Question

Two bodies $P$ and $Q$ have thermal emissivities of ${ϵ}_P$ and ${ϵ}_Q$ respectively. Surface areas of these bodies are same and the total radiant power is also emitted at the same rate. If temperature of $P$ is ${\theta }_P$ kelvin then temperature of $Q$ i.e. ${\theta }_Q$ is equal to

• A. ${\left(\frac{{ϵ}_Q}{{ϵ}_P}\right)}^{1/4}{\theta }_P$
• B. ${\left(\frac{{ϵ}_P}{{ϵ}_Q}\right)}^{1/4}{\theta }_P$
• C. ${\left(\frac{{ϵ}_Q}{{ϵ}_P}\right)}^{1/4}\times \frac{1}{{\theta }_P}$
• D. ${\left(\frac{{ϵ}_Q}{{ϵ}_P}\right)}^4{\theta }_P$

Answer 1

The correct option is B ${\left(\frac{{ϵ}_P}{{ϵ}_Q}\right)}^{1/4}{\theta }_P$

Both $P$ and $Q$ have same radiant power.
$\frac{E_P}{E_Q}=1$
and both have same surface area. So, $\frac{S_P}{S_Q}=1$
Now from Stefan-Boltzmann law $\frac{E_P}{E_Q}=\frac{{ϵ}_P{S}_P{\theta }_P^4}{{ϵ}_Q{S}_Q{\theta }_Q^4}$
${\theta }_Q={\left(\frac{{ϵ}_P}{{ϵ}_Q}\right)}^{1/4}{\theta }_P$