A very small hole in an electric furnace is used for heating metals. The hole nearly acts as a black body. The area of the hole is 200 mm2 . To keep a metal at 727 ° C , heat energy flowing through this hole per sec, in joules, is ( take σ =5.67× 10- 8 W m- 2 K- 4 )

Question

A very small hole in an electric furnace is used for heating metals. The hole nearly acts as a black body. The area of the hole is 200 mm2 . To keep a metal at 727 ° C , heat energy flowing through this hole per sec, in joules, is ( take σ =5.67× 10- 8 W m- 2 K- 4 ) (A) 22.68 (B) 2.268 (C) 1.134 (D) 11.34

Tardigrade Admin · Accepted Answer

By Stefan's law,
if the absolute temperature of a black body is

T

, then the energy

E

emitted per second per unit surface area of the body

E \propto T^{4}

E = σ A T^{4}

Given,

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

T = 727 + 273 = 1000 K

A = 200 \times 1 0^{- 6} m

E = 5.67 \times 1 0^{- 8} \times (200 \times 1 0^{- 6}) \times (1000)^{4}

= 5.67 \times 2 \times 1 0^{- 12} \times 1 0^{12}

= 11.34 J s^{- 1}

$22.68$

$2.268$

$1.134$

$11.34$

Q. A very small hole in an electric furnace is used for heating metals. The hole nearly acts as a black body. The area of the hole is 200mm2 . To keep a metal at 727∘C , heat energy flowing through this hole per sec, in joules, is ( take σ=5.67×10−8Wm−2K−4 )

22.68

2.268

1.134

11.34

By Stefan's law, if the absolute temperature of a black body is T , then the energy E emitted per second per unit surface area of the body E∝T4 E=σAT4 Given, σ=5.67×10−8Wm−2K−4 T=727+273=1000K A=200×10−6m E=5.67×10−8×(200×10−6)×(1000)4 =5.67×2×10−12×1012 =11.34Js−1

$22.68$

$2.268$

$1.134$

$11.34$