A cubical block of mass `1.0kg` and edge `5.0cm` is heated to `227^(@)C` . It is kept in an evacuated chamber maintained at `27^(@)C` . Assming that the block emits radiation like a blackbody, find the rate at which the temperature of the block will decreases. Specific heat capacity of the material of the block is `400Jkg^(-1)K^(-1)` .

Explanation

Rate fo cooling
`(dT)/(dt)` = `(esigmaA)/(ms)(T^(4)-T_(0)^(4))`
Given, `sigma=1`, Area,
`A = 6a^(2)= 6 xx (5xx10^(-2)^(2)=150 xx 10^(-4)m^(2)`
Temperature, T = 227 + 273 = 500K
`T_(0) (27 + 273) = 300K
Given, m = 1kg, s=400J/kg.K
`=`6 xx10^(-8) xx 150 xx 10^(-4)[(500)^(4)-(300)^(4)]`
`=2.25 xx 10^(-12)xx(625 -8)xx10^(8) = 0.12^(@)C//s`