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Question: The rate of emission of radiation of a black body of temperature 27<sup>0</sup>C is E<sub>1</sub>. I...

The rate of emission of radiation of a black body of temperature 270C is E1. If its temperature is increased to 3270C, the rate of emission of radiation is E2. The relation between E1& E2 is –

A

E2 = 24 E1

B

E2 = 16 E1

C

E2 = 8 E1

D

E2 = 4 E1

Answer

E2 = 16 E1

Explanation

Solution

T1 = 27 + 273 = 300 K

T2 = 327 + 273 = 600 K

By Stefan's law

E1E2\frac{E_{1}}{E_{2}}= T14T24\frac{T_{1}^{4}}{T_{2}^{4}} = (300600)4\left( \frac{300}{600} \right)^{4}

\ E2 = 16 E1