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 the rate of loss ofheat by the block is given as
`(dQ)/(dt) = sigma A(T^(4)-T_(5)^(4))`
` = 5.67 xx 10^(-8) xx 150 xx 10^(-4) xx [(500)^(4)-(300)^(4)]`
[surfacearea ofcube is `6a^(2) = 150 cm^(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) (1 xx 400) xx 38.556`
`= 0.115^(@)C//S`