Two bodies P and Q have thermal emissivities | vedclass

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(B) According to the Stefan-Boltzmann law,the radiant power $E$ emitted by a body is given by $E = \varepsilon \sigma A T^4$,where $\varepsilon$ is em

Vedclass · Accepted Answer

${\left( {\frac{{{\varepsilon _P}}}{{{\varepsilon _Q}}}} ight)^{1/4}}\,{	heta _P}$

Vedclass · Answer

(B) According to the Stefan-Boltzmann law,the radiant power $E$ emitted by a body is given by $E = \varepsilon \sigma A T^4$,where $\varepsilon$ is emissivity,$\sigma$ is the Stefan-Boltzmann constant,$A$ is the surface area,and $T$ is the absolute temperature.Given that the radiant power $E$ and surface area $A$ are the same for both bodies $P$ and $Q$,we have:$E_P = E_Q$$\varepsilon_P \sigma A_P 	heta_P^4 = \varepsilon_Q \sigma A_Q 	heta_Q^4$Since $A_P = A_Q$ and $\sigma$ is constant,the equation simplifies to:$\varepsilon_P 	heta_P^4 = \varepsilon_Q 	heta_Q^4$Rearranging for $	heta_Q$:$	heta_Q^4 = \frac{\varepsilon_P}{\varepsilon_Q} 	heta_P^4$$	heta_Q = \left( \frac{\varepsilon_P}{\varepsilon_Q} ight)^{1/4} 	heta_P$Thus,the correct option is $B$.

Two bodies $P$ and $Q$ have thermal emissivities of $\varepsilon_P$ and $\varepsilon_Q$ respectively. Surface areas of these bodies are same and the total radiant power emitted is also at the same rate. If the temperature of $P$ is $\theta_P$ Kelvin,then the temperature of $Q$ i.e.,$\theta_Q$ is:

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