NEET-XII-Physics
28: Heat Transfer
- #5Two bodies A and B having equal surface areas are maintained at temperature 10°C and 20°C. The thermal radiation emitted in a given time by A and B are in the ratio
(a) 1 : 1.15
(b) 1 : 2
(c) 1 : 4
(d) 1 : 16digAnsr: aAns : (a) 1 : 1.15
From Stefan-Boltzmann law, energy of the thermal radiation emitted per unit time by a blackbody of surface area A is given by
`` u=\sigma A{T}^{4}``
Here, `` \sigma `` is Stefan-Boltzmann constant.
The thermal radiation emitted in a given time by A and B will be in the ratio
`` \frac{{u}_{\,\mathrm{\,A\,}}}{{u}_{\,\mathrm{\,B\,}}}=\frac{{{T}_{\,\mathrm{\,A\,}}}^{4}}{{{T}_{\,\mathrm{\,B\,}}}^{4}}``
`` \frac{{u}_{\,\mathrm{\,A\,}}}{{u}_{\,\mathrm{\,B\,}}}=\frac{(273+10{)}^{4}}{(273+20{)}^{4}}``
`` \frac{{u}_{\,\mathrm{\,A\,}}}{{u}_{\,\mathrm{\,B\,}}}=\frac{1}{1.15}``
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