Laws of heat transfer - class-XI online quiz

According to Stefan-Boltzmann law, the total energy radiated per unit surface area per unit time is E = sigma * T^4. Given E = 459.27 * 10^4 W/m^2 and taking Stefan's constant sigma approx 5.67 * 10^-8 W/m^2/K^4, solving for T gives T = (459.27 * 10^4 / 5.67 * 10^-8)^(1/4) = (8.1 * 10^12)^(1/4) = 3000 K.

According to Planck's quantum theory, radiation is emitted in the form of small packets of energy and is not continuous. Quantum of energy can be expressed by the relation E=hv, where h is Planck's constant and v is the frequency.

Prevost postulated that  radiative equilibrium is the condition where a steady state system is in dynamic equilibrium, with equal incoming and outgoing radiative flux and negligible heat transfer by conduction and convection.
So by extended logic, at $0K$ equilibrium cannot be reached.

Radiation emitted per second depends on the temperature of the body.
Stefan's law states that the rate of emission of radiant energy by unit area of perfectly black body is directly proportional to the fourth power of its absolute temperature.
      $E \propto AT^4$
or   $E \propto r^2$
($\because A= \pi r^2$ and T is same for both the spheres)
where r is radius of sphere.
$\frac{E _1}{E _2} = \frac{r^2 _1}{r^2 _2}$
$=\left(\frac{2}{1}\right)^2=\frac{4}{1}$
$=4:1$
Note : A black body at absolute temperature T surrounded by another black body at absolute temperature $T _0$ not only loses an amount of energy $\sigma T^4$, thus the amount of heat lost by the former per unit time is given by 
$E=\sigma (T^4-T _0^4)$
This law is stefan Boltzmann's law.

Energy radiated E is proportional to T^4. If E' = 2E, then (T'/T)^4 = 2, so T' = T * 2^(1/4). T = 400 + 273 = 673 K. T' = 673 * 1.189 = 800 K. Converting back to Celsius is not required as the options provide Kelvin.

Power P = sigma * A * T^4. P1 = sigma * 4 * pi * (4)^2 * (2000)^4. P2 = sigma * 4 * pi * (1)^2 * (4000)^4. Ratio P1/P2 = (16 * 2000^4) / (1 * 4000^4) = 16 / 16 = 1. The ratio is 1:1.

The rate of radiation emitted by a body is given by Stefan-Boltzmann law: P = e * sigma * A * T^4 = e * sigma * (4 * pi * r^2) * T^4. Thus, P is proportional to r^2 * T^4. For sphere 1: r1 = 1 m, T1 = 4000 K. For sphere 2: r2 = 4 m, T2 = 1000 K. The ratio P1 / P2 = (r1/r2)^2 * (T1/T2)^4 = (1/4)^2 * (4000/1000)^4 = (1/16) * (4^4) = (1/16) * 256 = 16:1.

Given: The intensity of the incident radiations is $0.5\ W/m^2$.

For approximate calculation of the time taken,

Heat Q = n * Cp * deltaT. Q = 5 kmol * 6.994 kcal/kmol K * (305 - 295) K = 5 * 6.994 * 10 = 349.7 kcal.

Water has a very high specific heat capacity, meaning it can absorb a large amount of heat energy for a relatively small increase in temperature, making it an excellent coolant.

$\begin{array}{l} \dfrac { { \lambda { m^{ 1 } } } }{ { \lambda m } } =\dfrac { T }{ { { T^{ 1 } } } }  \ \dfrac { { \lambda { m^{ 1 } } } }{ { 5000A } } =\dfrac { { 1227+273k } }{ { 2727+273k } }  \ \lambda { m^{ 1 } }=2500A \end{array}$