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-

Given: Temperature of body A, T_A=10^∘C=273+10=283 K Temperature of body nbsp;B, nbsp;T_B=20^∘C=273+20=293 K we know from Stefan's law, u_Au_B=σ A T_A^4σ ...

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Standard XII

Physics

Blackbody

Two bodies A ...

Question

Two bodies $A$ and $B$ having equal surface areas are maintained at temperatures $10^{\circ} C$ and $20^{\circ} C$ . The thermal radiation emitted in a given time by $A$ and $B$ are in the ratio,

1 : 1.15

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1 : 2

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1 : 4

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1 : 1.6

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Solution

The correct option is A $1 : 1.15$
Given:
Temperature of body A, $T_{A} = 10^{\circ} C = 273 + 10 = 283 K$

Temperature of body B, $T_{B} = 20^{\circ} C = 273 + 20 = 293 K$

we know from Stefan's law,
$\frac{u_{A}}{u_{B}} = \frac{σ A T_{A}^{4}}{σ A T_{B}^{4}} = \frac{T_{A}^{4}}{T_{B}^{4}} = {(\frac{283}{293})}^{4}$

$\Rightarrow \frac{u_{A}}{u_{B}} = \frac{1}{1.15}$

Hence, $(A)$ is the correct answer.

$\begin{matrix} Why this question ? Caution : when using the Stefan's law, remember to convert temperature into SI unit, i.e., Kelvin \end{matrix}$

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