Question
Two bodies A and B having equal surface areas are maintained at temperatures 10°C and 20°C. The thermal radiation emitted in a given time by A and B are in the ratio:
- 1 : 1.15
- 1 : 2
- 1 : 4
- 1 : 16
✓
Answer
-
1 : 1.15
Explanation:
From Stefan-Boltzmann law, energy of the thermal radiation emitted per unit time by a blackbody of surface area A is given by,
$\text{u}=\sigma\text{AT}^4$
Here, $\sigma$ is Stefan-Boltzmann constant.
The thermal radiation emitted in a given time by A and B will be in the ratio.
$\frac{\text{u}_\text{A}}{\text{u}_\text{B}}=\frac{\text{T}_\text{A}^4}{\text{T}_\text{n}^4}$
$\frac{\text{u}_\text{A}}{\text{u}_\text{B}}=\frac{(273+10)^4}{(273+20)^4}$
$\frac{\text{u}_\text{A}}{\text{u}_\text{B}}=\frac{1}{1.15}$
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