Question

If temperature of a black body increases from $-{73}^{o}C$ to ${327}^{o}C$, then ratio of emissive power at the two temperature is

• A. 27 : 1
• B. 81 : 1
• C. 1 : 27
• D. 1 : 81

Answer 1

The correct option is D 1 : 81

Given,

$T_1=-{73}^{0}C={273}^{0}C-{73}^{0}C=200K$

$T_2={327}^{0}C={273}^{0}C+{327}^{0}C=600K$

According to Stephen's law of radiation,

$E\propto T^4$

Ratio, $\frac{E_1}{E_2}=(\frac{T_1^4}{T_2^4}{)}^4=(\frac{200}{600}{)}^4$

$\frac{E_1}{E_2}=\frac{1}{81}$

$E_1:E_2=1:81$

The correct option is D.