Question

The surface area of a black body
maintained at 127°C radiating energy
at the rate of 1459.2 Js- is ... (o
(o=
5.7 x 10-8]m-2 s-4k-4)
a) 4m2
c) 2m
b) 3m2
d) 1m2​

Answer 1

d option is the answer

10thx=follow

Answer 2

Answer:

The surface are of a black body is  $1 m^{2}$

Explanation:

Given:

The surface area of a black body maintained at 127°

Radiating energy Q is 1459.2 Js

To find: The surface area of the black body

Formula used: Q = σA$T^{4}$

Solution:

Q = 1459.2T  = 127 c

σ = 5.7 x 10-8

Convert celcius to kelvin

Kelvin = Celsius + 273.15.

T = 127 + 273

T = 400 K

Surface area of the black body a = ?

Stefan-Boltzmann law

The Stefan-Boltzmann law states that a surface's overall radiant heat output is inversely correlated with its absolute temperature. The Stefan-Boltzmann equation is the term for the statement Q = ρ AT4.

Q = σA$T^{4}$

A = Q /σA $T^{24$

A = $\frac{1459.2}{5.7 * 10^{-8} * (400)^{4} *1 }$

A = $\frac{1459.2}{5.7 * 10^{-8} *(4 *(10^{2} )^{4} }$

A = $\frac{1459.2}{5.7 *(4)^{4} }$

A = $\frac{1459.2}{5.7 * 256}$

A = $\frac{1459.2}{1459.2}$

A =  $1 m^{2}$

Final answer:

The surface are of a black body is  $1 m^{2}$

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