Question

A metal ball of surface area 200 square cm, temperature 527°C is surrounded by a vessel at 27°C. If the emissivity of the metal is 0.4, then the rate of loss of heat from the ball is approximately σ = 5.67 × 10⁻⁸ $\frac{joule}{m^{2}\times\ sec\ \times\ K{2}}$(a) 108 joule(b) 168 joule(c) 182 joule(d) 192 joule

Answer 1

By the stefan boltzmann law,

Rate of heat loss from the ball will be

E = seA(T4 -Ts4)

where s =Stefan-Boltzmann constant = 5.67 e-8 units (SI units, I dont remember the units exactly, but you should always right the correct units)

e = emissivity = 0.4

A = 200 cm2 = 200 * 10-6 m2 = 2e-4 m2

T = Temp. of ball = 527 C = 527+273 K = 800 K

Ts = Temp of surroundings = 27 C = 300 K

Putting in all the values,

E = (5.67e-8)(0.4)(2e-4)(8004-3004) = 1.85 J/s

Answer 2

Answer:

182 joules

Explanation:

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

tQ=σA(T14−T24)×e

so,T1=273+527K=800K

T2=273+27K=300K

and A=200×10−4m2

so,tQ=5.67×10−8×2×10−2[(8004)−(3004)]×(0.4)=(182.12) Joule