A sphere of aluminium of 0.047 kg placed for sufficient time in a vessel containing boiling water, so that sphere is at `100" "^(@)C`. It is then immediately transferred to 0.14 kg copper calorimeter containing 0.25 kg of water at `20" "^(@)C`. The temperature of water rises and attains a steady state at `23" "^(@)C`. Calculate the specific heat capacity of aluminium.

Explanation

Mass of sphere `m_(1)=0.047` kg
Initial temperature of sphere `T_(1)=100^(@)C`
Final temperature of sphere `T_(2)=23^(@)C`
Change in temperature `DeltaT_(1)=T_(1)-T_(2)`
`=(100-23)" "^(@)C`
`=77^(@)C`
Sphere specific heat of sphere `=S_(1)`
Heat lost by aluminum sphere,
`Q_(1)=m_(1)S_(1)DeltaT_(1)`
`=0.047xxS_(1)xx77`
`:.Q_(1)=3.619S_(1)` . . .(1)
Now, mass of water `m_(2)=0.25` kg
Mass of calorimeter `m_(3)=0.14` kg
Specific heat of water,
`S_(2)=4.18xx10^(3)" Jkg"^(-1)K^(-1)`
Specific heat of copper calorimeter, `S_(3)=0.386xx10^(3)" Jkg"^(-1)K^(-1)`
Heat gained by water and calorimeter,
`Q_(2)=m_(2)S_(2)DeltaT_(2)+m_(3)S_(3)DeltaT_(2)`
`=0.25xx4.18xx10^(3)xx3+0.14xx0.386xx10^(3)xx3`
`=3.135xx10^(3)+0.16212xx10^(3)`
`:.Q_(2)=3.29712xx10^(3)` . . .(2)
For steady state,
{Heat lost by sphere} = {Heat gained by water} + {Heat gained by calorimeter}
`Q_(1)=Q_(2)`
`3.619S_(1)=3.29712xx10^(3)` [from equation (1) and (2)]
`:.S_(1)=(3.29712xx10^(3))/(3.619)=0.911xx10^(3)`
`:.S_(1)=0.911" kJ kg"^(-1)=911" Jkg"^(-1)K^(-1)` (specific heat of aluminium)