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InterviewSolution
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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. |
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Answer» 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)` |
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