Question

A black metal foil is warmed by radiation from a small sphere at temperature ′ T ′ and at a distance ′ d ′ . It is found that the power received by the foil is P. If both the temperature and distance are doubled, the power received by the foil will be :
A 16 P
B 4 P
C 2 P
D P

Answer 1

Black-body radiation is the thermal electromagnetic radiation released by a black body when it is in thermodynamic equilibrium with its surroundings (an idealised opaque, non-reflective body). It has a particular spectrum of wavelengths that are inversely linked to intensity and are only dependent on the body's temperature, which is considered to be uniform and constant for the sake of calculations and theory.

Formula used
$\dfrac{{{P}_{2}}}{{{P}_{1}}}=\dfrac{T_{2}^{4}}{T_{1}^{4}}$

Complete step by step solution:
In terms of temperature, the Stefan–Boltzmann law defines the power emitted by a dark substance. The Stefan–Boltzmann law says that the total energy emitted per unit surface area of a black body across all wavelengths per unit time j* (also known as the black-body radiant emittance) is proportional to the black body's thermodynamic temperature T to the fourth power.
${{j}^{\star }}=\sigma {{T}^{4}}$
The Stefan–Boltzmann constant, which is derived from other known physical constants, is a constant of proportionality.
Multiply the surface area of an item by its total power emitted, A:
$P=A{{j}^{\star }}=A\varepsilon \sigma {{T}^{4}}$
From Stefan' s law $\dfrac{P_{2}}{P_{1}}=\dfrac{T_{2}^{4}}{T_{1}^{4}}=\left(\dfrac{2 \times T}{T}\right)^{4}=\dfrac{16}{1} ;$
The power received by blackmetal foil = Intensity $\times$ Area of foil Intensity $=\dfrac{\mathrm{P}}{4 \pi \mathrm{d}^{2}}(\mathrm{P}=$ power of point sorce, $\mathrm{d}=$ distance from point source)
$\dfrac{\mathrm{I}_{2}}{\mathrm{I}_{1}}=\dfrac{\mathrm{P}_{2}}{\mathrm{P}_{1}} \dfrac{\mathrm{d}_{1}^{2}}{\mathrm{~d}_{2}^{2}}=\dfrac{16 \times \mathrm{d}^{2}}{1 \times(2 \mathrm{~d})^{2}}=4$
The amount of energy received by foil will quadruple.
Radiation energy is proportional to the temperature to the fourth power. The radiation energy absorbed by T will be 16 times that of the initial amount if it is doubled. The intensity of energy received by foil is inversely proportional to the square of the distance d from the sphere, resulting in a $\dfrac{1}{4}$ the reduction in intensity. As a result, the intensity will grow by a factor of four times the original, making B the right answer.

Note:
The radiant emittance $j*$ has energy flow dimensions (energy per unit time per unit area), and the SI units of measure are joules per second per square metre, or watts per square metre, respectively. The kelvin is the SI unit for absolute temperature T. The emissivity of the grey body is $\varepsilon$; if it is a perfect blackbody, $\varepsilon =1$. The emissivity of a material relies on its wavelength in a more broad (and realistic) scenario $\varepsilon =\varepsilon (\lambda )$