A sphere of alumininum of mass 0.047 kg placed for sufficient time in a vessel containing boling water, so that the 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 aluminum. Specific heat capacity of copper = `0.386 xx 10^(3) J kg^(-1) K^(-1)`.
Specific heat capacity of water = `4.18 xx 10^(-3) J kg^(-1) K^(-1)`

Explanation

In solving this example we shall use the fact that at a steady state, heat given by an aluminium sphere will be equal to the heat absorbed by the water and calorimeter. Mass of aluminium sphere` (m_(1)) = 0.047 kg`
Initial temp. of aluminium sphere = `100 ^(@)C`
Final temp. = `23 ^(@)C`
Change in temp `(DeltaT ) = (100^(@)C - 23^(@)C) = 77^(@)C`
Let specific heat capacity of aluminium be `s_(Al)`.
The amount of heat lost by the aluminium sphere = `m_(1)s_(Al)DeltaT=0.047kgxxs_(Al)xx77^(@)C`
Mass of water `(m_(2)) = 0.25 kg`
Mass of calorimeter `(m_(3)) = 0.14 kg`
Initial temperature of water and calorimeter = `20^(@)C`
Final temperature of the mixture = `23^(@)C`
Change in temperature `(DeltaT_(2))=23^(@)C–20^(@)C=3^(@)C`
Specific heat capacity of water `(s_(w))=4.18xx10^(3) J kg^(-1) K^(-1)`
Specific heat capacity of copper calorimeter `=0.386 10^(3)Jkg^(–1)K^(–1)`
The amount of heat gained by water and calorimeter = `m_(1)s_(w)DeltaT_(2)+m_(3)s_(cu)DeltaT_(2)`
`(m_(2)s_(w)+m_(3)s_(cu))(DeltaT_(2))`
=`( 0.25kg xx4.18xx10^(3) J kg^(–1) K^(–1)+0.14kgxx0.386xx10^(3)J kg^(–1) K^(–1))(23^(@)C–20^(@)C)`
In the steady state heat lost by the aluminium sphere = heat gained by water + heat gained by calorimeter.
So, `0.047 kgxxs_(Al)xx77^(@)C`
` = (0.25 kgxx4.18xx10^(3) J kg^(–1) K^(–1)+ 0.14 kgxx 0.386xx10^(3) J kg^(–1) K^(–1))(3^(@)C)`
`s_(Al)=0.911 kJ kg^(–1) K^(–1)`