A solid copper sphere of dimater `10mm` is cooled to temperature of `150K` and is then placed in an enclousure at `290K` Assuming that all interchange of heat is by radiation, calculate the initial rate of rise of temperature of the sphere The sphere may be treated as a black body `rho_(copper) =8.93xx 10^(3)kg//m^(3)`
`s = 3.7xx10^(2) Jkg^(-2) K^(-1) , sigma = 5.7 xx 10^(8) Wm^(-2) K^(-4)` .

Options

A. `0.68 K//s`
B. `0.68 K//s`
C. `0.34 K//s`
D. `0.034K//s`

Answer

B

Explanation

Using Stefan 's law the rate of increase of temperature is `(dT)/(dt) = (sigma)/(rho) (A)/(V) = ((T_(0)^(4)-T^(4)))/(s)`
where `rho = 8.93 xx 10^(30 kg//m^(3)`
`s =3.7 xx 10^(2) J//kg//K A//V` area/volume ratio
`(A)/(V) = (6)/(d),d` dimeter of the sphere
`T_(0)` = temperature of the surrouding `=290K`
`T=` temerature of the body ` =150K`
`:. (dT)/(dt) = (6sigma)/(rhosd) (T_(0)^(4) -T^(4)) = 0.068Ks^(-1)` .