The rate of radiation by a black body is R at | vedclass

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

Question

(C) The rate of radiation for a black body is given by the Stefan-Boltzmann law: $R = \left(\frac{dQ}{dt}\right)_1 = e A \sigma T^4$. Since it is a bl

Vedclass · Accepted Answer

$16.2$

Vedclass · Answer

(C) The rate of radiation for a black body is given by the Stefan-Boltzmann law: $R = \left(\frac{dQ}{dt}ight)_1 = e A \sigma T^4$. Since it is a black body,emissivity $e = 1$,so $R = A \sigma T^4$. For the second body,the rate of radiation is: $\left(\frac{dQ}{dt}ight)_2 = e' A' \sigma (T')^4$. Given $e' = 0.2$,$A' = A$,and $T' = 3T$: $\left(\frac{dQ}{dt}ight)_2 = 0.2 	imes A 	imes \sigma 	imes (3T)^4$. $\left(\frac{dQ}{dt}ight)_2 = 0.2 	imes A 	imes \sigma 	imes 81 T^4$. $\left(\frac{dQ}{dt}ight)_2 = 16.2 	imes A \sigma T^4$. Substituting $R = A \sigma T^4$,we get: $\left(\frac{dQ}{dt}ight)_2 = 16.2 R$.

The rate of radiation by a black body is $R$ at temperature $T$. Another body has the same area but an emissivity of $0.2$ and a temperature of $3T$. Its rate of radiation is: (in $R$)

Similar Questions

$A$ 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 $x$.

Two spheres made of the same material have radii in the ratio $1 : 2$. If both are at the same temperature,what is the ratio of the energy emitted per second by them?

$A$ black body surface with an area of $8 \ cm \times 4 \ cm$ emits energy at a rate of $E$ per second at a temperature of $127^{\circ}C$. If the length and width are halved and the temperature is increased to $327^{\circ}C$,find the new rate of energy emission.

If the temperature of the Sun gets doubled, the rate of energy received on the Earth will increase by a factor of

Vedclass Test Series

Exam Paper Generator

Online Exam Module