The coefficients of reflection and transmission of the surface of a thin plate are 0.22 and 0.04 respectively. If 250 calories of radiant heat is incident on the surface of the plate, how much heat is absorbed by the surface?

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Solution

We haave
$a + r + t = 1$
where
$a =$ Coefficient of absorption
$r = 0.22$ =Coefficient of radiation
$t = 0.04$ =Coefficient of transmission
$a + 0.22 + 0.04 = 1$
$a = 0.74$
Now,
$Q_{a} = a Q = 0.74 \times 250 = 185 J$
Therefore radiant heat absorbed is 185 J

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The coefficients of reflection and transmission of the surface of a thin plate are 0.22 and 0.04 respectively. If 250 calories of radiant heat is incident on the surface of the plate, how much heat is absorbed by the surface?

We haavea+r+t=1wherea=Coefficient of absorptionr=0.22=Coefficient of radiationt=0.04=Coefficient of transmissiona+0.22+0.04=1a=0.74Now,Qa=aQ=0.74×250=185 JTherefore radiant heat absorbed is 185 J

We haave
$a + r + t = 1$
where
$a =$ Coefficient of absorption
$r = 0.22$ =Coefficient of radiation
$t = 0.04$ =Coefficient of transmission
$a + 0.22 + 0.04 = 1$
$a = 0.74$
Now,
$Q_{a} = a Q = 0.74 \times 250 = 185 J$
Therefore radiant heat absorbed is 185 J