In an experiment on 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`. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal ?

Explanation

Mass of metal piece `m_(1)=0.2` kg
Temperature of metal piece `theta=150^(@)C`
Calorimeter + temp. of water `theta_(2)=27^(@)C`
Final temp. of mixture `theta=40^(@)C`
Volume of water `V=150" cm"^(3)=150xx10^(-6)" m"^(3)`
Mass of water `m_(2)=Vrho=15xx10^(-5)xx10^(3)` kg
`:.m_(2)=15xx10^(-2)` kg
Equivalent mass of calorimeter `W=0.025` kg
Specific heat of metal piece = `C_(1)`
Specific heat of water `C=4.2xx10^(3)" J kg"^(-1)K^(-1)`
Heat lost by metal piece `Q_(1)=m_(1)C_(1)Deltatheta`
`:.Q_(1)=0.2xxC_(1)xx(150-40)`
`=0.2C_(1)xx110`
`:.Q_(1)=22C_(1)` . . . (1)
Heat gained by water and calorimeter,
`Q_(2)=(m_(2)+W)CDeltatheta.`
`=(150xx10^(-3)+25xx10^(-3))xx4.2xx10^(3)xx(40-27)`
`=(150+25)xx10^(-3)xx4.2xx10^(3)xx13`
`:.Q_(2)=175xx4.2xx13` . . (2)
From principal of calorimetar,
`Q_(1)=Q_(2)`
`22C_(1)=175xx4.2xx13` [from equation (1) and (2)] If asked in calory, then
`:.C_(1)=(175xx4.2xx13)/(22)`
`:.C_(1)=434" J kg"^(-1)K^(-1)`
`C_(1)=(434)/(1000xx4.2)`
`=0.1033" cal g"^(-1)K^(-1)`
`:.C_(1)=0.1033xx4.2=0.433" Jd"^(-1)K^(-1)`
This value is smaller than actual value of specific heat of the metal.