A point source of heat is placed at the centre of a spherical shell of mean radius 10 cm. the material of the shell has thermal conductivity 9.5`xx10^(-2)`kcal/`m-s^(@)C`. Calculate the power of point source of heat, when the temperature difference between the outer and inner surface of the shell is `50^(@)C` in steady state. the thickness of shell is 2 cm.

Explanation

Consider a concentric spherical shell of radius r and thickness dr and shown in figure.
The radial rate of flow of heat through this shell in steady state will be ltBrgt `Q=-`(dH)/(dt)=-kA(dT)/(dr)`
the negative sign is used as with increase in r, T decreases.
Now,
`A=4pir^(2)`
`thereforeQ=-4pir^(2)k(dT)/(dr)`
or `(dr)/(r^(2))=-(4pik)/(Q)dT`
Leat a and b be inner and outer radius of shell and `T_(1) and T_(2)` be temperature of inner and outer surface.
`therefore int_(a)^(b)(dr)/(r^(2))=-(4pik)/(Q)int_(T_(1))^(T^(2))dT`
or `[-(1)/(r)]_(a)^(b)=-(4pik)/(Q)[T_(2)-T_(1)]`
`(1)/(a)-(1)/(b)=-(4pik)/(Q)(T_(2)-T_(1))`
or `Q=k(4piab(T_(1)-T_(2)))/((b-a))`
here `T_(1)-T_(2)=50^(@)C and k=9.5xx10^(-2)kcal//m-s^(@)C`
Now in steady state no heat is absorbed, rate of loss of heat by conduction is equal to that of supply i.e.,
Q=power of the source
`thereforeP=k(4piab(T_(1)-T_(2)))/(b-a)`
Now, mean radius of shell is 10 cm
`thereforea=9cm,b=11cm and b-a=2cm`
`thereforeP=(9.5xx10^(-2)xx10^(3)xx4pixx(99)xx50xx10^(-4))/(2xx10^(-2))`
`=5.90xx10^(4)cal//s`