The top of an insulated cylindrical container is covered by a disc having emissivity 0.6 and conductivity 0.167 `"WK"^(– 1)m^(–1)` and thickness 1 cm. The temperature is maintained by circulating oil as shown in figure. Find the radiation loss to the surrounding in `"Jm"^(–2)s^(–1 )`if temperature of the upper surface of the disc is `127^(@)"C"` and temperature of the surrounding is `27^(@)"C"`.

Answer

(i) `595 W//m^(2)` (ii) `162.6 ^(@)C`

Explanation

(i) Rate of heat loss per unit area due to radiation
`I = esigma(T^(4) - T_(0)^(4))`
Here . `T = 127 + 273 = 400K`
and `T_(0) = 27 + 273 = 300K`
`therefore I = 0.6 xx (17)/(3) xx 10^(-8)[(400)^(4) - (300)^(4)] = 595 W//m^(2)`
(ii) Let `theta` can be the temperature of the oil.
Then , rate of heat flow through conduction = rate of heat loss due to radiation
`therefore ("temperature difference")/("thermal resistance") = (595)A`
`implies ((theta - 127))/(((l)/(KA))) = (595)A`
Here A = area of disc , K = thermal conductivity and l= thickness (or length) of disc
` therefore (theta-127) (K)/(l) = 595`
`therefore theta = 595((l)/(K)) + 127`
=`(595xx10^(-2))/(0.167) + 127 = 162.6^(@)C`