A solid sphere of radius R is made of a material of density rhoand specific heat c. The sphere is initially at 220 K and is later suspended inside a chamber whose walls are at 0K. What time is required for the temperature of the sphere to drop to 160 K?

A solid sphere of radius R is made of a material of density rhoand spe

1.	A solid sphere of radius R is made of a material of density rhoand specific heat c. The sphere is initially at 220 K and is later suspended inside a chamber whose walls are at 0K. What time is required for the temperature of the sphere to drop to 160 K?
Answer» Solution :The energy radiated by the sphere is given by Stefan.s law , ` (dQ)/(dt) = sigma AT^4 = sigma (4pi R^2) T^4` ` = 4pi sigma R^2 T^4`.....(i) Due to energy loss, the temperature of the sphere is reduced. Rate of energy loss due to rate of decrease of temperature is `(dQ)/(dt) = mc (dT)/(dt)` `m= V rho = 4/3 pi R^3 rho ` ` therefore (dQ)/(dt) =- 4/3 pi R^3rho c ( (dT)/(dt) )`..(ii) Comparing (i) and (ii) ` 4pi sigma R^2 T^4 = - 4/3 pi R^3 rho c ((dT)/(dt) )` `rArr 3 sigma T^4 = - rho RC ((dT)/(dT))` ` rArr (rho R c)/(T^4) dT= -3sigma dt` INTEGRATING both sides, we get `rho Rc int_(T_1)^(T_2) T^(-4) dT = - 3 sigma int_(0)^(t) dt` where `T_1` and `T_2`are initial and final temperature of the sphere, respectively, and t is the time taken for the sphere to drop its temperature from ` T_1 "to" T_2` ` rArr t = (rho Rc)/(9 sigma) [ (1)/(T_2^3) - (1)/(T_1^3) ]` ` =(rho Rc)/(9 (5.67 XX 10^(-8)) ) [ (1)/(160^3) - (1)/(220^3) ]` = 0.3`rhoRc`