Question 1: Thermodynamics & KTG

Question

Two spherical bodies A (radius 6 cm ) and B (radius 18 cm ) are at temperature T
1
and T
2
, respectively.
The maximum intensity in the emission spectrum of A is at 500 nm and in that of B is at 1500 nm.
Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that
of B?

Accepted Answer

According to Wien's displacement law, $\lambda$
m
T = constant
$	herefore$ ${({\lambda _m})_A}{T_A} = {({\lambda _m})_B}{T_B}$
or, ${{{T_A}} \over {{T_B}}} = {{{{({\lambda _m})}_B}} \over {{{({\lambda _m})}_A}}} = {{1500\,nm} \over {500\,nm}}$ or ${{{T_A}} \over {{T_B}}} = 3$ ...... (i)
According to Stefan Boltzmann law, rate of energy radiated by a black body
$E = \sigma A{T^4} = \sigma 4\pi {R^2}{T^4}$ [Here, $A = 4\pi {R^2}$]
$	herefore$ ${{{E_A}} \over {{E_B}}} = {\left( {{{{R_A}} \over {{R_B}}}} ight)^2}{\left( {{{{T_A}} \over {{T_B}}}} ight)^4} = {\left( {{{6\,cm} \over {18\,cm}}} ight)^2}{(3)^4}$ (Using (i))
$ = 9$

Thermodynamics & KTG

Integer / Numeric Answer Type