The rate of radiation of a black body at $0^{0}$ C is EJ/sec. The rate of radiation of this black body at $273^{0}$ C will be

16 E

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

8 E

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4 E

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
16 E

$\frac{E_{1}}{E_{2}} = {(\frac{T_{1}}{T_{2}})}^{4} \Rightarrow \frac{E}{E_{2}} = {(\frac{273 + 0}{273 + 273})}^{4} \Rightarrow E_{2} = 16 E .$

Suggest Corrections

Similar questions

The rate of radiation of a black body at $0^{o} C$ is E J/sec. The rate of radiation of this black body at $273^{o} C$ will be

^{0}

^{0}

0^{\circ} C

E

273^{\circ} C

273^{\circ} C

E

0^{\circ} C

0

^{o}

E

273

^{o}

Join BYJU'S Learning Program