Question

The power received at distance d from a small metallic sphere of radius r < < d and at absolute temperature T is p. If the temperature is doubled and distance reduced to half of the initial value, then the power received at that point will be:

(A) 4p

(B) 8p

(C) 32p

(D) 64p

Answer 1

Hint: When a metallic body is heated, it radiates heat in the form of electromagnetic radiation (mostly infrared). This heat energy received at a point in free space per unit time is known as power of that heated body per unit area at a distance $r$. The power received at a distance r is inversely proportional to the distance squared. 

Formula used:

\[P = \varepsilon \sigma A{T^4} \]

\[I = \dfrac{P}{{4\pi {r^2}}} \]

Where, $\varepsilon $ is the emissivity of the body (for black body it is 1), $\sigma $ is the Stefan-Boltzmann Constant, $A$ is the surface area of the body and $T$ is the absolute temperature of the body.

Complete step by step answer:

Radiation is a form of transfer of heat via electromagnetic radiation. This radiation, mostly infrared, carries heat energy in the form of photons from one place to another. A metallic body when heated does the same. Radiation does not require a medium of propagation like convection and conduction. Therefore, we are able to receive heat energy from the sun.

Now, as we move further away, the heat energy flux remains the same for the whole surface area of the sphere (imaginary) at that point but the number of flux lines per unit area decreases. This means that the energy per unit time (power) also decreases at farther distances. So, the intensity of the power at farther point is described as:

\[I = \dfrac{P}{{4\pi {r^2}}} \]

\[\Rightarrow I = \dfrac{{\varepsilon \sigma A{T^4}}}{{4\pi {r^2}}} \]

\[\Rightarrow I = c\dfrac{{{T^4}}}{{{r^2}}}  \]

Except T and r, the rest of the terms are constants, so we have combined them in a single constant c.

Initially at point d distance away, we have,

\[I = c\dfrac{{{T^4}}}{{{r^2}}} = p\]

Now the distance is halved and temperature is doubled, so,

\[{I_{new}} = c\dfrac{{{{(2T)}^4}}}{{{{\left( {\dfrac{d}{2}} \right)}^2}}} \]

\[\Rightarrow  {I_{new}} = {2^6}c\dfrac{{{T^4}}}{{{d^2}}} \]

\[\therefore {I_{new}} = 64\,p \]

The power received at the new point when distance is halved and temperature is doubled is 64 times the power received in initial conditions.

Therefore, the correct answer is option (D).

Note: The area given in the Stefan-Boltzmann law is the surface area of the given radiating body while at the area in the intensity formula is the area of an imaginary sphere which indicates that at every point on that sphere the power per unit area received is the same. So, don’t get perplexed.