black colored solid sphere of radius "R" and mass "M" is inside a cavity with vacuum inside.The walls of the cavity are maintained at temperature `T_(0)` .The initial temperature of the sphere is `3T_(0)` .If the specific heat of the material of the sphere varies as `alpha T^(3)` per unit mass with the temperature of the sphere,where `alpha` is a constant,then the time taken for the sphere to cool down to temperature `2T_(0)` will be ( `sigma` Stefan Boltzmann constant)

Options

A. `(M alpha)/(4pi R^2 sigma) ln (3/2)`
B. `(M alpha)/(4pi R^2 sigma) ln (16/3)`
C. `(M alpha)/(16 pi R^2 sigma) ln (16/3)`
D. `(M alpha )/(16pi R^2 sigma) ln (3/2)`

Answer

C

Explanation

In the given problem, fall in temperature of sphere,
` dT = (3T_0 - 2T_0) = T_0`
`T_("surr") = T_0`
`T_("initial") = 3T_0`
Specific heat of the material of the sphere varies as,
`C = alpha T^3` per unit mass (`alpha` = constant)
Applying formula,
` (dT)/(dt) = - (sigma A)/(MC) (T^4 - T_(surr)^(4))`
` (dT)/(dt) = - (sigma 4pi R^2)/(M alpha (T)^3) (T^4 - T_(surr)^(4) ) `
`int_(3T_0)^(2T_0) (T^3dT)/((T^4 - T_(surr)^(4) ) =- (sigma 4pi R^2 )/(M alpha) int_(0)^(t) dt `
Let ` (T^4 - T_(surr)^(4) ) = x `
Differentiating w.r.t. x
` rArr 4T^3 (dT)/(dx) - 0 = 1`
` rArr 4T^3 dT = dx`
` T^3 dT = (dx)/(4)`
when , ` T = 3T_0 rArr x = (3T_0)^4 - (T_0)^4 = 80T_0^4`
when `T = 2T_0 rArr x = (2T_0)^4 - (T_0)^4 = 15 T_0^4`
Now , substituting in integral
`int_(80T_0^4)^(15T_0^4) (dx)/(4x) =- (sigma 4pi R^2)/(M alpha) int_(0)^(t) dt`
` 1/4 [ln x ]_(80T_0^4)^(15T_0^4) = - (sigma 4pi R^2)/(M alpha) [t]_0^t`
`1/4 {ln (15T_0^4) - ln (80 T_0^4) } = - (sigma 4pi R^2)/(M alpha ) (t)`
` - 1/4 ln ( 15/80) = (sigma 4pi R^2)/(M alpha) (t)`
` t = (M alpha)/(16pi R^2 sigma) ln (16/3)`