Question

The temperature at which a black body of unit area loss its energy at the rate of 1 joule/second is:
(A) $-65^{\circ} \mathrm{C}$
(B) $65^{\circ} \mathrm{C}$
(C) 65 K
(D) None of these

Answer 1

Hint: We know that Planck’s radiation law, a mathematical relationship formulated in 1900 by German physicist Max Planck to explain the spectral-energy distribution of radiation emitted by a blackbody (a hypothetical body that completely absorbs all radiant energy falling upon it, reaches some equilibrium temperature, and then reemits. Blackbody, in physics, a surface that absorbs all radiant energy falling on it. The term arises because incident visible light will be absorbed rather than reflected, and therefore the surface will appear black.

Complete step by step answer
We know that black-body radiation has a characteristic, continuous frequency spectrum that depends only on the body's temperature, called the Planck spectrum or Planck's law. As the temperature increases past about 500 degrees Celsius, black bodies start to emit significant amounts of visible light. A black body is one that absorbs all the EM radiation light that strikes it. To stay in thermal equilibrium, it must emit radiation at the same rate as it absorbs it so a black body also radiates well.
Area is unity so $\mathrm{so} \mathrm{E}=\sigma \mathrm{T}^{4}$ which is $1 \mathrm{J} / \mathrm{s}$
So, we have ${{\text{T}}^{4}}=1/\sigma $
After we put the values we get:
$1 /\left(5.67 \times 10^{-8}\right)=1763.66 \times 10^{4}$
So $\mathrm{T}=64.8 \mathrm{K}$

Option c is correct.

Note: Thus, we conclude that blackbody refers to an opaque object that emits thermal radiation. A perfect blackbody is one that absorbs all incoming light and does not reflect any. At room temperature, such an object would appear to be perfectly black. A blackbody is defined as one that absorbs all incident radiation so that all the radiation that comes from its surface is its own emission. Since the intensity of the energy at any temperature and wavelength can be determined using the Planck Law of radiation. A blackbody radiation source with a known temperature, or, whose temperature can be measured, is usually used for calibrating and testing the radiation thermometers.