Two bodies P and Q have thermal emissivities | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

If the temperature of a black body increases from $7^oC$ to $287^oC$,then the rate of energy radiation increases by a factor of:

Two spheres $P$ and $Q$,of the same color,having radii $8 \; cm$ and $2 \; cm$ are maintained at temperatures $127^{\circ}C$ and $527^{\circ}C$ respectively. The ratio of energy radiated by $P$ and $Q$ is:

The total radiant energy per unit area,normal to the direction of incidence,received at a distance $R$ from the centre of a star of radius $r$,whose outer surface radiates as a black body at a temperature $T \ K$ is given by

Obtain the equation for the rate of emission of heat for a perfect black body by using the Stefan-Boltzmann law.

$A$ black body at $227^{\circ}C$ radiates heat at the rate of $20 \, cal \, m^{-2} \, s^{-1}$. When its temperature is raised to $727^{\circ}C$,the rate of heat radiation will be $x \, cal \, m^{-2} \, s^{-1}$. Find $x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module