A cube of mass 1 kg and volume `125cm^2` is placed in an evacuated chamber at `27^@C`. Initially temperature of block is `227^@C`. Assume block behaves like a block body, find the rate of cooling of block if specific heat of the material of block is `400 J//kg-K`.

Explanation

In this case ther rate of loss of heat by the block is given as
`(dQ)/(dt)=sigmaA(T^$-T_0^4)`
`=5.67xx10^-8xx150xx10^-4xx[(500)^4-(300)^4]` (surface area of cube is `6a^2=150cm^2`)
If `(dT)/(dt)` is rate of cooling of block then we have
`(dQ)/(dt)=ms(dT)/(dt)`
or `(dT)/(dt)=(1)/(ms)(dQ)/(dt)=(1)/(1xx400)xx38.556=0.115^@C//s`