First slide

Heat transfer-Radiation

Question

Energy is being emitted from the surface a black body at $127^{\circ} C$ temperature at $1 \times 10^{6} {Js}^{-1} m^{-2}$ , temperature of the black body at which the rate of emission is $16 \times 10^{6} {Js}^{-1} m^{-2}$ will be

Moderate

A

$254^{\circ} c$
B

$508^{\circ} c$
C

$527^{\circ} c$
D

$272^{\circ} c$

Solution

$\begin{array}{rcl} {given temperature of black body is T}_{1} =127+273=400K \\ T_{2} =? \\ e n e r g y e m i t t e d b y b l a c k b o d y i s E_{1} =1 \times 10^{6} {Js}^{- 1} m^{- 2} \\ E_{2} =16 \times 10^{6} {Js}^{- 1} m^{- 2} \\ f r o m s t e f a n' s l a w, \\ {E =σ T}^{4} \\ \Rightarrow & {E α T}^{4} \\ \Rightarrow & \frac{E_{1}}{E_{2}} = {(\frac{T_{1}}{T_{2}})}^{4} \\ \Rightarrow & \frac{E_{1}}{E_{2}} = {(\frac{400}{T_{2}})}^{4} \\ \Rightarrow & {(\frac{E_{1}}{E_{2}})}^{\frac{1}{4}} = (\frac{400}{T}) \\ \begin{array}{rcl} \Rightarrow & {(\frac{10^{6}}{{16×10}^{6}})}^{\frac{1}{4}} = (\frac{400}{T}) \end{array} \Rightarrow \frac{\begin{array}{rcl} 1 \end{array}}{2} \begin{array}{rcl} = \end{array} \begin{array}{rcl} (\frac{400}{T}) \end{array} \begin{array}{rcl} \Rightarrow \end{array} \begin{array}{rcl} T \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} 800 \end{array} \begin{array}{rcl} K \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} 800 \end{array} \begin{array}{rcl} - \end{array} \begin{array}{rcl} 273 \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} 527 \end{array} \begin{array}{rcl} ° \end{array} \begin{array}{rcl} C \end{array} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App