A solid copper shere of density `rho` specific heat c and radius r is at temperature `T_(1)` It is suspended inside a chamber whose walls are at temperature 0 `K` The time required for the temperature of sphere to drop to `T_(2)` is `(rrhoc)/(xesigma)((1)/(T_(2)^(3))-(1)/(T_(1)^(3)))` Find the value of x? Take the emmissivity of teh sphere to be equal to e .

Answer

9

Explanation

The rate of loss of energy due to radiation `P = eA sigmaT^(4) …T` is rate must be equal to mc `(dT)/(dt)`
Hence `-mc (dT)/(dt) eAsigmaT^(4)`
Negative sign is used at temperature decreases with time In this equation `((4)/(3) pir^(3)) rho` and `A =4pir^(2)`
`:. (dT)/(dt)=(3esigma)/(rhocr)T^(4)or,-underset(0)overset(t)intdt=(rrhoc)/(3esigma)underset(T_(1))overset(T_(2))int(dT)/(T^(4))`
Solving this, we get `t=(rrhoc)/(9esigma)((1)/(T_(2)^(3))-(1)/(T_(1)^(3)))` .