The power of a black body at temperature 200 K is 544 W. Its surface area is (σ=5.67 × 10-8 Wm -2 K -4)

Question

The power of a black body at temperature 200 K is 544 W. Its surface area is (σ=5.67 × 10-8 Wm -2 K -4) (A) 6 × 10-2 m 2 (B)  6 m2  (C)  6× 10-6 m2  (D)  6× 102 m2

Tardigrade Admin · Accepted Answer

According to Stefan's law (Q/A l)=σ T4 Here (Q/t)=544, T=200 K σ=5.67 × 10-8 W m-2 K-4 =A=(Q/σ t T4)=(544/5.67 × 10-8 ×(200)4)=6 m2

Q. The power of a black body at temperature $200 K$ is $544 W$ . Its surface area is
$(σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4})$

$6 \times 1 0^{- 2} m^{2}$

$6 m^{2}$

$6 \times 1 0^{- 6} m^{2}$

$6 \times 1 0^{2} m^{2}$

According to Stefan's law
$\frac{Q}{A l} = σ T^{4}$
Here $\frac{Q}{t} = 544, T = 200 K$
$σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}$
$= A = \frac{Q}{σ t T ^{4}} = \frac{544}{5.67 \times 1 0 ^{- 8} \times ( 200 ) ^{4}} = 6 m^{2}$

Q. The power of a black body at temperature 200K is 544W. Its surface area is (σ=5.67×10−8Wm−2K−4)

6×10−2m2

6m2

6×10−6m2

6×102m2