Question

A black body is heated from ${27}^{\circ }$ C to ${127}^{\circ }$ C. The ratio of their energies of radiations emitted will be

• A. 3: 4
• B. 9 : 16
• C. 27 : 64
• D. 81: 256

Answer 1

The correct option is D 81: 256

$\frac{Q_1}{Q_2}=\frac{T_1^4}{T_2^4}={\left(\frac{273+27}{273+127}\right)}^4={\left(\frac{300}{400}\right)}^4=\frac{81}{256}$