In an experiment to determine the specific heat of a metal,a `0.20 kg` block of the metal at `150 .^(@) C` is dropped in a copper calorimeter (of water equivalent `0.025 kg`) containing `150 cm^3` of water at `27 .^(@) C`. The final temperature is `40.^(@) C`. The specific heat of the metal is.

Explanation

Let specific heat of brass `=C_(1)`
Mass of brass `m_(1)=0.2 kg`
Fall in temp. of brass `Delta theta_(1)=150-40`
`=110^(@)C=110 K`
Mass of water `m_(2)=0.150 kg`
Rise in temp. of water `Delta theta_(2)=40-27`
`=13^(@)C=13 K`
Water equivalent of calorimeter `w=0.025 kg`
According to the principle of caloriemetry,
Heat lost = Heat gained
`m_(1)C_(1)Delta theta_(1)=m_(2)C_(2)Delta theta_(2)+wC_(2)Delta theta_(2)`
`C_(1)=((m_(2)+w)C_(2)Delta theta_(2))/(m_(1)Delta theta_(1))` ...(i)
`=((0.150+0.025)xx4.2xx10^(3)xx13)/(0.2xx110)`
`=(0.175xx4.2xx10^(3)xx13)/(0.2xx110)`
`=0.434xx10^(3)J kg^(-1) K^(-1)`
From eqn. (i), the maximum permissible error in sp. heat is given by
`(deltaC_(1))/C_(1)=(deltam_(2))/((m_(2)+omega))+(delta(Deltatheta_(2)))/(Delta theta_(2))+(Delta m_(1))/m_(1)+(delta(Delta theta_(1)))/(Deltatheta_(1))`
`=0.001/0.175+0.2/13+0.001/0.200+0.2/110`
`=0.0057+0.0153+0.005+0.0018`
`=0.0278`
`:. DeltaC_(1)=0.0278xx0.434xx10^(3)J kg^(-1) K^(-1)`
`=0.012xx10^(3) J kg^(-1) K^(-1)`
thus, specific heat of brass is
`C_(1)=(0.43 pm 0.01)xx10^(3) J kg^(-1) K^(-1)`