Question

Two bodies A and B having equal surface area are maintained at temperatures 10°C and 20°C. The energy of thermal radiations emitted in a given time by A and B are in the ratio:
1. 1 : 1.15
2. 1 : 2.5
3. 1 : 4.2
4. 1 : 5.6

Answer 1

Correct Answer - Option 1 : 1 : 1.15

Concept:

The thermal energy emitted by a body is given as, Q = σAT⁴

where, T is absolute temperature, and σ = (5.67 X 10-8 W/m2/K4) is the Stefan-Boltzmann constant. 

From the given condition, we have,

\(\frac{{{Q_A}}}{{{Q_B}}} = \frac{{T_A^4}}{{T_B^4}}\)

Calculation:

Given:

T_A = 273 + 10 = 283 K, T_B = 273 + 20 = 293 K

\(\frac{{{Q_A}}}{{{Q_B}}} = {\left( {\frac{{283}}{{293}}} \right)^4} = \frac{1}{{1.149}}\)