In the exp of finding sp heat capacity of an unknown sphere `(S_(2))` mass of the sphere and caloreter aro `1000m` and `200gm` respectively and sp heat capacity of calorimeter is equal to `(1)/(2)cal//gm^(@)C` The mass of liquid 9water) used is `900gm` Initially both the water and the calorimeter were at room temp `20.0^(@)C` while used is `900gm` Initially both the water and the calorimeter were at room temp `20.0^(@)C` while the sphere was at temp `800^(@)C` initially If the steady state temp was found to be `40.0^(@)C` estimate sp heat capacity of the unknown sphere `(S_(2))` `(use S_(water) = 1 cal//g^(@)C)` Also fin the maximum permissible error in sp heat capacity of unknown sphere `(S_(2))` mass end specific heats of sphere and calorimater are correctly known) .

Explanation

To determine the specitle heat capacity of unkonw soild
We use`s_("solid")=(m_(1)s_(1)+m_(2)s_(2))/(m_(1))((0_(s)-0_(2))/(0_(1)-0_(s)))and" get "s_(solid)=1//2cal//g// .^(@)C`
`((Deltas)/(s))_(max)=20Deltatheta((1)/(0_(s.s)-2_(2))+(1)/(0_(1)-0_(s s)))=2(0.1^(@)C)((1)/(40.0-20.0)+(1)/(80.0-40.0))=1.5%` .