Heat loss takes place from a body maintained at a temperature of `400^(@)C` to the surrounding air at `30^(@)C` by convection and to the surrounding surfaces at `30^(@)C` by radiation. The Newton's cooling coefficient is 20 `W//m^(2)` K and the Stefan-Boltzmann constant is `5.67 xx 10^(-8) W//m^(2)K^(4)`. If the rate of heat loss by convection is equal to the rate of heat loss by radiation, the emissivity of the body surface is

Options

A. 0.35
B. 0.46
C. 0.55
D. 0.66

Answer

D

Explanation

Given, rate of heat lost by convection = Rate of heat lost by radiation
`rArr h A (T-T_(0)) = eA sigma (T^(4) - T_(0)^(4))`
`rArr 20 xx 370 = e xx 5.67 xx 10^(-8) [(400)^(4) - (30)^(4)]`
`rArr e = (59)/(90) ~~ 0.66`