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(C) According to the Stefan-Boltzmann law, the thermal energy radiated per second $(Q)$ by a black body is given by $Q = \sigma A T^4$, where $\sigma$

Vedclass · Accepted Answer

$16: 81$

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(C) According to the Stefan-Boltzmann law, the thermal energy radiated per second $(Q)$ by a black body is given by $Q = \sigma A T^4$, where $\sigma$ is the Stefan-Boltzmann constant, $A$ is the surface area, and $T$ is the absolute temperature in Kelvin.Given that the surface areas of both bodies $A$ and $B$ are equal $(A_A = A_B = A)$, the ratio of the energy radiated is:$\frac{Q_A}{Q_B} = \frac{\sigma A T_A^4}{\sigma A T_B^4} = \left(\frac{T_A}{T_B}\right)^4$Convert the temperatures from Celsius to Kelvin:$T_A = 27^{\circ} C = 27 + 273 = 300 \ K$$T_B = 177^{\circ} C = 177 + 273 = 450 \ K$Substitute the values into the ratio:$\frac{Q_A}{Q_B} = \left(\frac{300}{450}\right)^4 = \left(\frac{2}{3}\right)^4$$\frac{Q_A}{Q_B} = \frac{16}{81}$Thus, the ratio of the thermal energy radiated per second by $A$ to that by $B$ is $16: 81$.

Two black bodies $A$ and $B$ have equal surface areas and are maintained at temperatures $27^{\circ} C$ and $177^{\circ} C$ respectively. What will be the ratio of the thermal energy radiated per second by $A$ to that by $B$?

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