First slide

Heat transfer-Radiation

Question

A body coated black at 600Ksurrounded by atmosphere at 300K has cooling rate $r_{0}$ the same body at 900K surrounded by the same atmosphere will have cooling rate

Moderate

A

$4 r_{0}$
B

$16 r_{0}$
C

$\frac{16 r_{0}}{3}$
D

$\frac{81 r_{0}}{16}$

Solution

$Stefan's law is E=σ (T^{4} {-T}^{4}_{s}), here E is energy radiated by a black body, σ i s s t e f a n' s c o n s \tan t, T i s t e m p e r a t u r e o f t h e b o d y$
$f o r g i v e n t e m p e r a t u r e T_{1}, T_{2} \frac{E_{1}}{E_{2}} = \frac{{T_{1}}^{4} - {T_{s}}^{4}}{{T_{2}}^{4} - {T_{s}}^{4}} {substitute the given temperature and cooling rate E}_{1} {=r}_{0} in the above equation$
$\frac{r_{0}}{E_{2}} = (\frac{600^{4} - 300^{4}}{900^{4} - 300^{4}}) ⇒ \frac{r_{0}}{E_{2}} = (\frac{6^{4} - 3^{4}}{9^{4} - 3^{4}}) = (\frac{1296 - 81}{6561 - 81}) = (\frac{1215}{6480}) = \frac{3}{16}$
$E_{2} = \frac{16 r_{0}}{3}$

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