1.

A body at a temperature of 727^(@)C and having surface area 5 cm^(2), radiations 300 J of energy each minute. The emissivity is(Given Boltzmann constant =5.67xx10^(-8) Wm^(-2)K^(-4)

Answer»

e = 0.18
e = 0.02
e = 0.2
e = 0.15

Solution :As `Q =esigmaT^4` At
So `e=Q/(sigmaT^4 "At")=300/((5-67xx10^(-8))xx(1001)^4XX(5xx10^(04))xx60)`
= 0.18


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