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 .

A solid copper shere of density `rho` specific heat c and radius r is

1.	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» Correct Answer - 9 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)))` .

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