Question 759: Thermodynamics & KTG

Question

Rate of radiation by a black body is ' R ' at temperature 'T'. Another body has same area but emissivity is 0.2 and temperature 3T. Its rate of radiation is

Accepted Answer

To determine the rate of radiation for a black body and another body with the same area but different emissivity and temperature, let's review the calculations.
Rate of Radiation for a Black Body
The formula for the rate of radiation from a black body is given by:
$ R = \left(\frac{dQ}{dt}\right)_1 = e A \sigma T^4 $
Since the emissivity ($e$) of a black body is 1, the equation simplifies to:
$ R = A \sigma T^4 $
Rate of Radiation for Another Body
Now, consider another body with the same surface area but emissivity $e'$ of 0.2 and temperature $3T$. The rate of radiation formula for this body is:
$  \left(\frac{dQ}{dt}\right)_2 = e' A \sigma (T')^4  $
Substituting the given values:
$  \left(\frac{dQ}{dt}\right)_2 = 0.2 A \sigma (3T)^4  $
Simplifying the equation:
$  \left(\frac{dQ}{dt}\right)_2 = 0.2 A \sigma \cdot 81 T^4  $
$  \left(\frac{dQ}{dt}\right)_2 = 16.2 A \sigma T^4  $
Since $A \sigma T^4$ is equivalent to the rate of radiation $R$ from the black body, we can write:
$  \left(\frac{dQ}{dt}\right)_2 = 16.2 R  $
Thus, the rate of radiation for the second body is $16.2 R$.

Answer

$(162) R$

Answer

$(81) \mathrm{R}$

Answer

$(16.2) R$

Answer

$(8.1) \mathrm{R}$

Thermodynamics & KTG

Rate of radiation by a black body is ' R ' at temperature 'T'. Another body has same area but emissivity is 0.2 and temperature 3T. Its rate of radiation is

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