At 273° C, the emissive power of a perfect black body is R. What is its value at 0° C, ?

Question

At 273° C, the emissive power of a perfect black body is R. What is its value at 0° C, ? (A)  (R/4)  (B)  (R/16)  (C)  (R/2)  (D) None of the above

Tardigrade Admin · Accepted Answer

From Stefans law, the total radiant energy emitted per second per unit surface area of a black body is proportional t6 the fourth power of the absolute temperature (T) of the body.

∴ E = σ T^{4}

where,

σ

is Stefan's constant.
Given,

E_{1} = R, T_{1} = 27 3^{\circ} C

= 273 + 273 = 546 K

T_{2} = 0^{\circ} C = 273 K

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

\Rightarrow E_{2} = \frac{T _{2}^{4}}{T _{1}^{4}} E_{1}

\Rightarrow E_{2} = \frac{( 273 ) ^{4}}{( 546 ) ^{4}} R

∴ E_{2} = \frac{R}{16}

Q. At $27 3^{\circ} C,$ the emissive power of a perfect black body is R. What is its value at $0^{\circ} C,$ ?

$\frac{R}{4}$

$\frac{R}{16}$

$\frac{R}{2}$

None of the above

Q. At 273∘C, the emissive power of a perfect black body is R. What is its value at 0∘C, ?

4R​

16R​

2R​

None of the above

Q. At $27 3^{\circ} C,$ the emissive power of a perfect black body is R. What is its value at $0^{\circ} C,$ ?

$\frac{R}{4}$

$\frac{R}{16}$

$\frac{R}{2}$