Question

A metal ball of surface area 200 square cm, temperature 527ºC
is surrounded by a vessel at 27ºC. If the emissivity of the
metal is 0.4, then the rate of loss of heat from the ball is
approximately σ = 5.67 x 10⁻⁸ joule/m² x sec x K²
(a) 108 joule (b) 168 joule
(c) 182 joule (d) 192 joule

Answer 1

Answer:

option C is the correct answer ...........................

Answer 2

Answer:

From stefan's law we know that,rate of loss of heat is given as,

Q/t=σA(T⁴1−T⁴2)×e

so,T1=273+527K=800K

T2=273+27K=300K

and A=200×10^−4m^2

so,Q/t =5.67×10^−8×2×10^2−2[(800^4)−(300^4)]×(0.4)=(182.12) Joule

therefore option (c)

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