Question

All bodies emit heat energy from their surfaces by virtue of their temperature. This heat energy is called radiant energy of thermal radiation. The heat we receive from the sun is transferred to us by a process which, unlike conduction or convection, does not require the help of a medium in the intervening space which is almost free of particles. Radiant energy travels in space as an electromagnetic spectrum. Thermal radiations travel through vacuum with the speed of light. Thermal radiations obey the same laws of reflection and refraction as light does. They exhibit the phenomena of interference, diffraction and polarization as light does. The emission of radiation from a hot body is expressed in terms of that emitted from a reference body (called the back body) at the same temperature. A black body absorbs and hence emits radiations of all wavelengths. The total energy E emitted by a unit area of black body per second is given by $E = \sigma {T^4}$where T is the absolute temperature of the body and $\sigma$ is a constant known as Stefan’s constant. If the body is not a perfect black body, then $E = \varepsilon \sigma {T^4}$where $\varepsilon$is the emissivity of the body?
When a body A at a higher temperature ${T_1}$is surrounded by another body B at a lower temperature${T_2}$, then the rate of loss of heat from body A will be proportional to:
A. ${T_1}^4$
B. ${({T_1} - {T_2})^4}$
C. $({T_1} - {T_2})$
D. ${T_1}^4 - {T_2}^4$

Answer 1

To solve this question we have to know about the Stephan- Boltzmann equation. We are going to use that in this question. We know that the rate of heat energy is dependent on the emittance, area of the surface and the difference of the temperature of the body.

Complete step by step answer:
In thermodynamics, Boltzmann constant is the physical constant relating average kinetic energy of the gas particles and temperature of the gas represented by k or kB. The value of the Boltzmann constant is measured using J/K or m2Kgs-2K-1. Which is mostly observed in Boltzmann’s entropy formula and Planck’s law of Black body radiation.

We know that, according to the Stephan – Boltzmann equation, the rate of heat energy radiated is $({T_b}^4 - {T_a}^4)eA\sigma$
Where $e$ is the emittance of the body, and sigma in Stephan- Boltzmann constant. $A$ is the area of the surface, so all the options are the right options.

Note: We know that the emittance depends on the nature of the surface. We have to keep that in our mind that all these options are correct here. We can get confused when we see this huge question but this is an easy one.