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 that 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 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 black 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 a 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 $E=\epsilon \sigma T^4$where $\epsilon$ 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

The correct option is D $(T_1^4-T_2^4)$

From rate of transfer of heat, we can say that

$\frac{dQ}{dt}=eA\sigma (T_1^4-T_2^4)$

where $e$-emissivity of body

$A$- surface area

$\sigma$- stefan's constant

$T_1$- higher temperature

$T_2$- lower temperature

Rate of heat transfer $\alpha (T_1^4-T_2^4)$