If temperature of a black body increases from -73° C to 327° C, then ratio of emissive power at these two temperatures is

Question

If temperature of a black body increases from -73° C to 327° C, then ratio of emissive power at these two temperatures is (A) 27: 1 (B) 81: 1 (C) 1: 27 (D) 1: 81

Tardigrade Admin · Accepted Answer

According to Stefan's law, the radiant energy emitted by a perfectly black body per unit area per second (i.e., emissive power or radiancy) is directly proportional to the fourth power of its absolute temperature.
i.e.,

E \propto T^{4}

or

E = σ T^{4}

where

σ

is Stefan's constant whose value is

5.67 \times 1 0^{- 8} W / m^{2} K^{4}

As

T K = t^{\circ} C + 273

∴ T_{1} = - 7 3^{\circ} C + 273 = 200 K

;

T_{2} = 32 7^{\circ} C + 273 = 600 K

∴ \frac{E _{1}}{E _{2}} = (\frac{T _{1}}{T _{2}})^{4}

= (\frac{200 K}{600 K})^{2} = (\frac{1}{3})^{4} = \frac{1}{81}

Q. If temperature of a black body increases from $- 7 3^{\circ} C$ to $32 7^{\circ} C$ , then ratio of emissive power at these two temperatures is

27 : 1

81 : 1

1 : 27

1 : 81

Q. If temperature of a black body increases from −73∘C to 327∘C, then ratio of emissive power at these two temperatures is

27 : 1

81 : 1

1 : 27

1 : 81

Q. If temperature of a black body increases from $- 7 3^{\circ} C$ to $32 7^{\circ} C$ , then ratio of emissive power at these two temperatures is