dQ = -Mc dT$,where $c = \alpha T^3$.
Equating the two expressions: $-M(\alpha T^3) dT = \sigma (4\pi R^2) (T^4 - T_0^4) dt$.
Rearranging to solve for $dt$: $dt = -\frac{M\alpha T^3 dT}{\sigma (4\pi R^2) (T^4 - T_0^4)}$.
Integrating from $T = 3T_0$ to $T = 2T_0$: $t = \int_{2T_0}^{3T_0} \frac{M\alpha T^3}{4\pi R^2 \sigma (T^4 - T_0^4)} dT$.
Let $u = T^4 - T_0^4$,then $du = 4T^3 dT$,so $T^3 dT = du/4$.
$t = \frac{M\alpha}{4\pi R^2 \sigma} \int_{T=2T_0}^{T=3T_0} \frac{du/4}{u} = \frac{M\alpha}{16\pi R^2 \sigma} [\ln(u)]_{T=2T_0}^{T=3T_0}$.
$t = \frac{M\alpha}{16\pi R^2 \sigma} [\ln(T^4 - T_0^4)]_{2T_0}^{3T_0} = \frac{M\alpha}{16\pi R^2 \sigma} \ln \left( \frac{(3T_0)^4 - T_0^4}{(2T_0)^4 - T_0^4} \right)$.
$t = \frac{M\alpha}{16\pi R^2 \sigma} \ln \left( \frac{81T_0^4 - T_0^4}{16T_0^4 - T_0^4} \right) = \frac{M\alpha}{16\pi R^2 \sigma} \ln \left( \frac{80}{15} \right) = \frac{M\alpha}{16\pi R^2 \sigma} \ln \left( \frac{16}{3} \right)$.

(A) According to $Stefan's\,law$, the power radiated per unit area is given by $E = \sigma T^4$.
Given:
Temperature $T = 3000\,K$
Stefan's constant $\sigma = 5.7 \times 10^{-8}\,W\,m^{-2}\,K^{-4}$
Time $t = 1\,hour = 3600\,s$
The heat radiated per unit area $(Q/A)$ in time $t$ is:
$Q/A = E \times t = \sigma T^4 \times t$
Substituting the values:
$Q/A = (5.7 \times 10^{-8}) \times (3000)^4 \times 3600$
$Q/A = 5.7 \times 10^{-8} \times 81 \times 10^{12} \times 3600$
$Q/A = 5.7 \times 81 \times 3600 \times 10^4$
$Q/A = 1662120 \times 10^4 = 1.66212 \times 10^{10}\,J \approx 1.7 \times 10^{10}\,J$.

(B) According to Stefan-Boltzmann Law,the rate of heat radiation $E$ is proportional to the fourth power of absolute temperature $T$,i.e.,$E = \sigma T^4$.
Given:
$E_1 = 2 \times 10^5 \, J/s \cdot m^2$
$T_1 = 127 + 273 = 400 \, K$
$E_2 = 32 \times 10^5 \, J/s \cdot m^2$
Using the ratio formula:
$\frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4$
$\frac{32 \times 10^5}{2 \times 10^5} = \left(\frac{T_2}{400}\right)^4$
$16 = \left(\frac{T_2}{400}\right)^4$
Taking the fourth root on both sides:
$2 = \frac{T_2}{400}$
$T_2 = 800 \, K$
Converting back to Celsius:
$T_2 = 800 - 273 = 527 \, ^\circ C$.

(C) According to Stefan-Boltzmann law,the power radiated is given by $P = \sigma \varepsilon A T^4$.
Here,$P = \frac{1.58 \times 10^5 \ cal}{1 \ hour} = \frac{1.58 \times 10^5 \times 4.2 \ J}{3600 \ s} \approx 184.33 \ W$.
Given: $A = 10^{-4} \ m^2$,$\varepsilon = 0.80$,$\sigma = 5.67 \times 10^{-8} \ W/m^2K^4$.
Substituting the values: $184.33 = 5.67 \times 10^{-8} \times 0.80 \times 10^{-4} \times T^4$.
$T^4 = \frac{184.33}{5.67 \times 0.80 \times 10^{-12}} \approx 4.06 \times 10^{13} \approx 3906250000000$.
Taking the fourth root,$T \approx 2500 \ K$.

(C) According to Stefan-Boltzmann Law,the rate of energy emission per unit area $E$ is proportional to the fourth power of the absolute temperature $T$,i.e.,$E \propto T^4$.
Given:
$T_1 = 127^{\circ}C = 127 + 273 = 400\,K$
$E_1 = 1.0 \times 10^6\,J/s\cdot m^2$
$E_2 = 16.0 \times 10^6\,J/s\cdot m^2$
Using the ratio formula:
$\frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4$
Substituting the values:
$\frac{16.0 \times 10^6}{1.0 \times 10^6} = \left(\frac{T_2}{400}\right)^4$
$16 = \left(\frac{T_2}{400}\right)^4$
$2^4 = \left(\frac{T_2}{400}\right)^4$
Taking the fourth root on both sides:
$2 = \frac{T_2}{400}$
$T_2 = 800\,K$
Converting back to Celsius:
$T_2 = 800 - 273 = 527^{\circ}C$.

(D) According to the Stefan-Boltzmann law,the total energy radiated per unit area per unit time by a black body is proportional to the fourth power of its absolute temperature $(E \propto T^4)$.
The area under the spectral energy density curve $E_\lambda$ versus $\lambda$ represents the total emissive power of the black body.
Given the ratio of the areas under the curves is $\frac{A_1}{A_2} = 16 : 1$,where $A_1$ corresponds to temperature $T$ and $A_2$ corresponds to $2000 \ K$.
Using the relation $\frac{A_1}{A_2} = \left(\frac{T}{2000}\right)^4$,we have:
$16 = \left(\frac{T}{2000}\right)^4$
Taking the fourth root on both sides:
$2 = \frac{T}{2000}$
$T = 2 \times 2000 = 4000 \ K$.

(A) According to the Stefan-Boltzmann law,the power radiated by a black body is proportional to the fourth power of its absolute temperature: $E \propto T^4$.
Initial temperature $T_1 = 27 + 273 = 300\,K$.
Final temperature $T_2 = 127 + 273 = 400\,K$.
The ratio of radiation emitted is given by $\frac{E_2}{E_1} = \left( \frac{T_2}{T_1} \right)^4$.
Substituting the values: $\frac{E_2}{E_1} = \left( \frac{400}{300} \right)^4 = \left( \frac{4}{3} \right)^4$.
Calculating the value: $\frac{4^4}{3^4} = \frac{256}{81}$.
Therefore,the radiation increases by a factor of $\frac{256}{81}$.

(A) Given:
Area $A = 1\, cm^2 = 10^{-4}\, m^2$
Temperature $T = 1000\, K = 10^3\, K$
Time $t = 1\, s$
Stefan's constant $\sigma = 5.67 \times 10^{-8}\, W\, m^{-2}K^{-4}$
According to the Stefan-Boltzmann law,the energy $E$ radiated by a body is given by:
$E = \sigma A T^4 t$
Substituting the values:
$E = (5.67 \times 10^{-8}) \times (10^{-4}) \times (10^3)^4 \times 1$
$E = 5.67 \times 10^{-12} \times 10^{12} \times 1$
$E = 5.67\, J$
Thus,the amount of energy radiated is $5.67\, J$.

(D) According to the Stefan-Boltzmann law,the energy radiated per unit area per unit time by a black body is proportional to the fourth power of its absolute temperature: $E \propto T^4$.
Let $E_1 = E$ at temperature $T_1 = T$.
Let $E_2$ be the energy radiated at temperature $T_2 = T/2$.
Using the ratio: $\frac{E_2}{E_1} = \left( \frac{T_2}{T_1} \right)^4$.
Substituting the values: $\frac{E_2}{E} = \left( \frac{T/2}{T} \right)^4 = \left( \frac{1}{2} \right)^4 = \frac{1}{16}$.
Therefore,$E_2 = E/16$.

(B) According to Stefan-Boltzmann law,the rate of energy radiation $E$ from a black body is proportional to the fourth power of its absolute temperature $T$,given by $E = A\sigma T^4$,where $A$ is the surface area and $\sigma$ is the Stefan-Boltzmann constant.
Given initial temperature $T_1 = 7^{\circ}C = 7 + 273 = 280\,K$.
Given final temperature $T_2 = 287^{\circ}C = 287 + 273 = 560\,K$.
The ratio of the rate of energy radiation is given by $\frac{E_2}{E_1} = (\frac{T_2}{T_1})^4$.
Substituting the values: $\frac{E_2}{E_1} = (\frac{560}{280})^4 = (2)^4 = 16$.
Therefore,the rate of energy radiation increases by $16$ times.

(A) According to the kinetic theory of matter,all bodies at temperatures above absolute zero $(0 \ K)$ possess thermal energy and emit electromagnetic radiation. Thus,the assertion is correct.
According to the Stefan-Boltzmann law,the total energy radiated per unit surface area of a black body per unit time is directly proportional to the fourth power of its absolute temperature,given by $E = \sigma T^4$. This law explains why the rate of radiation depends on the temperature of the body. Therefore,the reason is correct and provides a valid explanation for the assertion.